Okay, I will start. I hope that everybody will be there soon. So we had started to talk about fields and field extensions. And so we finished by talking about, so we had done the degree theorem and introduced a degree of field extension and proved the degree theorem. And then we finished by studying simple algebraic extensions. Maybe I state the last theorem that we stated and started to prove. So theorem, so we are in the situation that k, large k is a field, let's say it is a field and small k in k is a subfield. So large k is a field extension of small k. And so we take an element a in k, which is algebraic over the smaller field. And we want to look at the simple algebraic extension generated by it. So we assume the minimal polynomial of a polynomial f a. And then we have the following statements. First, the simple extension generated by a can be written is isomorphic to the polynomial ring in x divided by the ideal generated by the minimal polynomial. And second, the degree of k a over k is equal to the degree of the minimal polynomial. And in fact, if the m is equal to this degree, then the elements one a, a squared and so on until a to the m minus one are basis, k a. So basis of k as a k vector space. Of k a as a k vector space. Okay, so we had basically done the proof of one. Maybe I can go through it very quickly again. So we have the evolution homomorphism from the polynomial ring in x, which maps to k a, which sends any polynomial g to g of a. Obviously this goes to the field generated by a because we just have polynomials in a in the image. So this is a ring homomorphism. And our definition of the minimal polynomial was that the kernel is the ideal generated by f a. So if we take l to be the image, then we know for one thing that, so this kernel is an ideal in the principle ideal ring k x and principle ideal domain to x. So the kernel is a maximal ideal. So f a is a maximal ideal in the principle ideal domain k x. And thus we have that the quotient k x model of f a is a field. And by the homomorphism theorem, the quotient is isomorphic to l by the homomorphism theorem. No, if you have a subjective ring homomorphism, then the image is isomorphic to what the source divided by the kernel. So in particular, l is a subfield t a. And we have to basically show however that it is equal to k a because our statement is that this. So the first point is that it contains k as the image is cons of the constant polynomials. And l also contains a because if we take the polynomial x it is mapped to a a is equal to the evaluation at a of the polynomial x. So we have that l is a subfield of k a which contains a and k. But this means it is equal to k because k a is the smallest field which contains k and k a. So thus, so this describes this thing. Now the second statement is that we want to show to compute the degree and so we have so two in one. We have in particular shown that the evaluation map at a from k x to k a is surjective because after all l was just the image. So we have a subfield of k x to k a because after all l was just the image and l turns out to be k so it's a surjective map. So therefore we can write k a just as the image of the evaluation map that means it's the set of all g of a where g is a polynomial in k x. That's what it is. Now we want to show but maybe we don't need all polynomials we want to show that first show that so if m is the degree of this minimal polynomial that these elements here spend k a as a k vector space. So one a until a m minus one generate k a is k vector space. So let's take any element in k a we can write it like this. Let's k of g of a in element in k of a. So any element in k of a looks like this. So it's a sum g in k x. So we want to show that you can find another polynomial of lower degree which gives us the same element in k a and obviously we do this by division with rest. Maybe I write it differently. So just to make it logically clear that b being element in k of a then but by what we have seen here then it exists a polynomial g in k x. So we want to show that we can find another polynomial in k x such that b is equal to g of a. So if the degree of g is bigger equal to m the degree of the minimal polynomial we can make division with rest. So right do we have by division with rest the g is equal to q times f a plus r with q and r in k x and r is equal to zero or the degree of r is smaller than m. Okay and now the point is obviously that r of a is equal to g of a. So then we have because f a of a is equal to zero we have that g of a is equal to q times f a of a which is equal to zero plus r of a so q of a which is equal to r of a but that means that we can write this is equal to r of a and so if you write r as sum i equal to zero to m minus one because the degree is at most m b i x to the i we find that b is equal to sum i equal to zero m minus one b i a to the i that means it is in the linear span of a of one up to a. Okay so this shows that these elements generate k a as a k vector space and now we also want to show that they are linear dependent so that they are basis. So assume one a m minus one are linear dependent so then this kind of the obvious thing then there exist say elements say b one b zero to b m b m minus one in k such that you know this linear combination is zero so sum i equal to zero to m minus one b i a to the i is equal to zero but obviously this means we have a polynomial of this degree which vanishes at a so thus for g equal to sum i equal to zero to m minus one b i x to the i we have g of a is equal to zero so we know that the so that means that g lies in the kernel of this evaluation morphism of a so it is a multiple of f a is in the kernel of the evaluation at a which is the ideal generated by f a so we have the g is equal to so it means that g is in f a but you know the degree of g is at most m minus one the degree of f is m so this is only possible if g is equal to zero so that means that g is equal to zero so these coefficients were all zero zero is equal to b one m minus one is equal to zero that means these elements were linearly independent so we have not found a non-trivial linear combination which is zero so one a until a to the m minus one are linearly independent so we have seen it somehow I think this proof is quite straightforward maybe most of you would have found it not so exciting from a certain point because one only does the tries the obvious things and they work so now in particular we can if you think of it this gives you a completely explicit description of this field k a as this portion k x model then s k x model of the f a you can also maybe make it slightly more ever so slightly more explicit as follows so if k a over k is a simple algebraic extension and c of degree m and we have f is the minimal polynomial or maybe I still write f a of a then we can explicitly say describe this as follows can describe k a as follows so as a set or as a k vector space k a is equal to the set of all kind of polynomials evaluated a sum I equals zero to m minus one say bi a to the I where the bi okay bi elements of k so as I said it's just this vector space of linear combinations of the ai's and the addition and the multiplication are essentially the obvious ones as polynomials so addition and multiplication defined in the obvious way as if these were polynomials no you just s for polynomials in a so remember the rules you add them component wise you multiply by this product formula with the additional rule so so with the extra relation so if you write f a equal to sum i equals zero to m say c i x to the i then we want if we multiply two polynomials we can get we get the polynomial in a whose degree might be bigger than m bigger equal to m and so we have to see how to get lower and we do this by applying this namely with the extra relation that a to the m is equal so this is a monic polynomial so we have that c m is equal to one equal to minus sum i equals zero to m minus one c i a to the m to the i and this is used to eliminate all polynomials in a of degree bigger equal to m or whenever we have a power of a which is bigger than m we can replace the m by this until we get only terms of lower degree so it's a very simple description of this ring of this field so just as an example for instance you can view the construction of the complex numbers out of the real numbers joining i as a as a simple algebraic extension so so you have that c is equal to r i is a simple algebraic extension where i is the square root of minus one in the complex numbers and the minimal polynomial of i is x squared plus one which means i squared is equal to minus one and so then if we just apply this rule we find that c so thus according to this remark we have that c can be written as a set of all a plus where a and b are real numbers as it should be and the addition is the obvious one so a one plus b one a i plus a two plus b two i is just component wise so it's a one plus a two plus b one plus two times i and the multiplication is given according to this rule so if I take say a one plus b one i times a two plus b two i according to this we first just multiply them out as polynomials in i so this is a one a two plus a one b two plus a two b one times i plus i squared times say b one times i squared and then we apply this thing that the i squared is replaced by minus what we have here so i squared is replaced by minus one so this is equal to a one a two plus a one b two plus a two b one so i squared is minus one minus b one b two plus this term here a one b two plus a two b one times i the rule of computing and the complex numbers that you know can be viewed as a special case of this okay so much for the moment about these simple extensions so now we want to characterize finite field extensions you remember that the finite field extension is one where the degree of the field extension is finite and so the claim is that the field extension is finite if and only if it is algebraic and is obtained by adjoining by you know and is generated by finitely many elements so it's a following theorem say let L over K be a field extension then L over K is a finite field extension if and only if first it is algebraic and second there exist finitely many elements so a one whatever a n in L such that L is the field extension generated by these finitely many elements so if I alright you know it was not clear that a finite extension has to be algebraic because for extension to be algebraic it means that every element of L is algebraic over K if we take for instance the extension generated by just one element contains many other elements it could be some of them aren't but we now easily see that this is not the case the proof is not particular difficult but so we assume we have a finite field extension we want to show it's algebraic and it's generated by finitely many elements so let M be the degree of the field extension which is some finite number so then if I take any element a it's kind of the same trick we have used before any element a in L we find that the elements one a until a to the M so these are M plus one elements in L they must be linear dependent over K in the K vector space so linearly dependent over K so in the K vector space L you know we have one more than this dimension so it follows there exist P1 P0 to Pm in K such that that's obviously the same thing as before that sum I equals 0 to M A to the I is equal to 0 and the Bi are not all 0 so that means if I take the polynomial which has these as coefficients then this vanishes at A G equal to sum I equals 0 M Bi X to the I which is the polynomial with coefficients in X and it's not the 0 polynomial satisfies G of A is equal to 0 so A is algebraic A is algebraic over K so we even find that we have a bound on the degree of any polynomial which vanishes and on the other and we can also simply how do we find these finding many elements let us just take any basis of L as K vector space so if A1 until A in this case the degree is actually M is a basis of L as a K vector space then we can write any element in L as a linear combination of of these AIs in particular the field generated by these elements is the whole field so this is