So we have been discussing about this mixed flow reactor and of course this is the diagram which we have given and then we just want to know what is the concept of mixed flow and if you do not want to use your brain like you have done in your b-tech you know simply saying that perfect mixing is one where everywhere inside you have same concentration and temperature, okay. So that will create some cobwebs, some you know not properly understanding things like for example. So the moment we say that perfect mixing most of us feel that you know it is instantaneous reaction it is not, right instantaneous mixing that is correct instantaneous the moment you have a concentration of C A not just entering it should be instantaneously mixed to a concentration of C A f or conversion wise X A f C A f is safe because C A not is coming in C A f is is the concentration due to perfect mixing but when I looked at that concentration has come that concentration in the reactor is prevailing please listen to this one carefully that concentration C A f is prevailing due to elements of various residence times inside the reactor and you know that because it is a perfect mixer by definition of perfect mixing itself the outlet residence time distribution and also the inside residence time distribution is exactly same here also you will have 0 to infinity residence time distribution that means okay I will take a small sample outside here and then I will look into what are the residence times I may see there molecules or packets entered infinite time back that means very long time back infinite means not really infinity okay then it also would have average residence time is 10 minutes just for easy discussion so all those molecules will not be exactly at 10 minutes so we will have 100 minutes even 1000 minutes 1 or 2 molecules, right 500, 150, 10, 1 almost 1 second also why all this is happening again because of perfect mixing when you are continuously sending this C A not there may be one molecule which may be very quickly coming out of the reactor and then joining this string so practically that molecule is not able to react but on the average when I look because of the mixing there are molecules now spent you know at 10 minutes 100 minutes 20 minutes 1 minute and 500 minutes all that combination and that will give you the average concentration inside the reactor which is also reflecting in the outside you know the outlet so that is the concentration what we see here C A F is the concentration because of mixing of various elements of various residence times just extend your imagination from here to plug flow so the moment you have plug flow I think this is a tubular reactor like this and we are only talking about exit here also exit concentration is C A F but what about inside concentrations very seldom yeah so if I draw this this may be C A not this may be C A F and of course here I have F A not F A and all that right so we are talking about here the concentration C A F at the exit but the moment I go inside the C A F is different than the higher than the next one then you move somewhere inside that will be higher like that if you come back here then you will have C A not concentration that means here what do I have I have the concentration decreasing slowly right and this is C A not this is C A F okay now it is the same thing if I plot here C A not yeah how do I plot this this is C A not okay this is C A this is C A not because it is instantaneous mixing concentration will decrease to C A F which is this there is no slow decrease here this is C A F okay sorry this is only spatial okay no problem this is C A not this is C A F but now the difference why even writing without writing equation why plug flow reactor should give you more conversion for a given volume for a given volume in this and here you will get less conversion for a given volume is here the concentration is changing slowly and then what about rate here here here do I have different rates why I have different rates concentration are different we know that at any point we have K into C A if it is first order reaction or correspondingly whatever reaction we have so that means what I see at the outlet is the accumulation of all that I do not say not accumulation or not material accumulation we are talking the changes in reaction and the concentration slowly that means number of rates I have inside the thing and what I see here at the outlet is some kind of average of all those rates whereas here how many rates I have only one because throughout the system I have only one one concentration so that is and that rate corresponding to C A F outlet concentration so is it more than or less than inlet concentration definitely much lesser right so if I I mean example if I say that C A not equal to 10 moles entering 10 moles per liter and outlet is 1 mole C A equal to 1 mole per liter okay what is the conversion now we are talking about constant density system conversion equal to 90 percent you know the definition C A not minus C A by C A not so I think for some people I think also better right C A not minus C A F by C A not this will be point nine that is the concentration okay that is the conversion which I get here in the outlet and which is same throughout right so corresponding to this C A C A F the rate if I calculate minus R A equal to K into C A F only one rate which is very low that means that is almost the exit rate right so now in our design expressions we will have always this rate in the denominator by R A okay by something by R A it will come now here I have only one rate and the rate is very less okay I mean very less means it is not I mean it is corresponding to exactly this rate only okay here also if I have 90 percent conversion here also I will have 1 mole 1 mole per liter only right I mean for comparison sake please remember that so here also I have 1 mole coming out okay here also we have 1 mole per liter and here also I have 1 mole per liter only but before coming here as 1 mole per liter it would have gone through so many rates so rate means already some conversion would have occurred whereas here there is only one rate and one conversion and one concentration so this rate is because the average of that is only one only one rate whereas here there are so many rates and then all of them giving me more conversion when compared to this because I have only one rate here and I have here theoretically how many rates you can put inside imagine infinite because that is the reason why at every cross section again we are thinking that we have infinite mixing that means each cross section now depends on one I mean sorry is equivalent to one tank so infinite number of tanks you can put there so that is why you have here more conversion for given volume and here you have less conversion for a given volume because the rate is only one that r a is small now but only one rate and that is the reason why you have you know larger volume for a given conversion larger volume for a given conversion or for given volume higher conversion right so that is what you have to remember I think you know this should go out now I think after this course definitely at least the basic definitions must be very clear in the mind right why plug flow is efficient what is the definition of plug flow what is definition of mixing I could have completed this long time back if I do not discuss much about you know simply saying that assume complete mixing but still you will have many many doubts in your mind that what is this perfect mixing and instantaneous reaction you feel there is no instantaneous reaction there is still an average rate right average rate due to various residence times inside the reactor whereas here I have all the residence times of all the molecules are exactly same here we are talking about outlet here it is also outlet only but outlet and inlet exactly same here right whatever is happening also reflect in this one if you have doubts you have to raise and ask me right okay so that is how it happens that P f r is definitely you know having more conversion for a given volume if someone is asking you in the interviews or in some examinations this question probably you know you have to definitely imagine that this residence time distribution that is coming into picture that is why by definition of plug flow we say that each individual molecule should spend exactly same time and there we can prove that that is not so in a mixed system where the residence times are varying from 0 to infinity right that means the molecules are packets which are spending 1 minute less conversion 2 minutes slightly more conversion 1,000 minutes 100% conversion but there is no use of that being inside because it is like it is already converted there is nothing will happen but still that is occupying some volume right so that is why to compensate that volume if you want to get the same conversion as plug flow what you do here you increase the total volume that is why for a given conversion the volume of plug flow will be smaller and the volume of the mixed plug flow will be larger because you want to compensate the rate lost due to residence times so see here every particle is spending exactly same means the conversion is same in each and every packet right so even if I take the average of that definitely I will have same conversion whereas here I have 100 packets I mean here also 100 packets but equal concentration here also 100 packets but starting from CA0 to CAF that I am again averaging right so that will definitely give me yes less conversion but what I am trying to do is I am now trying to provide more volume so that some more time all these packets will be on the average okay all these packets again are going to spend time so that is why that will compensate this conversion yeah higher conversion here by increasing the volume anyway even if you derive the equations and then able to calculate you will definitely see there are many graphs I think Levenspiel book has given that you know for given conversion what is the volume for first order reaction second order reaction third order reaction you will just go take 90 percent line and then read this side then you will know what is the ratio of volume of CSTR single CSTR and volume of PFR that ratios are given right so directly some people may give always in chemical engineering we use lot of dimension less numbers the volume of mixed flow by volume of plug flow also is a kind of dimension less number okay yeah and conversion is anyway dimension less number but it will be depending on again what is the order of reaction so you have one line for one order of reaction then you just read and then okay for first order so much difference second order so much difference okay so that is why one can also easily prove it so if you have still further any doubts about mixed flow and plug flow now I think we are going to stop that now right so please tell me please ask me please do not keep those cobwebs again okay how do you operate CSTR okay to see we are talking about only steady state system now but how do you come to that level where you have zero to infinite residence time under steady state condition so that means you can slowly fill up the entire empty take empty and then slowly fill up so level raises and then after that it comes out okay so then what is the concentration when it is about to come out because our steady state starts only at time t equal to zero yeah but I think you know we have to be specific here so means how slow you are talking 20 hours also we can down here 20 hours but we are talking about average residence time right yeah see very slow reaction means it may take more time for the reaction to happen and may be much more time the moment you want to go to 99 percent conversion see these are all vague things in our mind that is what exactly I am asking you know that cobwebs so why should I use a reflex in this reactor that is boiling of the reactor what should I do under what conditions yeah there is perfect mixing in the system when you are talking why should I boil and why why should I replace there are some the extent that doesn't take place in a normal temperature like what I mean we should also be clear there like what it has to be taken up to the boiling point of the mixture okay then body flux will come back we have to put a what for you are doing the question is that I have a liquid way the boiling point is only 60 example but the reaction is taking place may be around 100 degree centigrade okay it will not happen normally but you know you can conduct in gas phase problem comes when you have two reactants where once one boiling point is higher another boiling point is you know lower and if the reaction is somewhere above or in between then the low boiling point one will be boiling you have to keep the number of moles same that is why what you do is either you yeah either you put a condenser so that it evaporates immediately gets cool and then falls there but that is not the reaction problem okay it is only just it is only disting there and then falling there so then no question of taking out okay that is the easiest one otherwise you can also put a outside condenser and then again feedback under steady state conditions the number of moles will be kept same okay so that is only you are only trying to see that the concentration are not changing due to evaporation that is all you know that is only the physical thing it is not the actual reaction because if I allow that to go out then you will have one reactant yeah you lose one reactant right so that is why just close it perfectly otherwise you know I told you in our villages when they are cooking rice I told you know they put water just above and then cook because that is also same thing water should not get evaporated yes here will not change you know I think you have perfect mixing there and I close this okay here I have and then you have the outlet somewhere here right so put a condenser and again send it back that is only the engineering problem as a mechanical engineering problem where how do you put that whole where the vapor is escaping where it has to condense and it has to come down that is all okay and for that condenser again you should have some coolant entering coolant coming out and all that that is just only engineering problem it is not really CRI problem there CRI problem means that you should maintain number of moles yeah not uniformly the way it has to be in normal reactor so that is all but you know other than that there is nothing special in that kind of thing you know not only that there I think this question is good now because he is asking only reflex but I can extend this question to heat transfer if the reaction is highly exothermic right simply putting a external jacket may not give me that kind of area where I can control the temperature highly exothermic I want to control around may be 75 degrees but with providing only that jacket and also whatever is available you know coolant right still I may be getting around 90 what will do so people take out a stream there put an external heat exchanger again send it back that is one way other way is directly put the coils inside there are many ways or again inside also you have all the time instead of wall you have only coils these are all again mechanical engineering problems how do you remove the heat I mean heat if the area is not sufficient even for batch it is same it is not only for C S T even in batch reactor if only jacket is not providing sufficient area for removal then you have to go to various means various means of removing putting internal coils because sometimes internal coils you cannot put because they may be highly corrosive liquids okay so then you have to take out and then you know effective volume will be good effective volume will also decrease and mixing may not be proper because inside coils if you put near the coils again mixing these are all general engineering problem that is what exactly the difference between science and engineering a scientist will not bother about all these things you somehow remove the heat okay so that is what I have been telling you the difference between science engineering and technology technology is the one where it is ultimate telling you that use this temperature use these flow rates use this volume you will get this much product and this technology doesn't need any brain at all if you want to just copy one technology that is all that is why I don't know you may not anyone has worked in pharmaceuticals oh you work it now which pharmaceutical is that here yes okay good so earlier when when we were doing also our b-tech and m tech వారధిందింద్టి మారం పూడికెత్ కందేలిడికటారికిసందికంది. totally that was destroyed by the people who are working there inside. You know the employers, the, sorry, employees. What they did was, they just you know duplicated the technology, there was some software drags, this, I also took training at the time. In REC Varangal was one place where we had what is called chemical plant engineering, in BITEC. So one semester we have to go for industrial training. One semester, it is not in between the semesters where you have three weeks, four weeks like that. So one week, I mean one semester we were there in Hyderabad, some people went to fact Kerala FACT, FACT and some people went to Bombay, Noseal and because at that time our strength was only 30, now also maybe I do not know 30, 45 they increased, okay in REC Varangal I am telling, now NIT Varangal. So when we went there, at that time it was running okay. But after 10 years later when we saw, then in the news and all that, that entire company was totally destroyed. What happened was the people who are working there, they simply copied the technology because we know only temperature, pressure, what is the size of the vessel, simply duplicated, outside, very near to I D Pell only. You