one direction and now the other one so we prove this so assume that L is algebraic and we have this with this number N generated by N elements we make induction on N in fact we don't even so if the case N equals 0 is trivial because then we just have our field L is equal to K and the degree is 1 we make the induction step so we assume that our L can be written as K A1 to An plus 1 for some elements AI in L and we put L prime generated by the first N of them then obviously there is some degree so we have that L prime over K is some finite number M by induction so it's finite which I can call M by induction and so I have to see by the degree theorem we have to say see only what L prime what L over L prime is so we know that An N plus 1 is algebraic over K and thus over L prime because we know that it's algebraic over K and so every element in L is algebraic over K particular also this thing so if we take G the minimal polynomial of An plus 1 over L prime and then we know that the degree of L over K is equal by the degree theorem to the degree of L over L prime times the degree of L prime over K which this one was M and this degree is equal to the degree of the minimal polynomial because L is equal to L prime from L L prime adjoint we just have adjoint one more element so this is also finite so we see that every finite algebraic every finite field extension is algebraic and is kind of obtained by joining finitely many algebraic elements and vice versa so we have a few we have a small corollary so if you have any field extension we can look at a bigger field which are algebraic over the smaller field and then the claim is these themselves form a field and they form an algebraic field extension of the smaller field so let K over K be a field extension and at L be the set of all elements say A in K such that A is algebraic over the smaller field K then L over K is an algebraic field extension L is a subfield of K so we know obviously K is contained in L and so to see that this is an algebraic obviously because every M K is algebraic over K now we want to see that L we have to see that L is a subfield of K because obviously every element by definition every element of L is algebraic over K so the field extension will automatically be algebraic so to show L is a subfield of large K well that's actually quite simple let us just take two elements in L and these are two we have to see that the sum, the product whatever the difference or the quotient is contained in L I mean unless one of them is 0 so maybe we can assume that they are not 0 so to show that to say A plus B A minus A A times B and B A to the minus 1 are all elements in L then it is a field the point is we can look so if we look at KAB the field extension generated by A and B this is a field extension generated by two algebraic elements so this is a finite according to the theorem this is a finite extension so it is algebraic thus algebraic so therefore KAB is contained in L because all elements in KAB are algebraic and therefore obviously the sum and the product and so on are also in KAB ok so that's quite simple and as I said that this is an algebraic extension so as an example we use this kind of abstract but we can for instance look at Q bar this is supposed to be the set of all elements all complex numbers which are algebraic over Q so we can look at this set and this is an algebraic field extension of Q so Q bar over Q is an algebraic field extension but obviously it's not a finite extension you can take nth roots numbers for any n so of infinity ok that's simple we can also see that if we take a composition of algebraic field extensions it will also be algebraic so if we have a smallest field we make an algebraic field extension of it and then we make an algebraic field extension of the bigger one then the biggest field or the smallest field is also algebraic and in fact yeah so it is independent of whether these field extensions are finite or not so let say k over L and L over small k the algebraic field extensions then the extension of the largest field over the smallest is also extension is also algebraic so let's see so I don't make the assumption that these field extensions are finite but I still want to use the fact that how you do that finite field extensions are how they are related to algebraic extension that finite extensions are always algebraic so how do we do that so proof so let U any element of k we have to show that k is that this element U is algebraic over small k first we know it's algebraic over L so then there exists a polynomial say in L of x which is not the 0 polynomial such that f of U is equal to 0 so we can write f sum i equal to 0 n whatever n is a i x to the i where the a i are some elements of f now we want to so we want to know whether the thing is algebraic over k so see where the i so we have that this a i are elements in L and therefore the algebraic over k so we take m make another field m which is the smallest field k and we join the element the coefficients a 0 to a n the a i are elements in L so they are algebraic over k and they are finally many of them in this case n plus 1 so this m over k is a finite extension a i thus m over k is a finite extension and we also have if I take m of U over m so if U is an algebraic element so it's algebraic over L but it's also it's also algebraic over m because the coefficients of the polynomial are all out of m so this extension m U over m is also an algebraic extension and as we have only joined one element it's a finite algebraic extension and so we find that the degree of m of U over k is again by the degree theorem the degree of m of U over m times the degree of m over k and these numbers are both finite so therefore this is finite so this extension here is a finite extension so as it's a finite extension we know it's an algebraic extension every element here is algebraic over k U is algebraic over k it's finite thus algebraic therefore U is algebraic over k and so we indeed find that U was an arbitrary element in k so every element in k is algebraic over small k so now we will talk about extensions of field homomorphisms so basically field homomorphisms so we will later do Galois theory which means to study field extension via their Galois group and the Galois group is the group so if L over k is a field extension the Galois group will be the group of k isomorphisms k automorphisms of L so somehow therefore as a something one needs to therefore one somehow needs to study how field homomorphisms from a from a smaller field extend to a field extension and so we want to do this here so so the thing basically show at this point is we want to show that if so I have again say k over k field extension we want to show that if say we have two elements A and B in the larger field k which have the same minimal polynomial over the smaller field field small k then there is an automorphism then there is a small k automorphism of large k which sends A to B and there is a unique one then there exists a unique k automorphism not of from k sends so phi maybe with phi of A is equal to B so if two polynomials if two elements have the same minimal polynomial then there is an isomorphism between the fields automorphism then there is an isomorphism from the field the field extension generated by this element the field extension generated by that which sends this element to that and there is actually a unique one there is only one such thing so we prove however and this will be important when we want to do this Galois theory as I try to hint and we will however need a slightly more general result where the difference is that we even here we don't just start with one field k but we have two different k's we have an isomorphism between them and we want to extend it so let me formulate it so definition so let say phi from k to k prime we have field isomorphism maybe I and let say L over k and L prime over k prime B field extensions so an isomorphism say large phi from L to L prime is called an extension of small phi if it's restriction to k is this given map it's in particular if we restrict phi to the small k it maps to k prime okay so let's now we want to study for this simple algebraic extension the existence and uniqueness of such things so maybe I for need this trivial observation if I have a field isomorphism so if phi from k to k prime is a field isomorphism then this gives me also an isomorphism between polynomial rings in the obvious way then I could call this phi star from kx to k prime x which sends any polynomial some i 0 n ai x to the i to the polynomial where we apply phi to the coefficients so this is a ring isomorphism so this is essentially completely obvious you can see that it sends the sum to the sum and it's easy to check it sends the product to the product and obviously as this is an isomorphism if I take the inverse isomorphism here and apply it in this way this gives us me the inverse map between the polynomial rings so now we want to make the statement about the extension of polynomial rings so let again phi from k to k prime be a field isomorphism and we have again our field extension l over k l prime over k prime field extensions and now we take a in l which is algebraic over k and we take so with the specified polynomial polynomial f a and now we take the element a prime in l prime to be a 0 of the image of the minimal polynomial so then there exists a unique extension of this field isomorphism so maybe I will a unique extension of say phi from k a to k prime of a prime of our isomorphism phi with so this means that the extension means that phi restricted to k is equal to k prime and the thing which makes it unique is that phi of a is equal to a prime so there exists a unique extension of this original map phi which sends a to a prime as usual if something exists and is unique the normal approach is always that you first prove the uniqueness you just see that the conditions determine it uniquely and this will actually tell you what it is and then you check for that thing that it satisfies what you want so that occurs all the time okay so if we start with the uniqueness so let phi from k a to k prime of a prime be such an extension so with phi of a is equal to a prime now we have to see whether we can and am I so note we have seen that k of a can be written as the set of all g of a with g is an element in kx in the same way k prime of a prime is equal to the set of all whatever g prime of a prime h of a prime where h is an element in k prime of x so every element in k a so it can be written as a polynomial evaluated on a and in the same way well and so and then the ring structure is in particular involves that you additional modification of polynomials we know if so so if b is an element of k a then you can write b equal to g of a for a some polynomial for g in kx so if you write g for instance equal to i equals 0 so n bi x to the i bi elements in k well then what is phi of our element b the same as phi of g of a that is phi of some i equals 0 to n and 8 to the i now this is a ring homomorphism so it so it we know that phi is supposed to send a to a prime and you know and on the bi which elements of k it's supposed to be small phi so this is equal to the sum i equals 0 to n like first write it again phi of bi phi of a to the i because it's a ring homomorphism and phi of bi is equal to phi so this is equal to phi small phi of bi and phi of this is equal to a prime so this thing is equal to some which in other words is nothing else than g prime or was it phi prime of a prime so we find that we have you know we have actually computed the element so that it's uniquely determined because just from the definition we were able to compute it so now we have to check that this is a ring a field isomorphism maybe we should have checked it but anyway I don't existence so we define phi k a to k prime of a prime by phi of g of a and k of equal to g of a is equal to phi star of g of a prime so by this so yeah so in principle I should have checked that this is well defined so it's easy this is well defined so if two polynomials g and h give me the same element in k a then the difference is a multiple of the minimal polynomial and then you can easily see that the difference maps to zero and so gives me the same so we get the same element here you can easily check that same as before and but on the other hand it's clear from the definition that this is a homomorphism I mean once it's well defined it's because phi is a homomorphism no because the multiplication is precisely given by and it is clearly and phi is also clearly subjective because we know that this map phi star is an isomorphism from kx to k prime so obviously I get if I apply this if I have all polynomials applied to a by applying phi star I get all polynomials applied to a prime so the map is subjective and now we have a subjective morphism between two fields then it is an isomorphism is an isomorphism I mean in the notes there's something more complicated but because you know the kernel of a field isomorphism is an ideal so it must be either 0 or the whole of the field but it cannot be 0 because this map is subjective it cannot be the whole of the kernel cannot be the whole of the field so the map cannot be the 0 map because it's subjective so it is an injective map so p is injective as a non-trivial homomorphism of fields so we find that this is an isomorphism of fields and finally it is clear by definition where do I have a thing so for an element b in k we have that b is a constant polynomial so we apply this you know if you apply to a constant polynomial this map it's just the map phi the constant polynomial is just if I have the constant polynomial b then it's b times a to the 0 and the map is by sending it to phi of b is equal to phi star of the constant polynomial b which is just phi of b so that means that phi restricted to k is the small phi and by definition we also have that if we take phi of a you know remember how do I get a by as evaluating a polynomial in kx at a we do this by taking the polynomial x so this is phi of the polynomial x evaluated at a so according to our definition we are supposed to apply phi star to this but you know the coefficient here is 1 so the phi star maps x to itself so this is you know the map here is obtained by evaluating it at a prime so this is x evaluated at a prime which is a prime and so we have precisely the statement that we have a unique extension of this map from k of a to k prime of a prime which sends which is the map yeah this is obviously a misprint phi restricted to k the original map phi and phi of a is equal to a prime okay and the thing which we are most interested in is the thing which I mentioned in the beginning namely if we send set k equal to k prime and let the map and let the map phi that we have here to be the identity that we get in particular the following corollary so let say L over k here be field extension and we take two elements in L which have the same minimal polynomial so algebraic with the same minimal polynomial so the algebraic with the same minimal polynomial then there exists a unique k isomorphism phi from k of a to k of a prime which sends a to a prime so somehow you if you have a field extension you have several zeros of the same polynomial then you find that there is an extension there is a field isomorphism which sends one and the unique one so we will later see that this is an important tool for us this is up and you know this is a special case where we have to prove nothing so before going on I wanted to just make a few remarks about the algebraic closure of a field we are not really going to use it in the future it's just you know for completeness so that you have seen what the algebraic closure is I also will not prove things I will just state stated and state the theorem about the existence of an algebraic closure so so we want to introduce the algebraic closure of a field so this will be so this will be a field extension L over K an algebraic field extension such that every polynomial with coefficients in the smaller field K has a zero in L so all polynomial equations in one variable with coefficients in K can be solved in L but the field extension is algebraic for instance it's certainly not true that C is the algebraic closure of Q because it's not an algebraic extension so let me write the definition once again completing precisely although it's already written so let L be a field we say L we say L is algebraically closed here I actually wanted well whatever what I said is also correct but it's it's maybe not what I will say later it's maybe it would be an exercise for you to check that it's equivalent to what I would say now so I take a field we say that L is algebraically closed if every polynomial F with coefficients in L has a zero in L if well I wanted to actually say differently if the following equivalent statement holds so the first one is that every polynomial F which coefficients in L has a zero in L and the second statement is that every polynomial L with coefficients in L can be written as a product of linear factors so it splits into linear factors so every F in Lx splits over L into linear factors by this I mean the following I can write I mean I mean I'm obviously supposed to mean that I should say it so I mean that F can be written as say B times X minus A1 times say X minus An so there exist B and A1 to An elements in our field L such that F looks like that so F is a product of linear factors I need to B because it might not be money now obviously these two statements are equivalent 1 and 2 equivalent 2 implies 1 is completely trivial because you can see the zeros F has zeros A1 to An in this case which are elements in L and the other direction is not much is by trivial induction so 1 2 so so let A be a zero so A and L be a zero of an element F in Lx so then we can write F is equal to X minus A times G where G is an element in Lx and the degree of F of G is equal to the degree of F minus 1 and so then easily you can see that you can deduce the statement of 2 by induction on the degree of F on the degree of F I mean that's clear because in C that Q is not algebraically closed for instance X squared minus 3 has no zero in Q and R is not algebraically closed because the typical polynomial X squared plus 1 has no zero in C in R obviously it has zero in C but I claim that this Q bar which I have introduced before is algebraically closed this does not I mean completely directly follow from the definition but you can easily check it because by definition it consists of all the elements which are algebraic over Q so every polynomial with coefficient has a zero but I claim it follows from that that also polynomial with coefficient Q bar has a zero in Q bar that's okay and then I want to introduce this algebraic closure and I think with that I will stop so definition a field over K is called an algebraic closure so a field extension is an algebraic closure of K by closure if L is algebraically closed and the field extension is algebraic if L over K is algebraic is algebraically closed and you might check as an exercise that this is equivalent to what I wrote here so one can prove that every field has an algebraic closure it's a statement of very much of the same kind that every vector space has a basis in fact it's proven in the same way with the lemma of so on so with the action of choice it's a kind of standard application of the action of choice so I will just try it out because we don't really care so much so theorem so the first statement is every field K has an algebraic closure and it's also say and it's also essentially unique it's unique up to isomorphism so second if K and L are two just K then K and L are small K isomorphic so there's an isomorphism between them which fixes K in particular they are isomorphic and K are K isomorphic so then I should write it out then there exists K isomorphism phi from K to L as I said it's quite simple and so one can check it's kind of clear that so as an example you could easily check that for instance Q bar is an algebraic closure isomorphism is a unique algebraic closure of Q and obviously it's also and Mark and you know that C is algebraically closed I mean I don't know whether you have had it yet this is proven in complex analysis I don't know whether you have already covered it but that's the way and it's therefore C is the algebraic closure is an algebraic closure unique after isomorphism of the real numbers maybe as my time is essentially up I will stop here so we see each other the day after tomorrow are there questions comments okay so thank you


Adaptation is the capacity of a system to alter its state in response to some event within its environment. This capacity of adaptation is something we more often associate with biological systems as opposed to the technologies we design. The industrial world we have engineered is, in many ways, a relatively static one. We produce things like electrical power grids, buildings, and chairs, and then they sit there, specifically designed not to change. Every day we walk by the same advertisement on the street. It will present the same information to thousands of people that day, but it will only be of any relevance to a very small percentage of them, and because it is static, they will only take note of it the first time before tuning it out to become simply background noise. Now imagine if that advertisement changed every day, that is to say it was dynamic. Instead of a poster, we put in a screen that could be updated. It would be more relevant, more functional. Now let's go even a step further. Imagine if this screen could receive information about the profiles and preferences of the users that were in its vicinity and dynamically deliver content relative to their interests. This would be the world of complex adaptive engineered systems, and as information technology provides us with the tools for building smarter technologies, it is increasingly the world we have to design for. To understand this transition from static to adaptive systems, let's take the history of the web as a good example. Web 1.0 was a very static system where web developers hard-coded web pages. When you visited a site, the server just gave you the same page that had been written possibly two or three years earlier with no changes. Web 2.0, that we all know and love, leveraged new server-side scripting technologies to get information in and out of databases, and thus dynamically updating web pages, making them interactive and change over time. The emerging Web 3.0 uses semantic technologies and social networking to adapt content relevant to your specific profile and interests, thus making it not just dynamic but also responsive to the context. Outside of the web, the massive cost reduction in integrated circuits is leading to sensors and actuators being placed in many devices and objects as packages within supply chains, cars in traffic, and electrical power grids are becoming smarter. They can respond to events within their local environment through real-time mesh networks. But they can also feed data into large centralized systems for analysis, allowing for greater optimization through dynamic load balancing, as things like washing machines and streetlights begin to have the capacity to adapt their power demands to the current load on the system. There are essentially two levels to these complex adaptive engineered systems that we need to consider, the micro and the macro. On the micro level, we need elements with some form of control system. A control system is a mechanism for taking in information, processing it according to some set of instructions, and generating a response that alters the state of the component. Of course, all living creatures have this, from the simplest single-celled organisms to the most complex, the human brain. But increasingly we are using what we call cyber-physical systems to enable all kinds of technology to have this adaptive functionality as they become part of networks of technology that can communicate and respond to the changes in state of other technologies in real-time, as is the case in automated production lines, aircrafts, and mass transit systems. On the macro scale, when we are designing these adaptive systems, we can no longer rigidly control the