know that Balanagar area, I do not know what, you are familiar with thing? Yeah, there were many, many pharmaceutical companies that have come there. All of them, you know normally we have a partner called wife, so wife's name, that company because this fellow cannot start on his name. He is working still there, he is getting salary here but he is producing the same thing in some other unit and then he was trying to send it to other, not export but other companies where most of the pharmaceutical companies if they are producing headache tablet, they collect material here and there and then only mix and then tablet and then give it to you. Putting Navaljan name or Disprin name, okay. Now I do not know, now Chinese are dumping I think, lot of those things here, right. So but unfortunately what has happened, or fortunately we do not know, what has happened there was that if there is some problem there, they were not able to solve because they know only that temperature, that pressure, that volume of the vessel and that is all, that kind of studying. And many units have been closed may be after 5, 6, 7, 8 years because the product was not coming the way they expected, most of the time the colour is changing because you do not know how to control that. There are no fundamentals, that is what you know, technology state but even then few companies have survived because now they used, they started using their brain, that means they went to technology to engineering how to solve that colour removal for example. And then from there to science if it is required, right, that is one way. So technology, engineering and science, in the beginning itself you understand all the science in the lab with the help of chemists, right. And then try to find out under what conditions you will get very purest form of product when you do not get the colours, all that you understand and then start. And you know engineering principles are heat transfer and heat removal for example or heat addition or stirring or you know collecting the filtration, tabulating, all these are engineering applications, I mean engineering principles. So then finally Navaljan tablet will have this condition, this condition, this condition, this condition if you want to produce Navaljan that is what is technology because again you know I am just giving you, many people lifelong we may not remember what is the difference between or we may not know what is the difference between science, engineering and technology. This Saturday I was there in Guntur one college, RVR, JC college or so and because my friend is a vice chancellor now in Vigyan, we both did PhD here in IIT Madras, so one V0, Professor V0, so he was also there in Vigyan, Vignan University, he also just asked me just to come there because I had morning afternoon there was lecture, lecture is not for students but for faculty only, they are faculty, not chemical engineering, all faculty about research that is why I went there, what is research because there are many, they have 240 faculty members, out of that may be I think around 30 or 35 or may be around 30 have the PhDs, okay. You may think that so what, I think there are many people with PhDs who do not know anything still, okay but I think in academic institution that gradient must be as much as possible, gradient means PhD to B Tech teaching, now if you go to many private engineering colleges B Tech teaching to B Tech, okay he would have just finished it now finally here and would have got the result and immediately he is asked to teach again B Tech people may be third year, second year or whatever, so where is the gradient that delta C concentration gradient because I think this fellow is in equilibrium that fellow also may be how much time may be one year before or six months before, so that accumulation of knowledge is not that easy to do, so that is why I just went and told them that you know unless at least for your self-confidence even if you do not teach well that confidence you will have at least, okay I can teach if you do PhD and when you are doing PhD a good PhD make you think, that is all. So wonderful quotations are there, Arbindo told that no one can teach anyone, really no one can be taught, when you are listening to me that means your mind is already thinking about that, that is why you are listening, so but only thing is as a teacher I can only make you think, that is what all my classes last 15, 20 years, before that I do not remember what I have done, okay, first 5 years so, right, so that is why so many questions I raise even if you do not raise because those questions are supposed to make you think, that is all, why I was telling was that I asked the same question, in Vignan college I asked the same question what is the difference between science engineering and you know technology and I also asked them, I do not know whether I told you this particular thing, okay, in the last class, I mean in the beginning, I asked them I think there are chemical engineering students only I met in Vignan and I told them that you know you have written your examination that M-set entrance examination in Andhra, okay, they call M-set, still they call M-set or yeah, so that examination and your subjects were maths physics chemistry, correct no, only maths physics chemistry, then you join here in R-V-R J-C college in chemical engineering and do you know what is the degree you are getting, that is what I asked them, some people kept quiet, they do not even know poor fellows you know, nice, nice guys, they do not want to know what is the degree also, so some people said that sir we get technology, yeah that is what precisely I wanted to ask you, you started with basic sciences courses and then you passed them, then suddenly you came to R-V-R J-C and then depending on your rank and all that, then you are given chemical engineering, chemical engineering, okay, it is not chemical technology what they got and at the end of 4 years you are going to get chemical technology, have you any time thought what is the connection between these three, I tell the same thing everywhere, okay, did I ask this question, okay I also told you, yeah, so not even one could not and I saw there was some 6, 7 faculty and the faculty members, at least I observed some 3 that their face also has shrunk a little bit, that means they also do not know, that is not their fault, I am not finding any blame, okay, because I was interested in that so I first question myself, I also do not know when I first joined here what is the difference between science, engineering and technology, later I have to see many things because no book you find this, I tell you and some journal papers are you know about technology, you have to go to humanities journals and all that finding out what is the definition of technology, so when I am doing it only finally I thought that this is the logical explanation for and now I think many people say that you know this is the same thing what we are discussing, so that is why the technology is totally different but always you can start from technology to science or science to that is only left to you but there are some things where you cannot start with trial and error, for example aeroplane design, okay or racket design, these are all highly sophisticated, okay, they have to solve all the fluid mechanics equations, there are only two three forces, right buoyancy, lift and drag, that is all, you can make I think even 100 tons object to fly using only three forces but you should have the feel for that three forces, okay, then only you can solve that otherwise you caught and you do not have to worry drag, lift and what is the other one, buoyancy when you are designing a bullock cart, correct, no, in villages who is using computer to design bullock carts, no one, no, very happily they will do it, I think this is by experience, family, family wise it will come, one family in the village will be experts only in making bullock carts, so they know what is the size of the wheel but if you go to science again there is tremendous science and the science through technology those people have perfected and you know by wrongly designing the length wise and where you keep those, I am talking about bullock cart, where you keep those big wheels, why should there be first of all so big, there is beautiful theory in that and where in between you put, is it in between or somewhere slightly behind or slightly front, if you change wrongly the bull will die, then you have to pull it, okay, because you know its neck will be strained, that design also is there in that but by experience they got it, okay, so that is what is technology, for example brick I told you, another example, brick technology, try to make scientific bricks, then you have to go to materials, you know what is that, material science course what you have taken and you have to find out strength of this each grain and strength of multi grains, how do you put them, what