system and determine its functionality in the way we can, say, designing a bridge, as the end result is going to be more organic, like an ecosystem of products, devices, and people interacting and adapting within networks rather than the rigid mechanical systems we're used to. So if we take away this key feature of control that is central to our traditional conception of being able to design, where we see constraining the autonomy of the components as a prerequisite to design, how then can we engineer these adaptive systems at all? The answer is to work with the innate features of adaptive systems, not against them. That sounds nice, but what does this mean exactly? Adaptive systems, by definition, adapt to their environment. If we place a plant in a new pot or a child in a new school, they will adapt to that particular environment. This is part of the dynamics of what biologists and ecologists call homeostasis. They do this through the process of synchronizing their state with that of other elements that they are exposed to, and this is the key to designing adaptive self-organizing systems. We don't try to directly alter the state of the components. We indirectly influence them by designing the connections within the environment that the system operates in. If we want people to be more environmentally conscious, we don't tell them to do this or not do that. We connect them with the natural environment, expose them to the consequences of their actions, both negative and beneficial. When a consumer picks up an anonymous product in a supermarket in a city a few thousand kilometers from where it was produced, they are totally disconnected and disassociated with it. But when we tell them a story about the product's life cycle, make them feel engaged and a part of that process, they're more likely to take ownership and responsibility for their actions. We, as designers of the product or services, have not tried to manipulate them. We have simply connected them with the reality of their environment and left it up to them as to how they adapt and respond. Thus, designing these adaptive systems is about creating open platforms that connect components. If we want to design an urban environment that offers a better quality of life to its citizens, we need to build open spaces where people can interact and self-organize to develop the socio-cultural fabric. Inversely, when we disconnect things, if, say, we build an expressway through the center of our residential community and thus dividing it, well, people adapt to their immediate environment and use these boundaries to emphasize the differences that divide them, tensions and inequality can arise. But disconnecting things can also be used to our advantage, to dampen down the undesired activities in the system. This is a naturally occurring event within complex adaptive systems. Think of how society regulates their values and norms by ostracizing or disconnecting those who break them. YouTube integrated this into the design of their platform by allowing users to flag down videos that are thought inappropriate. Thus, we can see how designing complex adaptive systems requires us to look at both the micro level of the individual components to design their control systems or at least understand how they work and consider the macro scale where we build platforms that either amplify desired behavior through connections or dampen down undesired behavior through disconnecting them. In summary then, information technology is taking us into the world of complex adaptive technologies and systems, whether we are talking about the smart phones that connect people to people to make our social organizations more responsive, more adaptive and dynamic or the cyber physical systems that are creating the Internet of Things. This world does not follow our traditional mechanistic conception of engineering but results in a much more organic model where the job of us as designers is more about the orchestration of ecosystems and systems in order to achieve their overall functionality.


Everybody else calls her the queen of the bundle. One of the things that has occurred to me in the last couple of weeks is how struggles can turn into successes. When you think about sales, almost anything you've ever done in your life has involved some type of selling.  You are trying to show people a way to protect themselves. Everybody understands insurance is one of the necessities of our lives. However, it's not always the easiest thing for someone to buy into because their priorities are different than ours. So it's up to us to make them feel it. It doesn't matter if you work for someone or if you're an independent contractor or you're part of a career organization. You still own what you do. You are still responsible for your successes.  Everybody else calls her the queen of the bundle. She knows her stuff. She's been the insurance industry for a long time. Super knowledgeable. She's hilarious. She's fun to be around like at the VIP party in Dallas.  I've gotten to know her a lot over the last several months and I'm telling you, she is awesome. She's fun. She is high integrity, high work ethic, great with people. Everyone loves her and I don't know how you wouldn't. Okay, so she's built an incredible company. Super excited to have you a part of this thing and I'm looking forward to spending a lot more time with you even in Dallas. Please welcome to 8% nation virtual stage. My good friend, Miss Galen Hendricks. Buddy, I hope you're having a great day. My name is Galen Hendricks. I am the CEO of Senior Security Benefits and also one of the original founders. We are so excited to have been asked to be part of 8% nation this year. We're coming to you guys here from lovely Fort Worth, Texas. We know that a lot has changed in all of our lives but thank goodness that we can work virtually and keep trucking along and being successful. One of the things that has occurred to me in the last couple of weeks while sitting here and thinking about what I would talk to you about at 8% nation is how struggles can turn into successes.  I think one of the reasons why I've asked to be a keynote speaker at 8% nation. So I'm super excited to be with you guys today. And I'm excited to share with you some of my story and hopefully what you can expect in July.  This will be the third one. I didn't make it to the first one. I made a surprise visit at the second one and here I am gonna be a keynote speaker on the third one and I couldn't be more excited. You know, when you think about sales almost anything you've ever done in your life has involved some type of selling. Well, my story is no different. I started out in the insurance business, believe it or not. It'll be 35 years this May.  But there's been a lot of struggles along the way. There's been a lot of what to do, a lot of what not to do. But most interestingly enough, in the last couple of months I got asked because I'm a female CEO, if I would do a podcast on the glass ceiling situation and I really at first was like, sure, I'll do it. And then I got to thinking about it. I don't know that I've ever really had a glass ceiling. I don't know that there's ever been something that I didn't feel like I could do. And I think that's because I've had some amazing, amazing leadership, parenting, mentorship in this business and that's what I've tried to do. I have tried to educate, I've tried to mentor people and some days I sit there and think about what does CEO really stand for? And I don't believe it's chief executive officer. I really honestly believe it's chief educational officer.  You are trying to show people a way to protect themselves. Almost everybody at 8% nation that I know of is involved in some type of insurance. And that's even if you're in the real estate world. Everybody understands insurance is one of the necessities of our lives. However, it's not always the easiest thing for someone to buy into because their priorities are different than ours. So it's up to us to make them feel it, make them get it, make them understand why it's so important. Again, why you're the CEO of your business. It doesn't matter if you work for someone or if you're an independent contractor or you're part of a career organization, you still own what you do. You are still responsible for your successes. And so what I'd like to talk to you today is about how to turn those struggles into successes. So for me, going back to the glass ceiling conversation when I went back and tried to figure out what had really gone on in my 35 years, there were some definite struggles. There were times when I really didn't think I could make it. But the good thing about 8% nation and what we're gonna be talking to you guys about is you cannot fail if you don't quit. So it's up to you to keep pushing even when times get their toughest. If you are one of those women that may be having trouble with the glass ceiling, you know, there are so many people that will welcome you, welcome what you do. I have had the fortune of working with some of the greatest people in this business and primarily men. Men have been a big supporter of mine. Some of my best students have encouraged me. Part of the reason I'm part of 8% nation is because some of my really good friends introduced me to Cody Askins. And it wasn't within but a week that we discovered that Cody and I had a lot in common. We really liked to help people. And that's what I've really figured out. Sales is about, it's educating, it's helping. One of the biggest struggles you will find is once you become successful, you will have people in your corner cheering you on, excited about it, couldn't be happier for you. And then there's gonna be some people that end up becoming what we so commonly called as haters. People that are kinda jealous that you kept pushing, kept moving, kept going when they didn't. There's three important times in my life where perseverance and not quitting proved to be incredibly fruitful. And they were huge struggles, but they turned out to be incredibly huge successes. One was the standardization of Medicare. And you know, we're all still here. And you know, our company is a leader and distribution of Medicare supplements.   When it came into our business, we could have easily said, I quit, but we didn't. We stayed in, a lot of other people left, and we succeeded and we became a leader again. And then there was this little thing called Obamacare in 2010 that I'm sure you're very familiar with. But we stayed in. We didn't make money selling ACA plans, but we all of a sudden started seeing that we could show our clients one to four products and one call become a whole lot more efficient. And our customers were much more protected. And guess what? Out of that came my new title, which was Queen of the Bundle. And I've been able to take that and educate large distributions and show them how such a negative situation turned into be an amazing gift. So what I wanna encourage you all to do is to keep persevering. Don't give up when it seems like it's dark. You know, when you think about what everybody's going on in the world today with, you know, there's over 13 million people that have filed unemployment. The folks in sales, we get to keep working. And some people say, yeah, but Galen, some of my clients are in that 13 million. And I would say to you, they need health insurance more now than they ever have before. Even though the government will tell you that they're gonna pay for the uninsured people, the problem with that is there's gonna be a huge line. Having health insurance is really, really important. I get asked all the time, Galen, what is it that you do? And I always say, I'm an insurance agent. I didn't dream my whole life to become an insurance agent.  But my father was a debit man. And that basically means he worked as tell-off for very little money, but he was an amazing provider for our family. And what he taught me in all of that is that sometimes people don't look out for themselves. So you have to look out for them. And sometimes that means making them incredibly uncomfortable, you being incredibly uncomfortable as the salesperson. But in the end, what you're planning on doing is getting that customer very comfortable. There's a lot of people out there that say the word I was taught by my very first boss at 17 years of age, not to say can't.  And over time, I learned why. A lot of people will tell you that they can't when really what they mean to tell you is they won't. You have got to get that client understanding why what you're offering to them is so very important, not just to them, but to their family. When you offer insurance products or any product that you have to educate your client on, it's up to you to keep moving, to keep being that educator, to keep being that public servant, and know you're gonna have struggles. Know you're gonna have people that aren't pulling for you like they once did. You will have issues that come up that will cause you severe indigestion, severe frustration. But what happens is, is after you get through it and you move through it, and you get to see how you came out on the other side of that, or when your client calls you and says, hey, my husband passed away. I need to figure out how to get that life insurance money.   You see, my dad was diagnosed with carcinoma in the stomach at 52 years of age. He was given six weeks to three years to live. He lived five weeks and five days. What killed my father was financial stress because he had life insurance. He knew that we would be taken care of and he just wished the death would come sooner. With cancer policies, my customers get to have a living benefit because my hope is that when I show a client a product, they understand what that will do for them and their family. It's like a story that just happened to me recently with a client, not a client of mine, a client of someone else's. I serve as a director for the Ronald McDonald House here in Fort Worth, Texas, and she needed help getting a claim paid. I was able to call some friends at some insurance companies that have been very good friends and they were able to move that along for her. She asked me what I did. I said, I was an insurance agent. She said, what products do you offer? And I said, I have a cancer plan. I wished I would have gotten to you sooner, but here's what it does.  And she said words that still pierced me to this day. And it makes me sad that I didn't get to her insurance agent before he sold her a plan. And what it simply was is this. She said, how come I don't know about those plans? It's up to us as salespeople to always educate our clients on the policies that we have. It's a struggle, but it will turn into success. My desire for each and every one of you who are watching this today is that you will be at 8% Nation along with me and a lot of my friends so that we can share together, become more successful together, hear your struggles and give you some solutions. Instead of you sitting there figuring out how you're gonna make this bill or that bill, it's up to us to show you how to make this sale or that sale so you will never have to worry about that bill again. I am so incredibly excited about getting to July. I know you are too. If there is anything I can ever do to help you or if you need any assistance, I have a Facebook page called Queen of the Bundle. Go out there, like the page. I will be able to help you if I can and I think I can because don't remember. The word can't is not in my vocabulary and it shouldn't be in yours either. I wish you happy days, happy sailing and happy selling at the same time. Yes, with that Texas accent, sometimes sailing and selling sound the same but remember you will never get better as long as you are better and your clients will always not be interested if you are not interesting. So I'm gonna sign off today with this. Remember, you cannot fail if you don't quit. Have a great day and so very excited about seeing you again at 8% Nation. Have a good one. Ms. Galen, you go girl, awesome job. I told you she knew her stuff. I love her story. I love her passion for the industry specifically and I'm excited to see her on stage in Dallas. Unbelievable job. Hey, if you love this video and you wanna learn how to attract more customers immediately, all right? The next video is right here. It's for you. Click on there and I'll see you over there. In the insurance industry, if people can reach out to you through the business with you, it's a lay down sell. Yeah, I think if you really understand the concept, the concept of being famous, right? I live outside of.


So, now I am going to look at another example which will lead us to the second notion of convergence called convergence in probability. So, what is the example? So, let us again try to look at a sequence of random variables I have, I am going to draw one random variable here X1. Again these random variables are defined on my unit interval probability space. This X1 is going to be one throughout and then I am going to define two random variables X2 which is going to be, I am going to split this half which is going to be like this. So, let us call this random variable 1, let us call this 2, let us call this 3, let us call this 4, 5, 6 and 7. So, this random variable X1 is going to take one throughout. So, X2 is going to take 1 in the first half, X3 is going to take 1 in the second half. Now, X4 is such that it is going to take value 1 in the first quadrant, X5 is going to take value 1 in the second quadrant, X is going to take value 1 in the third quadrant and X4, X1 is going to take value 1 in the last quadrant. Now, I can keep on reading sequence like this, right? So, you see that like I have doubled up from, this is like a good going in a geometric sequence and I can generalize this and write it as this equations as all j and n prime equals to 2k plus j, ok. So, let us take from n and then let us allow, let k to be the logarithmic of that and then if you let j go to 0 to 2 to the power k, for all the indices which are between 0 to 2k, 0 included, the n prime index will have, it is going to take length 2 to the power minus k on this interval. So, just like this is generalization, but you can just let us just try to understand for the case n equals to 1. So, n equals to 1 what happens? This is going to be 0, right? And in that case what is the possible j's I have? It is only 0, right? Then I have 2 to the power 0, 1 and j is 0. So, I have n prime equals to 1. So, that corresponds to this case and what is this value? It is going to be of length k was what? k was 0, right? So, this is 1 and it is in the interval 0 to 1 and now if you go and take further for these two things, I mean you can continue to do this and you will see that in the in this first half, when k equals to 1, this is going to be 0 and half, you are going to get the value of 1 and when you go to the next index, you are going to get, this is going to take the value 1 in the next interval. So, you can like this, you can generalize this description of your x n's and then you have a sequence of x n. And now let us try to understand where this, if I have a sequence of random variables like this, where do they converge? So, let us say n equals to 1, then I can write it as defined like this. So, take any n that can be expressed as in this form for some k and some j, right? So, and for that, it is going to be of length 2 to the power minus k over this interval. For example, if you are going to take n equals to 3, what is the k and j corresponding to this? k is going to be 1 and j is going to be, so then 3 can be expressed in this form and for this, what is the value of this? k is equal to be 1, right? This is going to be in the half and what is that interval? So, this is going to be, this j is going to be 1, this is going to be 1 by 2 to 1. So, this is covered the entire thing. And now for this kind of x n, I have, where does it converge? So, first of all, any guess what should be the limiting random variable here?  You expect these random variables to put mass on a small, small intervals, right? It is going to take value 1 every time, but it is going to take that value 1 on a smaller, smaller intervals. So, what do you expect as n goes to infinity? What? It is shifting, it is like putting mass on a narrower intervals and it is not putting, so here it has put mass on the first quadrant, but rest of the quadrant, it has put 0 mass. Here it has put mass on the second quadrant and on the other part it is going to be 0, right? So, like that it is always continued to happen. So, due to which, okay, let us assume, you know that as n is going to tend to infinity, right? This interval is shrinking. It is only going to put the mass on a very, very small, but we are right now unable to imagine what is that small interval where it is going to put mass, but everywhere other part, maybe it is going to put a 0 mass. Let us say whether x n converges to some x where x is 0 for all, x of omega is 0 for all omega. Okay, now let us try to understand. So, for this what we need to do? We have to understand whether probability that omega xn of omega converges to x of omega is going to be 1. Let us see. So, take any value of omega. So, if I want to include this omega in this set, that xn of omega should converge to 0, right? Because that is what I have taken x to be 0. So, now let us say that omega is here. Let us say for the time being let us take that omega to be 0.3. So, here omega 0.3 got value 1 when it is x1 is giving value 1 to this. And what is x2 is giving it? 1. But what is x3 is giving it? And here what is x4 is giving? And what is x5 is giving? 1. So, if you are going to construct such a binary thing, so here it is 1 element, 2 element, 4 element, then I have 8, 16 like that, right? In each of this rows that omega will have 1 at one of the graphs. In all the other graph it is going to take 0 value, right? So, however deep you go, you will see that after some number is it possible that x of omega is going to always get 0 or if you go further it is going to take 1 again. So, for example, if you start it took 1, it took 1 here, it took 0 here, 0 here, it took 1 here. And if you come back and when we went in the this row, it took 1 again. If you go to the next row below, it again took sometime 1, right? So, however down you go you will end up with some n where there it is going to take value 1, right? And if that is happening, what is whether x of omega x of 0.3 is going to converge in this case at all? No, right? That violates our definition that however far I go, I will find an n such that after that my x of omega is going to be 1. So, that is why it cannot converge to 0. And this is true for any omega, right? Do you understand this point? So, there is no omega here such that x n of omega converges to 0. So, then what is this set? If my x is 0, this is a null set, right? And if this is null set then this probability is actually 0, not 1. So, because of that my x n is in this case x n is not converging to x almost surely and where x is 0, identically 0, okay? But as you see from here, as my x n goes, most of the time I am getting the value 0 for each of the omegas, but it only happens that, but after some time 1 happens, but rest of the time it is going to remain 0. So, it looks like most of the time my random variable is going to take value 0 for omega, I mean as I proceed in n, but it is for some n further I go, it gets violated, that is why it is not true. But still you want to like you will see that most of the time for different values of n it is going to take value 0, then it looks like maybe like I could think of this sequence of random variables convergence to random variable x which is identically 0, but this definition of almost sure convergence is not capturing that notion. Maybe I need to capture, I need to define something weaker than that. So, for that we have this another notion of convergence called convergence in, so what it says? Okay, so now we are going to say that a sequence of random variables converges to x in probability if for any epsilon greater than 0, now if you look at this sequence of probabilities, now notice that this limit is now outside. If you are going to look at the probability of this event being larger than epsilon, if this sequence