should be the amount of mud, what should be the amount of you know the solids, what is the particle size, how do you orientation, what is the temperature to centre, all those things will, that is science, how many people are using that, still many houses are safe you know, I think many many houses are safe, so that is why I think all these things are also important but you have to think about this, you have to discuss in your mind, you know finally there also I told all faculty most important thing is they, for all of us is that, this brain, okay, unless you have that good brain you can never deliver either in teaching or even in research and 3 hours I have taken continuously, okay, they have not slept, I can see that, okay, they have not slept, I think in between jokes and all that were there so I think I thought they have enjoyed, that to when Saturday evening 3, 2, 2, 5, very difficult to, as I told them, very difficult to keep you here but still try, that is all, I think except some 4, 5 where they have to go, I think all, there are almost around 150 people, only 4, 5 people went after last half an hour, that is all, maybe they have some work there, okay anyway, so this is the one, that is why I was trying to encourage you to take you know, to think and then ask me questions, if you do not get that any doubt now, when you are preparing for the exam you will get definitely doubts, so at that time you can ask me, okay, yeah, good, so this is the one. Now once we understand this, here we have developed an equation, right, so that equation was V by F not equal to these things you have to remember, whoever it is, DXA by minus r A, now let us also derive an equation for this which is very, very, very simple, okay, so for this mixture flow reactor also when you are writing an equation, it is isothermal, so we have to write only mass balance equation and mass balance equation M B for A, this is M F for design equation, yeah, M B for A will be universal equation tells me it is input of A equal to output plus, yeah, plus reaction plus accumulation, okay and as I told you that we are steady state people, so that will be zero and the, this mass balance you can write mass means mass only, no K G, right, I have to write K G entered, K G leaving and all that, right, but we can also convert that into moles, but why do you do that in chemical engineering, particularly in reaction engineering, see the moment you go for distillation column, do you use moles, how many entering moles coming, what do you use there, okay, no, you are not, you are Bala, no, okay, input I am talking about, input distillation column, mass only, normally standard question is 100 K G Bala, right, okay, yeah, so like that only we do not use and for example heat transfer, you know these are simple things where you would have not thought I say, that is what only I am trying to give those connections, right and I think I also found in some interviews when I asked, okay why are you using moles in the reaction engineering and in heat transfer and mass transfer when you are using only you know the actual mass K G in terms of K Gs, I do not get answers easily and the same problem even in mechanical engineering when you go, even for chemical reaction they use mass, they also use combustion for combustion reactions, they simply write you know 12 K Gs of carbon reacting with the easy to relate, easiest one is 1 mole, 1 mole, 1 mole depending on the stoichiometry, otherwise 1 mole needs 2 moles to give 5 moles, otherwise you just now see that you know maybe 100 grams here, 100 K Gs here, another 500 K Gs here giving me 600 K Gs, mass cannot be produced suddenly anyway, so that balance, so that is why instead of that, this is what exactly what you have also done in your that batch reactor design problem, if you convert everything into moles that is easy, what you calculated you know that 769 divided by number of batches, no no, that 10, 10 tons divided by number of batches is only mass, but if you convert that into moles, then so many moles have been produced, acetic acid no, ethyl acetate, to produce that how many moles of acetic acid is required, I think acetic acid is the key component there, so many moles, but you can calculate maybe 6 moles, 6 moles, but this 6 moles equivalent to only 30 percent because your conversion is only 30 percent, so that is why you have to know compensate that, you have to take that divided by 0.3 will give you more number of moles that should be in the reactor, what you have assumed is 100 percent conversion, with the first time when you made that mistake, so that is why very simple balance only, that is the reason why all the time we will go for moles, that is why here also we will have moles per second because it is flow reactor, so and I know here our units f A naught is moles, volume is volumetric flow rate, meter cube per second, C A naught either you can use kg moles or gram moles per unit, for later and X n naught equal to 0 anyway that does not have any units, so here I will simply write f A 0 number of moles that are entering per unit time and output is f A, f very simple here, then reaction is minus R A into V, V is the total volume, earlier we have taken only, yeah only differential element, that is all what the volume, this is d V, why? Again I am telling you, yeah, so this is the distributed parameter system, that means there is distribution of concentration along the line, it is not one concentration, it is not lumped, lumped means everything together, right, so that is the reason why whenever you have changes within the system, either temperature or concentration because you know we are talking about either energy balance or mass balance or momentum balance, these three, right, so there you have to assume a small element and finally when you write input, output and then expand that with boundary conditions, you will get a differential equation, whereas in lumped parameters, that means everything together, there is no change at all here, you do not get a differential equation, you simply get an algebraic equation, so that is why here minus R A, change is only in this V, total V, there is no change, if I take a small element here what do you write, I cannot write anything because concentration is same, yeah, yeah, concentration is same throughout that element, R A is also same throughout that element, not only that element, entire system R A is same, so I cannot write anything, you know it changes, so that is the reason why you take the whole volume R V, anyway that is equal to zero, right, so this if I write this, this will be F A naught minus F A f equal to minus R A into V, so in our normal terms, you know in conversions if you are able to express this, what is the conversion equation we have given for flow system, F A naught minus F A by F A naught, F A f here is, otherwise I have to put X A f here, okay, good, anyway we will put that one here, so now yeah, this F A naught and F A f, if I convert this into F A f, this is general definition of X, right, so then what do I get here, this is equation 1, equation 2, equation 3, equation 4, then replacing equation 4 in 3, please write there, replacing equation 4 in 3, that will be F A naught X A f because we are talking about, this is general expression, okay, or if you are getting confused, but I think, no, not required, that should be general expression and you should be able to extend that, okay, yeah, so X A f equal to minus R A into V, right, so in this format when I want to compare, this will be V by F A naught equal to X A f by minus R A, so this is the equation what you get, this equation is, right, what is the difference between this equation and this equation, what does they tell you physically? yeah, within that integral you have that R A, so what does that mean? different rates, there are so many different rates, where you are having some kind of average of rates all right, whereas here there is no average, there is average but only one, okay, so that is why you will get here simple algebraic equation, there it is an integral expression because all these rates at various points I have also average and that comes through mathematics under that integral, so that is the reason why if I plot this integral, I mean this equation I will get here, yeah, 1 by minus R A versus X A you will get this and this is the area under the curve, this equal to V by F A naught, same thing I also draw here, this is also 1 by minus R A versus X A, I get exactly, you know, rate will not, 1 by R will not change, it is same right, for a given conversion same X A f, it is X A f, here also X A f, what volume I have to take now? Not area under the curve because there is no integral there, so I have to have this integral, sorry that rectangle area, okay, so then this entire thing now will be V by F A naught and you see this is the area in terms of volume, what you have to spend or what you have to provide more to compensate all