converges to 0, then we are going to call this as convergence in probability and we are going to denote this as x n converges to x, put a p here or limit as n tends to infinity x n is equals to x in p. So, now let us see whether this example we had here satisfies this definition. So, let us again take the same thing. This example here, let us take x to be 0. So, to do this, to verify this I need to verify it for any epsilon greater than 0. So, let us take a fix on epsilon greater than 0. Now look at this, so x is anyway 0 for all the points. Now I am asking what is the probability that mod of x n is going to be greater than or equals to epsilon. So, mod of x n I can just take it as x n because x n's are all positive in my case, x n being greater than or epsilon equals to 0. If I look at the limit of this, where this limit is going for this example. So, recall that this x size here I have defined it on unit probability space. So, the probability that it takes a value in some interval is nothing but the length of their interval. So, now what is the probability that x n is going to take a value greater than or equals to epsilon. If you take n any n, what will be the probability of just mod x n itself? What is the probability of mod x n itself? It is just going into the length of that interval where it is taking value 1. And as n increases, these values are shrinking. So, the probability where it is going to put a mass positive mass is also shrinking very fast. And this epsilon here is strictly positive. So, if it is strictly positive, you should be able to find n sufficiently large such that after that you are going to put a value in a smaller, smaller intervals whose probability is going to come down below epsilon. So, then this event will never hold and this probability will be 0. So, because of that is it true that this is going to be 0 for any epsilon greater than 0. So, then does it satisfy my definition of convergence in probability? That is right. So, x n converges to x here in probability. So, what is the difference between then convergence in almost sure sense and the convergence in probability? So, in a way, look at this. When you are looking at this sequence at any n, you are looking at the joint distribution of your sequence, sorry, your limit n given x. It is looking at these distributions at any point. But whereas, if you looked into the definition of almost sure, there you looked at probability. So, here you are computing the probability on the entire sequence. So, entire joint distribution of the entire sequence is affecting the definition of almost sure. Whereas, the definition of the probability convergence in probability is only looking at a pair wise distribution. The joint distribution of this pair x n and x at any time. But whereas, this is looking at the joint distribution of the entire sequence. In a way, this almost sure. So, that is why this almost sure convergence is a much demanding requirement than convergence in probability. So, later we will see that that is indeed the case. Almost sure convergence is a much stronger property than convergence in probability because almost sure convergence implies convergence in probability, but not the other way around. So, now, before I do that, if you look at this definition, what I asked is the probability that x n going to take a value larger than epsilon to be converging to 0. But it may happen that so here the value that it is going to take greater than epsilon is on a very small mass, but it may happen that this value, it takes on this small mass itself could be very, very large. For example, in all the examples I just erased now, instead of letting my x i only taking up to value 1, I could let it take very large value. Instead of 1, I can make it 100, 200, whatever. So, in those case, yes, it is happening that they are taking this value on a smaller intervals, but still the value they are taking on this smaller interval could be very, very large. Now, when you are interested in finding the expectation of the random variable, you are just not interested in the probability, you will be all also interested in the value taken with their probability. So, probability will involve product of the value of the random variable and the probability term. So, because of that, it may happen that some sequence of random variable satisfies that, but on the region where they are taking positive values, the small probability that positive value could be very, very large. So, in some application, this may not be desirable for you. You want it to be take small values even in that small interval. So, this is like some cases like fine, it may happen that when I am going to put money in a game, the probability that I am going to lose could be very small or let us say lottery. The probability that I win is going to be very small, but if I win, the amount is huge. So, the value product is going to be very large. So, when you think about this in the risk sense, something failureing is very small, but if it fails, it is going to be too expensive for me. So, in that case, you are interested in both the value taken. It may be taking some huge value with small probability, but that is of a concern for me. So, when that is the case, instead of this, you may be interested in knowing whether it converges in some expected sense. For that notion, so for that to capture those things, we are going to define another notion called convergence in mean square sense. So, now instead of looking at the value of the random variable itself, what I am now looking at is the expected value and not just the expected value, but the mean squared error from the random variable X. So, we are going to say that the sequence of random variables going to converge to X in mean square sense, if I look into difference, square difference and take their expectation that goes to 0 as n tends to infinity. So, notice that I have put this condition that each of this random variable should be such that their second moment is finite. So, why is this? Now, the natural question comes is fine, we are looking at different notions of convergence. Is it true that one already implies the other? So, we already said that convergence in probability is a weaker notion of convergence in almost sure sense and now we are talking about converges in mean squared sense. Is it already some weaker notions of earlier one or it is a total independent convergence we are talking about? So, we will look into that. So, before that let us look into an example. Let us take this random variable. Again, this is defined on a unit interval probability space. So, now each of this random variable have defined such that if you fix an n, then it is going to vary like this in between 0 to 1 by n, it is going to take a constant value of a n and then it is going to take a value of 0 in the rest of the interval. So, what do you think about this? Let us try to understand whether it convergence in almost sure sense. So, let us say to understand almost sure sense I need to compute this probability. So, first what is the guess? What is your guess? What could be the limiting? So, as n is tending to infinity, it is going to shrink this interval and it is the kind of putting value, nonzero value and only some small interval which is shrinking rapidly as an increases. So, most of the values of omega are only going to take value 0, right? Because this 1 by n as an increase I should keep on shifting to right to my left and then most of the time. So, you expect guess is like this will converge to 0 random variable. Delta does not matter to me, right? Like 1 point does not matter. What matters is the entire interval. If it is 0 everywhere, I am going to just call it a 0 random variable. So, is it converges in almost sure sense, right? Like if I am going to take a limit of n tends to xr omega whichever omega I take.  But as n increases, this guy comes to the left of omega and then it is going to get a 0 value, right? So, because of that every omega is going to be taking 0 value. So, this is going to be 1. So, xn converges to 0 almost surely. Now, what about its convergence in probability sense? So, what I need to verify that? I need to verify whether limit has n tends to infinity, probability that xn minus 0 going to be epsilon is going to be 0. Is this true? So, again, so I have already substituted 0 for x here, right? Now, if you look into the probability of this interval xn, that is nothing but this interval, right? And that interval is shrinking. So, it cannot, after some time it cannot exceed epsilon. So, after that point it is always going to be 0. So, it is also true, converges in probability. This is greater than epsilon. What you are talking about is probability of mod xn, in this case simply xn is going to be simply what? 1 by n. Yeah, so what we are just saying that, yes, it could take value greater than epsilon but on what regions? That is a region which is shrinking, right? Small, small shrinking region, that probability shrinking. So, that goes to 0 as n goes to infinity. So, now let us compute converges in mean squared sense. So, what is this value then? So, to compute this I need to compute expected value of xn minus. So, x I have taken to be 0. So, can I compute this expectation? So, this is nothing but expectation of xn square, right? What is this? So, xn is taking what value? It is going to take value an and with what probability 1 by n and with the other probability it is going to take value 0. So, this is going to be an square by n. Now, I want this probability to go to 0, if I have to claim that xn converges to 0 in mean squared sense. Is this true? So, notice that an is a number now and I am dividing an square by n. Will this go to 0? For any n, for n tending to infinity an square by n always goes to 0, n. Yes, an is a function of n. So, okay. So, fine. Let us say if I can choose like, say I can define this an to be log n. So, in this case it goes to 0. If I am going to say an equals to square root n, it goes to 0, n square. So, square root n is become the ratio 1, right? So, this does not go to 0 in this case. So, in this case s go to 0, but in this case it does not go to 0. So, what is happening in this case? So, it looks like in this case the convergence in the mean squared sense here at least depends on how this amplitude is or how this the value is. Like if this value of an is going only logarithmically in n and this guy is shrinking much faster than this guy, then yes, then this guy is going to go to 0. But if its amplitude the value taken is going to be much, much larger that means it is then this interval or sorry when this whatever this value of n, then this product could be still much, much larger, right? So, for example, you could almost come very close to here, but let us say n equals to some 10 to the power 6, you are like almost 1 by n is like some 10 to the power minus 6. So, you are putting positive value only in the small interval, but you are going to take a value of also you are going to take the value 10 to the power 6. Let us say like your chances of winning a lattery is 1 in million, but if you win you are also going to get a million dollars or million rupees. So, in that case that value is going to be still high, right? So, here as you see whether it converges in mean square sense or not depends on the sequence a n. So, it depends on, so far we looked at the sequence of random variable converges to some x. Now, we will look at the distribution itself and if is it possible that like if I have a sequence of random variables with certain distribution, they will converge to random variable which has some limiting distribution. So, we are going to say that a sequence of random variable converges in distribution to a random variable x if we are going to look at their CDFs. So, f of xn denotes the CDF of random variable xn, which is if you look them at a point x and if it converges to a CDF of x at that point and this point x are such that they are the continuity point of the CDF of the limiting distribution. In that case we are going to call it as converges in distribution, so sorry continuity point of x, fine then. So, let us stop here.


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A fiery horse with a speed of light, a cloud of dust and a hearty howl silver, the lone ranger. With his faithful Indian companion, Tato, the masked rider of the planes led the fight for law and order in the early western United States. The stories of his strength and courage, his daring and resourcefulness have come down to us through the generations, and nowhere in the pages of history can one find a greater champion of justice. Return with us now to those thrilling days of yesteryear. From out of the past come the thundering hoofbeats of the great horse Silver, the lone ranger rides again. Come on Silver, faster boy, faster! Hail Silver! Twilight was slanting down into the gloomy depths of Boulder Canyon. The last rays of the setting sun created fantastic patterns of red and purple and gold on the canyon walls. The lone ranger and Dan Reed, together with their stalwart Indian friend, Tato, rode slowly along the rim-rock trail. Be very careful here, Dan. I sure will. Remember, always let your horse pick his own footing on loose rocks. Ah, this plenty-blant-blant place for a horse to fall. Golly, she looks awful far to the bottom. Nearly over the worst part. Trail widens out just ahead and we can look for a place to camp. I bet the horses will be glad when this ride's over. Yes, they realize a danger and sometimes get very nervous. Boy, sure feels good to be riding on solid ground again. Those rocks are too slippery for comfort. Good mountain stream ahead, good place to make camp. I don't see any stream. We can't see it from the trail. It lies hidden beyond the ridge of Jackpine. Oh, hey, the new and Tato must have been through this country before. Yes, we were here several years ago for the same reason we're returning now. Oh, you mean about that escape convict, Red Brewster? Yes, that's right. See, Dan, Tato and I were able to help Sheriff Ashley round up Red Brewster in his outlaw bunch. But why would Brewster come back here? Because he knows the country like a book and has friends here who will shelter him. Mainly because he swore to get revenge on the sheriff. Then the sheriff will be on the lookout for Brewster, won't he? I doubt it very much. There's no telegraph near here. There hasn't been time for the sheriff to be notified by mail that Brewster escaped from the territorial prison. Golly, that's a good thing we got here before. Listen, who's over here? Hello, the rifle shots. Shots come from back a ridge. We'd better ride over there and investigate. That's a pretty bad place for ambush. We know it is. Come on, Silver. Later that evening in the town of Navajo Flats, a strange meeting was about to begin in the office of Sheriff Joe Ashley. The small group of men present were grim-faced, waiting expectantly for something or someone. Finally, old Jim Ellis, owner of the rafter L Ranch, broke the silence. I don't know what it's all about, no more than you do. All I know is that Jake Nelson wanted us to meet here in the sheriff's office tonight. Jake ought to be along in a minute. Jake's finally decided to agree with you boys about sending for this fellow Rod Blakely. I sure hope so, Sheriff. Blakely has the reputation to be in the fight in his two-gun marshal in the south-west. Yep, he's an old Lord Brewster for sure, Reagan. Yeah, that's what we need. Me, I'm willing to admit this job's too big for me to handle alone. And if you fellas want to send for Rod Blakely, that's all right with me. Well, he's glad to feel that way about his years. Look, here comes Jake Nelson now. Who's this stranger with him? I don't know. Never saw him before. Evening, gents. Last time we got together, you men wanted to raise the money to send for Marshal Rod Blakely. Yeah, that's right. You still figured the marshal could stop this cattle stealing here in the valley? Oh, with the sheriff helping him, of course. It sure is with a tragedy. How about you, Sheriff? You want another lawman in here? I'm agreeable to anything. It'll get the job done, Jake. All right. Now then, I get a real surprise for you, gents. I reckon you've been wondering who this armbary is abroad with me. Well, gents, I want you to meet the toughest two-gun marshal in the Southwest Territory. Rod Blakely. Well, shame it sure is a surprise. Mighty pleased to meet you, Mr. Blakely. Right proud to know you, gents. Howdy, Sheriff. You're more than welcome in Sage Valley, Marshal. We got plenty of work for a troubleshooter. Oh, go on, Jake. Here all along, we figured you was again sending for Marshal Blakely. Now, here he is. Well, boys, I got to thinking it over and sent the marshal word to come running. I reckon him and the sheriff will get along fine. Sure.  Well, we don't. We'll be some of the ranchers folding up before very long. It's right, Marshal. We can't afford to lose any more stock. Hey, Sheriff, there's some Redskin out here. He wants to see you. What's he want? Says he wants to see you, personal. All right. Well, what's on your mind, Redskin? Uh, me bring message. Yeah? What message? Who from? Oh, here. You take it. Huh? What? Say what? Hey, just a minute there. Well, I'll be... Hey, what's wrapped up in that note? It looks to me like it can't reach me that is silver. Yeah. It's a silver bullet. What's it mean, Sheriff? Help! Help! It might mean that the sheriff's office is going to have more help than they've been figuring on. What do you mean by that, Sheriff? Excuse me, Marshal, while I read this note. Well, after midnight, when a pair of horsemen rode silently down the main street of Navajo Flats. Better leave the street down and come up to the sheriff's office from the rear. Ah, light burning in the sheriff's office. Yes. I won't be in here very long, Tonto. Oh, silver bullet. Oh, silver bullet. Keep a sharp lookout. Signal if you see anything suspicious. Ah, me watch. I'll try the door. Sheriff Ashley! I wonder if he could have... No. It looks as though a red booster got here before me. Shot right through the heart. I've got you, killer. Help with your hands.    Marshal Rod Blakely. Looks like I happened along just in the nick of time. You're, uh, Marshal Blakely? That's right. You're caught red-handed, killer. You're mistaken, uh, Marshal.  About what? About an escaped convict named Red Brewster. He was coming here to kill him. Too bad, stranger. You almost had a good alibi there. Guess what do you mean? Red Brewster was shot and killed two days ago. In a boulder canyon. Are you sure it was Brewster? Yep. Just like I'm sure that you did kill the sheriff. I can prove that I was his friend. Prove nothing, Mask Man.  Don't fire that gun. No. Staying right where you are, mister, and don't move a muscle. That'll bring him on the double-quick. Yes, and you can tell them I wasn't able to wait. Yeah, what? All right, Tuttle, say what? You drop gun. This knife, plenty sharp. Hey. Drop gun. Good night, Marshal. We'll meet again. Come on, Tuttle. They're getting away. Out the back. What about the sheriff being killed? Can't see a blame thing out there. Get to your horses. Is there any idea who it was, Marshal? I know who it was. It was a mast man who caused himself the lone ranger. Listen, man. There's too much of a job for Sheriff Ashley to clean up this valley alone, and there's too much for me. I call this meeting tonight because I need your help. Every one of you. Cut on the red drill, Marshal. All right. We know there's just one man at the bottom of all this, and that's that mast man.  Marshal's right. Let's get after that mast man. Let's get started right now. Take every man you've got and scour the hills. Say on that mast man's trail until you find him. Good night, boys. Mr. Nelson. Yeah? I'd like to see you a minute before you leave. All right, what's on your mind?  What do you want? You know, if I was a fortune teller, I'd say that someday you're going to be an awful rich man. Yeah? Yeah. And if I was a fortune teller, I'd say that about the time you get awful rich, I'm liable to turn up awful dead. Cut out that kind of talk. You'll get your money, and the job's finished. More like to get a 45 slug in the back.  Uh-huh. Careful too. The way I figure the sheriff's out of the way. Joe Ashley was getting suspicious too. Sure. Tomorrow every cowboy in the valley will be out chasing this Lone Ranger gent. When they get home, I don't reckon they'll find much, do you? I reckon you know my plans. There won't be a pound of beef in the valley nor a building left standing. Sure. Of course I know your plans. What's more important, I know where all that stolen beef is right now. You've been spying on me? I sure have, Mr. Nelson. Or would you rather I use your real name? Never mind that. I know where that box canyon is. I know all about them hired gun slicks on your payroll. I know who Jake Nelson really is. So you see, I sort of know too much, don't I? I didn't say that. You don't have to. I don't trust you, Jake. I want my money right now. I'll get it for you in the morning. No, I get a better idea. Let's you and me right out to your place and get it tonight. All right, come on. Fine. Just so you don't get any funny ideas, I'll just take charge of that shooting eye in your packing. Get going. That's too bad their answers couldn't have stayed for the last part of that meeting. Ah, then be plenty surprised. Yes, we'll have to act fast now. Here, go right to our camp where Dan is. Give him this note. And who note for? There's a regiment of cavalry stationed at Whitewater Bend on the Powder River. You go with Dan, but let him deliver the message to the commanding officer. You wait on the back trailer to lead the troopers here. And see that Dan stays at the army camp until we call for him. Whitewater Bend? That's long right from here. I know, but it's our only chance. We're going to need the army's help before we're through here. And what you do, Kimosabi? I'm going to see that our friend, the marshal, stays in the valley until the soldiers arrive. Come on, let's get to the horses. Steady, Silver. Right for Camp Tutto. Yet Dan started with that message. I'm heading for the Ellis Ranch. Come on, Silver.    