that residence time distribution, that is why you know again I can tell you that, yeah, because this volume, this is the extra volume where why this is happening, this is happening because I have more and more, okay, more and more pockets, spending less and less time inside the reactor, so to compensate that I should have this much area more and why should I take rectangle, this is another question which I ask many students may not answer, why should I take, sorry, tell me Anupriya, no idea, yeah, who else can say for only, why should I take here rectangle, there of course area under the curve is that integral what you are talking, so many people may not know, Devayan you have the right hand side tells us X A if I am plotting 1 by minus R A to X A X into Y that will give you the rectangle yeah, I think you know most of us may not get that because this is 1 by minus R A just assume 1 by minus R A equal to Y okay, simply 1 by minus R A equal to Y now this is Y into X A F so Y is 1, X into Y area of a rectangle that is all, I think but many people do not know that, many people in my experience I talk to many students they are not able to answer that, why that is coming, because simply instead of writing Y we write 1 by minus R A so that is what we are doing, wonderful things in mathematics, because mathematics course is mathematics totally segregated, we do not know how to bring that information to our engineering, that is why we started talking mathematics of chemical engineering, so that once more can we combine I think d Y by d X you know how to solve, d Y by d X C as a function of some X and Y, but the moment I write d C A by d T, no idea d C A by d T also is a first time differential equation, is a function of some concentration, so then you do not know how to do that, so that is why I think those mathematics are taught only for you to make comfortable in these subjects wherever these mathematics come so that is why this is simply, I think follow know all of you see whenever I have a graph like this this is Y, this is X and this is Y, this is X so what is the area of this X and Y okay, so that is what we are doing, instead of writing 1 by Y we have written 1 by minus R A which is a parameter like Y, Y into X will give me the total area which is coming from this equation I have not done something else it is coming from this equation whereas here this will tell me from calculus that this is the area which I have taken a small strip calculate what is the area what is the area there it is delta X multiplied by Y, what is Y there again 1 by minus R A so Y d X and then Y not Y d X Y this is what is this where we are averaging all this in this equation so you will get the average rate definitely more than the average rate of this okay, you understood know this average rate is more than the average rate of this so that is why this is more more means you are dividing this volume will be less for a given conversion that always you have to tell okay so because it is minus R A is small there that is why when you are dividing with smaller number this volume will be larger good so now when you compare this equation and this equation where is tau coming there why do we worry about tau there in fact tau is not correct know by writing the natural material balance material balance means moles you automatically get only V why should I calculate tau and tau equal to again volume by volume volume rate another subroutine in the mind so many things I have to remember so simply that is why straight away go for V by F A not but why that tau has come can you just think why that tau would have come by 0 which is molar flow rate so V by V A not and I know also you can calculate or concentrations also you can calculate because these two are known volumetric flow rate also you can calculate but why why this another headache or you know in Tamil why this colovary what is that colovary only correct pronouns I am here 35 years but I do not know how to speak Tamil I understand so why that colovary it is not required can you think why that that should be what time of operation it is steady state no is a steady state reactor why do you need time do you need time for steady state steady state means once you start it goes to how many years if there is no problem correct now time will not come here if you are batch reactor then I have to tell time was in was in direct I am getting was in direct I am getting was in direct I am getting was in direct I am getting but why do I want to know that because I do not mean for producing that much conversion I need that much volume it is steady state reactor and volume let us say 2 meter cubed I got why do I want to know again time I am not saying whatever you are telling right or wrong I am just only making you think I cannot teach you anything so I am just making you think flow rate MBA first I think management market market survey that is one thing which you know before starting the calculations correct now you should know what is the capacity of the plant what do you do either you do market survey or you will hire MBA guy and then you fellow you go and get all the data how many times how much time we are going to get this much it is a steady state one why do you want to know how much time continuously it is coming once it is once it attained that conversion under great conditions where is the time coming there answer is very simple because unfortunately we started derivation batch reactor because batch reactor has T by C not equal to now whatever we studied in LKG only that is strongly printed here correct know you do not forget alpha beta when LKG so that but this is LKG for us batch reactor was the first so there it is in terms of time now we thought always we should have other reactors also in terms of time which is not required so this time this time is the batch time batch reaction time so because I have this T by C A not I would also like to write this equation as tau by C A not so that you know I do not have to remember that is all only for that reason otherwise if you want to really calculate volume of the reactor you do not need that expression is sufficient I mean it is not a great thing what I am trying to tell here but you know some things which are not useful not required for us right you understood know only for similarity because that is tau right this is T by C A not which we started our B take you know the first reactor so that is why now our mind goes okay now let me also write in that same term so that same format so that I will remember what is uniform format you can convert as tau by C A not and write for mixed flow that also you can convert as tau by C A not and write because V by F A not equal to tau by C A not you know how that comes where tau equal to tau is defined as volume by volumetric flow rate


Hi, my name is Pugyu Wu and I am a recent doctoral graduate from Griffith University in Australia. My research focuses on understanding the role of state in preventing and responding to conflict-related sexual violence. Most international efforts to prevent CRSV relies on states to lead the process of prevention and response mechanisms at a domestic level. However, in most instances it is the state actors such as in the form of military actors who are responsible for this very act of violence. In my research I questioned can states play an effective role in preventing conflict-related sexual violence. A common international approach to responding to conflict-related sexual violence with signing the joint agreement between the state government and the United Nations, Myanmar was one of them. I traced the process of implementation in the context of Myanmar and explained why these approaches not only failed but may have created additional harm. It is also a message to policymakers there is a requirement to obtain more sophisticated understandings of domestic political complexities as well as institutional legacies that can have huge impact on implementing and preventing conflict-related sexual violence in Myanmar and the similar context.


Zebra mussels have been found in three more area lakes. The Minnesota DNR today confirmed reports of zebra mussels in Big Pine Lake in Aiken County, Sock Lake near Sock Center in Todd County, and Blandin Reservoir on the Mississippi River in Itasca County. Several lakes connected to the reservoir by a Mississippi River tributary are also being added to the infested waters list. Because of the connection Lake Pockegama, Jay Gould Lake, Little Jay Gould Lake, and the Mississippi River from Little from Lake Winnebogoshish to Mississippi Lake will also be added to the infested waters list. Now, no zebra mussels were found in recent surveys of Lake Pockegama and Jay Gould Lake, but they will be added because they are closely connected to waters with confirmed zebra mussel populations. If you enjoyed this segment of Lakeland news, please consider making a tax-deductible contribution to Lakeland PBS.


The more senior you become in an organization, you have this tyranny of distance between the ranks and it presents challenges to leadership and that's kind of what you guys are talking about and that's what kind of Mass Chief Compton is speaking about and when they say open door policies and all these kinds of things, they're trying to bridge that gap in the tyranny of distance between the ranks because communication is a critical, absolute critical element for any organization to be effective and when you open up your door and sailors see it open and you're sitting there and you got this positive attitude, maybe with a smile on your face, I don't know how Mass Chief Compton said that's how I do and sailors walk by and they may or may not have something on their mind and if they do and they see that positive leader sitting there, they're probably more inclined to stop and have a conversation and it's important for the leader to be mindful of sometimes it's okay to never say anything in that meeting, sometimes sailors just want you to listen to them, they just want to get things off the chest, whatever it may be and they may even have an issue and you can sit there and listen to them for 15, 20 minutes and by the time what they think is a conversation and really it's just them talking, they've worked them their way all the way through whatever the concern may be and they're leaving with a smile on their face and all we have to do as leaders is say thanks, glad I could be here for you and that young sailor goes out and they walk the deck plate now with a smile on their face with a positive attitude which is infectious and an organization and then they share hey guys got anything going on go see chief whomever go see the Mass Chief or go see the CO you know whoever you need to sit down and talk to or that divoth that helped mentor you along the way their pales are young you know I heard somebody say a long time ago and I believe it was quoted from Winston Churchill really they said courage is what it takes to stand up and speak but courage is also what it takes to sit down and listen and that's what I'm mindful of as a leader as we try to figure out this and continue to improve our communication across the ranks in our Navy and that's probably been you know one of the biggest challenges the more senior you become in an organization to ensure you're leveraging and communicating with with our our our leaders under us to make sure that the things that pales are young has on his mind and the questions that that he wants to ask and the concerns that he has for for himself and his fiancee Lindsey we're addressing so so those are kinds of you know some of the challenges we face as leaders


And you know what? Here's the golden rule. When you lie to yourself, you're done. Guys, two rules I want to tell you real quick about business. And I know that you know this because you've heard it, but I want to make it very clear.  Okay? If you sell cars, right? Yes. I sell cars. If I worked across the street from him, my goal would be to bankrupt his... I'm serious. Like, I want to torture him. Okay? I would rather be feared than liked to my competition. Yeah.   I know, but you're getting paid, right? Yeah. That's what you want me to do. Make sure that you make home, that you get home and get money. Okay? If this was an ice cream place and you came in and you ate ice cream, but you were broke, you probably mad at me, right? Yeah. Okay. Yeah. Yeah. So I'm not going to feed you ice cream, okay? Or I'm going to make sure you get paid. That's why you're in here. By the way, one day, one day you're going to thank me. One day your kids are going to thank you too for disciplining them. Okay? Listen to me.  I don't know. Never.  I wish I could send home a f***ing camera on everybody in this f***ing room. And we had a screen that showed all of you, right?  Hey guys, what's going on? It's Andy. A lot of you leave comments telling me that you need help. Do me a favor. I'm going to tell you the best way to get a hold of me. Shoot me a text message right now.<PHONE_NUMBER>.<PHONE_NUMBER>. I'll help you with whatever you need.  Let's get back to the video. And we would watch this f***ing guy go home, and the first thing he's going to do is turn on f***ing Netflix. Sorry about that. I know we want to do the Netflix though, right? He's like, f*** Andy? No, no, no. When they get the money, then you guys can turn on Netflix, right? We're not trying to f*** up your socks. No, but my point is we would watch them all. The picture, we watch them go home. Nobody's around.  Okay? You said you weren't going to eat after seven, but nobody's looking.  And you know what? Here's the golden rule. When you lie to yourself, you're done. You're better off done even making the promise instead of f***ing making the promise and break it. Because I'm going to tell you something, right? You wouldn't stay in a relationship with somebody that constantly lied to you. So from now on, I want all of you guys to make a promise. Whatever you say you're going to do, you do no matter what. Yes. No matter what. Listen to me.  I don't care if you f***ing do it.  You're not going to let you down. Okay? And by the way, have the f***ing courage to tell other people no. You want something? Hey, bro, you want to go have a drink? No, I'm good, bro. I'm good. Yeah, no. By the way, watch this. Hey, let's go f***ing eat. Let's go to lunch. I don't go to lunch no more. I'll bring my f***ing lunch to work. What are you doing, man? Listen, dude, that's the last conversation they're going to have with me. Okay?  There's two types of family.  Okay?  Find your chosen family. That's what will take you far. Anyways, good job, man. See you soon.


I'm back to the morning session of the workshop. So today, we have Mano Sanz from Udelar Uruguay, who's going to present us the price of Inguineras. Very nice title, looking forward. The four is yours, Manuel. Thank you very much. OK, first of all, I will warn you that I am heavily jet lagged, so I hope this goes OK. If I spontaneously start speaking in Spanish, it's because I had a little bit of a breakdown. You just save me, like, switch back to English, and I will do my best to do it. OK, so this is a collaboration with other three authors, two of which are here, John, and Marco, somewhere, maybe. And Tianqi, that couldn't make it, he's working in China. OK, the model I will be discussing is a model that already came up at least once during the talks, so hopefully you already have some kind of familiarity, and this will make everything a little bit smoother. So the general setting is a setting in which the statistician has some certain data available, which is structured in a matrix. This matrix is a full-ranked matrix, and for some reason, within this full-ranked structure, there is a low-ranked structure that is of interest for the statistician. So the idea is to try to extract this low-ranked structure from the full-ranked structure. There are many applications, for example, in community detection and group synchronization that I will not dive into, but this is in general a very recurring problem in this kind of topics. This is, in a way, a kind of inference model for PCA in the sense that the kind of task that you're trying to do is essentially the same kind of task that is solved by PCA. But the difference is that PCA does not incorporate prior information about the signal or the model in general into the estimator that it gives you, and the idea is to have PCA as a kind of comparison background to see the performance of these kind of estimators. So what I'm going to discuss is this task of obtaining a low-rank signal out of a full-ranked data matrix in the setting in which one assumes that the high-ranked noise, which is a kind of additive noise in the data matrix one is observing, is not the uncorrelated Wigner Gaussian matrix that is in the typical models, but the matrix instead has some kind of structure. So a little bit the logic of assuming this is that one has a certain problem in which one wants to apply these techniques and wants to extract this low-rank structure for some reason. But the data matrix that's the full-ranked data matrix that is available, the thing that one calls noise is, in fact, some kind of information. It's not really noise in the traditional sense. It's something in which you're not interested. So you want to, in a way, forget about it. But it has some kind of structure because the distribution it comes from is a highly structured distribution because this is a data set that comes from a real situation. And the kind of question we want to answer is, OK. You have this setting in which the data matrix you have is a data matrix that comes from some structure distribution, and you want to extract this low-rank signal from the high-rank data matrix. But you don't have detailed information about the distribution where the high-rank component comes. So in a way, one nice proposal would be to forget about the structure in this matrix and model it as a Gaussian