The curtain falls on the first act of our Lone Ranger drama. Before the next exciting scenes, please permit us to pause for just a few moments. Now to continue our story. The Lone Ranger, discovering the plot between Marshal Blakely and the ranch owner, Jake Nelson, sent Tutto and Dan to bring the cavalry from the outpost at Whitewater Bend. Meanwhile, as stride the great Whitehorse Silver, the masked rider of the planes races through the night to Jim Ellis' Rafter L Ranch. Easy, Silver. Steady, big fella. This is the Rafter L. Easy now. There's the main ranch house, but we'll find Mr. Ellis. Oh, Silver. Oh, boy. Easy. Stay here, Silver. Richard!  You, the masked man. What do you want here, you sneaking vomit? I think it will pay you to listen to what I have to say, Mr. Ellis. Ain't nothing you can say to me, you killer. I'll prove to you that I'm not the outlaw you think. But first, I'd like to ask you some questions. Go to Blazers. Ain't answering no questions for you. How long have you been losing cattle? Just as if you didn't know. I can help you, Mr. Ellis, if you'll let me. You know when we start losing our beef.  Listen to me, and you can recover all of your cattle within 24 hours. Oh. I know who's responsible for this wrestling. I know who killed the sheriff. What's more, I know who burned your barns the other night. You do. Strangely, you sure got a convincing way about you. I hope so, for your sake. Uh, what is it you want to know? When did the wrestling start? Uh, most a year ago. It was just in small dribbles at first, but lately, them night writers have been making off with some sizable herds. Yes, I know. Today, those same night writers are planning to strip this entire range. Are you and your men right to the hills looking for me? Yeah. How come you know all these things? I make it my business to know. Now, tell me, when did Jake Nelson come to the valley? Jake? Oh, it's been about, uh, three years, why? I suppose the wrestlers have struck at his place, too. Yep. Jake's diameter spread has been the heaviest loser so far. I feel sorry for Jake. Yes, and you're going to have reason to feel sorry for her. Uh, what do you mean? Jake's my friend. I suppose he's made you an offer for the Rafter El. Matter of fact, he has. But Chuck, she's been buying up range land ever since he came to the valley. I see. Now, just think a moment, Mr. Ellis. Doesn't it strike you as strange that a man would lose as much stock as Jake Nelson has and still be trying to buy up these other ranches in the valley? Why would a man want those other ranches when they're all losing money? Yeah. Are you trying to tell me that Jake is a... I have positive proof that Jake Nelson is out to crush every ranch owner in Sage Valley. Good day. Nope. That's ridiculous, man, where Jake is the one who put up the money to bring Blakely in here. Yes, and that's where our friend Jake made his biggest mistake. You see, Mr. Ellis, Nelson and the so-called Marshall are really partners in this plot. That's mighty strong talk, man. Can you prove what you're saying? Yes, I can prove it. Will you give me the chance? Well, I'm willing to be convinced of the truth. Good. Now, call out your men and hurry. At this hour of the night? Yes. There isn't a moment to lose. We've a lot of writing to do before daylight. All right, come on. Well, wait till I get the land in. This'll fetch me an hurry. Hey, man, you like it, boy? I sure hope I let myself get took in on a slick scheme, man. You'll see in a moment that I'm working for you. What's up, Jim? Anything wrong? Hey, who's that tall jigger with the boss? That's the masked man, the fellow that killed the sheriff. Well, rain up and walk back there, you wild collutes. Me and the masked man have been having a collab here. I reckon he's got something to say to you. Just one thing, masked man. Understand, I believe you're all right, and I'm willing to do anything to get my beef herds back.  I know just how you feel, and I don't blame you for being suspicious. Now, listen to me, man. Starting at daylight, every man in the valley is supposed to get out and scour the hills and mountains, looking for me and my friends. That's why I'm here, so you won't have to look for me. Never, Cassini. You man, leave this ranch today. When you get back, the last of the rafterl herd will be gone, and every building on the place will be in ashes. Believe me, it's the truth. The same thing will happen to every ranch in the valley that's left unguarded today. What do you figure we better do?  Post a strong guard around the buildings. Then send a rider to warn every rancher in the valley to do the same thing. Every rancher makes men? No. Warn every ranch in the valley except Jake Nelson at the Diamond Age. Wait a minute. How come we don't ride to warn Jake Nelson, too? Because Mr. Nelson knows all about it. In fact, he's better prepared than we are. Now get to your horse's men and ride. When Tonho left the Lone Ranger at Navajo Flats, he rode at a furious pace to the well-concealed camp near the canyon.  Tonho gave Dan the Lone Ranger's note and instructed him to remain at the Army post after delivering the masked man's message. Tonho would wait on the back trail and lead the cavalryman through the mountains to the Lone Ranger. After a breathtaking ride that would have killed an ordinary horse, young Dan Reid arrived at the Army outpost where he was taken at once to the commanding officer. Captain Kennedy. Yes, Corporal? This boy wants to see you, sir. Come in, young man. What's on your mind? Thank you, Corporal. You can wait outside. Yes, sir. Here's a message for you, sir. Say, who gave you this note? A very good friend of mine, sir. The Lone Ranger. What? I've heard of him. What's he doing in this part of the country? He ran into some trouble in Sage Valley. This friend, the sheriff, was killed, and now he's uncovered a plot to ruin all the small ranchers. Yes. So the Lone Ranger sent Tonho and me to get the cavalry. Tonho? That's the Indian who always rides with the mask man. Who are you? What's your name? My name's Dan Reid, sir. Am I to proud to know you, Dan? I'm supposed to stay here and camp until a trouble's over with. Tonho's waiting for you on a back trail. Why didn't he come here with you? He said he wanted his horse to be fresh when he met you. He's waiting where the trail branch is off the Boulder Canyon. This message says there's a man in Sage Valley posing as the United States Marshal. So I guess that makes it our business. Corporal Hayes. Yes, sir. Have the bugler sound boots and saddles. We're riding to Sage Valley. Yes, sir. Troop is ready. Make yourself at home until we return, Dan. Thank you, sir. Troop! It's been scattered over a thousand acres. And it looks like somebody might have been expecting us. Hey, here comes your friend, the Marshal. Something's going haywire for sure, boys.  Oh, oh, there, oh. Jake, can we be heard in the valley? It's been scattered. We sure can't drive no cattle out of here today. Maybe not. We can start some mighty big bonfires. Come on. That's right for the ranch house. You know what to do. Don't leave a single timber standing. Get up there. After you and Butch circle around, get that fire started. Me and the boys will take care of the rest of the buildings. They sure looks deserted all right. Yeah, we'll find... All right, boys! Hey, look out! Look out, it's a trap. Right out. Everybody take cover. They'll hold up on that house. We'll just keep it metal dark and burn them out like rats. Listen, what's that I hear? Sounds like riders, Jake. Lots of riding. It's a cavalry from Whitewater Valley. Chris Nelson, we've got to make a break. We've got to get out of here. Keep up. Not Jake Nelson, anyway. That yellow scut can't run fast enough to get away from old Betsy. Hold your head, Jake. Get dirty, poor cat. Betsy wants to talk to you. You got him, boss. Drill him blown center. Yeah, and I got one more for that. Marshall, Jake. Hold your fire, Ellis. I'll take care of the Marshall. Here, Silver. Is there anything for the... Come on, Silver. Marshall! Ma'am, get out of the saddle. Come on, that's the Marshall's head. It's all bloody. Bloody my foot. That's hair dye. Marshall's redheaded. So you're the lone ranger. I'm proud to know you, Mr. Thank you, Captain. Glad to know you. Muddy glad you got here in time. Just who is the water-soaked rat over there? That's the Red Brewster. He recently escaped from territorial prison. Brewster? Well, we heard he'd been killed just a few days ago. No, it was Red Brewster working with Jake Nelson who killed the real Marshall, Rod Blakely. And Brewster took the Marshall's credentials, disguised himself, and rode into Navajo flats. He killed Sheriff Joe Ashley before we could get there. What about Nelson? I think your records will show that Jake Nelson served time with Brewster in prison a few years ago. Captain, I just want to say if it hadn't been for the Mayesh man here, Jake Nelson and that redheaded scoundrel would control the whole valley by now. You're right, Mr. Ellis. Well, they sure got the needons. I'm here to say. The Mayesh man showed me where the stolen beef was being hailed.  There he goes. He and the Indian who led us here. They're going to sure want to think they don't, briefly. Hey, looks like they're riding away. They'll have be horned swuggles. Guess what kind of a fella is that Mayesh man? Well, personally, Jim, I'd say he's about the best friend you'll ever have. You have just heard as a copyrighted feature of the Lone Ranger Incorporated.