matrix, which is essentially a matrix that has no structure at all. And the idea will be to answer the question of, OK, by using this kind of naive model on this structured matrix, what are the things that you're losing by doing so? And that's why the title of the talk is the price of ignorance. OK. And in particular, we're going to study a low-rank matrix estimation. I forgot to mention the name of the problem, in which, as I mentioned, the noise matrix, whatever you call noise, is some kind of matrix that comes from a structure distribution. But the statistician models it wrongly as a Gaussian matrix. And in this setting, the main results of the talk will be a characterization of the Bayesian estimator that wrongly assumes the noise to be Gaussian. So it's a heavily mismatched estimator. And the characterization of A and P that's also wrongly assumes that the noise matrix is Gaussian. So there will be parts of the algorithm which are incorrect, in a sense. And from these two theoretical characterizations that we have, then we have a whole variety of phenomenology that appears, which have very interesting kind of implications, which I will discuss a little bit. OK, but before diving into the problem, the structure problem, I want to discuss a little bit the traditional model in which you want to extract a rank 1 matrix out of a full-rank matrix. And really, you have an additive noise, which is Gaussian. So I want to discuss a little bit this setting before specializing in the new results. OK, so being more precise, the kind of things we have, in this case, is that the statistician has some data matrix, which is this y, which is equal to this rank 1 matrix, which is the P. As you can see, the P is a rank 1 matrix built out of a random vector, which is the x star. We will call this the signal that we want to infer. And then there is an additive, a Gaussian noise. And this parameter lambda star, which is the signal to noise ratio. During the talk, there will be two signal to noise ratios, this one, which is lambda star. Keep in mind that not in this setting, but in the setting of the work we did, there will be another signal to noise ratio, which will be the lambda without the star. And this lambda star will be really the signal to noise ratio involved in the data generating mechanism, while the lambda will be the signal to noise ratio assumed by the statistician. And we will allow, in the model we discussed, not here, for these two values to be different. OK, so you want to infer this random vector x star. And you have this Gaussian additive noise here. In this setting, because the real distribution of the noise is bigger, you have that the distribution of this random matrix is proportional to this density over here, this factor exponential of trace 2 of z squared. And this implies that the log likelihood of the model, the log likelihood that will appear on the posterior distribution, will have this quadratic form. This quadratic form comes directly from the quadratic expression that is from the distribution of the noise. OK, here I write some simplifications in two special cases, in the cases in which the prior you're using for the posterior has a support over a fixed norm. So it can be two cases. Here I wrote in the slides that the prior is uniform on the sphere, because here, when you distribute the square, you remove, you add it in this constant, the factor that has the term of the norm. But it could also be a product of n-radimajor, the prior, which would give you a vector which is uniform in the hypercube. In that case also, the norm would be fixed and would be equal to square root of n, and it will be incorporated into these constants over here. OK, so this form of the log likelihood will be important for us, because here I am discussing the log likelihood of the model that where the data generating mechanism has a Gaussian noise, but also the model models the data as having Gaussian noise. In the setting we will consider later, the data generating mechanism will be different. So this matrix Z will not be a Gaussian, but still because the statistician is not aware of this discrepancy, the posterior proposed by the statistician will be the same. So we will still have this same kind of expression. OK, in this setting there is plenty of literature available. I just mentioned here a couple of them. Some of the people here in the audience have been authors of this kind of works. But in a nutshell, in this context in which you have a data matrix that really has an additive Gaussian noise and the model has no mismatch, in a sense, everything is known. So the characterization of the usual quantities one would be interested in, for example, for the Bayesian estimator characterizing the limiting mean square of the estimator or characterizing the evolution of A and P. This has essentially all been done. In the particular context of the Bayesian estimator, here the Bayesian estimator is, well, in the slides I didn't write the meaning of this bracket, but this bracket means mean expectation with respect to the posterior. So expectation with respect to the measure that has the low likelihood that was on the previous slide. So it would be the measure that has this expression in the exponent. This is not working. OK, has this expression in the exponent plus a prior. In this context, the Bayesian estimator is then taking expectation with respect to the posterior to the matrix constructed in terms, the Ranguan matrix constructed in terms of samples from the posterior. So it would be this expression over here. And what you have is that for characterizing the asymptotic error committed by this estimator, it's usually useful to analyze the asymptotics of this auxiliary quantity, which in Bayesian terms would be the log evidence or in more statistical physics terms, it would be the log partition function. The details of this quantity, they are not really important for the talk in a sense. We will characterize it. But the important thing is that this quantity, in a way, when you study the asymptotic value of this quantity, it characterizes the asymptotic value of certain order parameters that are important in these problems. For example, in this particular case in which the data generating mechanism includes an additive Gaussian noise, and the statistician is aware of this structure of the data and its model correctly, you have this relationship, which is, if you characterize the limit of this quantity scaled correctly, which is divided by n, and you take a derivative with respect to lambda, that is the assumed signal to noise ratio, then you have that this limit characterizes exactly the mean square error of the Bayesian estimator. OK, nice. And in this case, this kind of result exists. So we know what is the limit of this normalized log evidence or log partition function, which is in terms of this optimization potential. And because this is known, then the mean square error of the Bayesian estimator is also known. So one of the contributions, one of the things we did in this work, is to have an analogous result, but in the mismatch setting I was mentioning before. OK, so what is the model we really explore? The model we really explore is a model in which we have a data matrix which is similar to the one I was describing before. So this looks exactly the same, right? We have a data matrix which is an addition between a term proportional to a matrix of rank 1 plus a full rank matrix, which is z. The only difference in this data matrix is that now this z is not a Wigner matrix, so it doesn't have this exponential of trace of z squared kind of density, but it is a rotational invariant matrix. So it's constructed by making a sandwich between a diagonal matrix and two uniform orthogonal matrices, a sandwich in which the bread would be the orthogonal matrices and the cheese would be this diagonal matrix. And we will be always assuming that this diagonal matrix has some elements that if you take the empirical measure of the elements of this matrix, it converges weekly to some well-defined limits. That has some kind of regularity assumptions that we need for the proofs. For example, we need this to converge to a distribution that has a compact support. And we need for the eigenvalues, the largest and minimum eigenvalues of the distribution to converge to the limits of this support. And another technical condition that it's not really to the point. But the important part in our model, the important thing is that, OK, so now the noise is not a Wigner matrix. So the distribution of this is much more general, right? Because a Wigner matrix, you can think like a rotational invariant matrix that has a really specific diagonal matrix here. Because Wigner matrices are rotational invariants, but they have a spectral distribution which is very particular. Here, we are allowing for much more diverse spectral distributions in this diagonal matrix. But the thing is that, OK, although this is the data generating mechanism, so really the data is coming from this additive noise that has all this structure, the statistician is not aware of all this complexity in the structure. So one nice proposal of what to do in these settings is to say, OK, there is something complex going on. I don't know what it is. So a default hypothesis would be to just assume that the noise is Gaussian. Of course, this is not correct. But this is a typical assumption that in a lot of real practical situations is used, right? And the motivation for this is really this one. OK, you have this data matrix that could be structured, but maybe you're not aware of this. So if you're really implementing this mismatched model in which you're assuming this naive Gaussian hypothesis, what is the impact in the inference task when you do so? And the kind of results we have include a variety of estimators. We divided them in three families. The first family of estimators are the spectral estimators because they are estimators that all are constructed in terms of the eigenvector corresponding to the largest eigenvalue. This is in principle a kind of a family of natural estimators. The most typical of this is PCA, which is just constructing the rank one matrix corresponding to this largest eigenvector. This exactly corresponds to the maximum likelihood estimator in the case in which you have a prior which is uniform on the sphere or rather maher. But as you can see, there are other two estimators that we consider, which we call the Gaussian PCA and Optimal PCA. What is the meaning of this? So these other two estimators, they are essentially PCA, but with a constant in front. Why would be a good idea to put a constant in front of PCA estimator? OK, these constants will depend, at least, in the signal-to-nose ratio. Why is it a good idea? OK, imagine you are in the situation in which you have some data matrix, but the signal-to-nose ratio is really small. One knows by theoretical results that the correlation between this eigenvector corresponding to the largest eigenvalue and the signal you want to infer beyond certain thresholds is really small. It's vanishing with n. So in these cases, if you use a PCA estimator, because PCA estimator is a matrix that has norm 1, this is a bad estimator in the situations in which there is no signal inside your data matrix. Because if there is no signal inside your data matrix, your mean square error will be much smaller if your estimator is exactly 0. So in these cases, you would like for your spectral estimator to have a constant in front, which if the data matrix you have has no information about the signal, that constant should be 0. And in that case, you would be estimating much better than if you're using PCA. And when the signal is really strong, maybe you want the constant to be 1. And in middle situations in which the signal-to-nose ratio is large enough for the data matrix to have some information, but not a strong correlation with the signal, you want this constant to be something intermediate between 0 and 1. So then why do we have two PCA estimators with these adaptive constants that tune the amount of information your data set has? OK, because there is an optimal way of putting a constant when you have this structure additive noise, which depends on the structure, really, on the spectral distribution of the noise matrix. So if you have information about the real distribution of the noise matrix, you could put a constant here that depends on lambda that would be optimal, in a sense, that would really adapt to the real strength of the signal you have in your data matrix. But if you're not aware of this, oh, sorry. You could still do this of putting a constant. But if you're assuming that the noise is Gaussian, you would put another constant in front, which would be the optimal constant in the setting in which you really have a data-generating mechanism that has an additive Gaussian noise. So we consider these two scale PCA estimators, one with the optimal Gaussian constant and one with the real optimal constant that takes into account the real distribution of the data. Again, in this setting, we also want to characterize the patient estimator, which the definition of the patient estimator now is exactly the same as in the setting of Gaussian noise without a mismatch. That is, patient estimator is the expectation with respect to the posterior of the rank one matrix that comes from a sample of the posterior. So it's this expression over here. But the difference with the setting in which you have additive Gaussian noise without a mismatch is that now, OK, you're taking an expectation with respect to the posterior, but now the posterior is mismatched. So this is not the correct posterior in the sense that the posterior does not reflect correctly the data-generating mechanism. In particular, we allow for three kinds of mismatch here. One is that the posterior we are going to consider only has uniform spherical priors, but the signal might come from another distribution. We also consider the setting in which the statistician doesn't know the real signal-to-nose ratio of the data matrix. And so maybe the signal-to-nose ratio proposed inside the posterior is not the correct one. That's why we have the difference between lambda star and lambda. So remember, lambda star is the real signal-to-nose ratio used in the data-generating mechanism. And lambda is the signal-to-nose ratio implemented inside the posterior. And also, the main mismatch in our posterior will be that our posterior will be the posterior corresponding to Gaussian noise, but the data-generating mechanism comes from this rotational invariant distribution. OK, one important point is that because we are in a mismatch setting, there are a lot of analytical tools that allow to analyze base optimal settings that cannot be used in this setting. So you have to work around some technical problems when you have these mismatches. And there is also the problem that is the general problem with the Bayesian estimator that in high dimensions, it's non-trivial to approximate the Bayesian estimator. So you need efficient algorithms to try to implement it. And finally, the last setting of estimators we analyze are AMP algorithms. There are two AMP algorithms we analyze. One is Gaussian AMP. And the other one, we call it, alternatively, I think in some slides, it's referred as true AMP. And in other slides, it's referred as correct AMP. Don't mind the difference. They are the same, essentially. And the difference between them is that in Gaussian AMP, that is the AMP that the statistician that implements a model for the data which has an additive Gaussian noise. So this is a mismatch AMP would use. It's the AMP that has the entire corrections corresponding to the Gaussian setting. But the true or correct AMP will be another AMP that has on-sugger corrections that are fitting the real spectral distribution of the rotational invariant noise. Here there is a mistake. I was trying to write. It's not working exactly properly. OK, here there is a derivative missing, but the idea is for you to see that's the kind of on-sugger corrections that were in the Gaussian AMP. And the idea is to take all these estimators, characterize them in the high-dimensional limits with some theory, and then to, using these characterizations in the high-dimensional limits, to compare the performances of all of them. OK, one quick comment is that, as I was mentioning before, because in the Bayesian estimator, we're using a model in which the data is assumed to be Gaussian, then the log likelihood that will be used in the Bayesian estimator will be exactly the one we saw before. So it will be exactly the same log likelihood which is here, the one that will be appearing. And the good thing is that, because we will be analyzing cases in which we have a Gaussian uniform prior on the sphere, this expression over here can be interpreted as a rank one spherical integral. So this is something we will exploit for the high-dimensional characterization. OK, the first of the main results is the characterization of the model corresponding to the posterior in the previous slide. There are these two functions we correspond in statistical mechanics, jargon to magnetization and overlap. That's why we use the usual letters to designate these order parameters. We have for these order parameters asymptotic values, which are these two functions, see that the limit of these two order parameters depend on these two parameters, which are lambda and lambda star. Remember, lambda is the assumed signal to noise ratio, and lambda star is a real signal to noise ratio. So these two values appear in both limits of the order parameters. And using the limit of these two order parameters, you can compute the limits of the mean square error of this mismatched Bayesian estimator. OK, here is a little comment that this H bar and this R, this is still just transform of the real distribution of the noise. And the H bar is another associated parameter. So in these two expressions, the real distribution of the noise matrix appears through this function and this quantity here. OK, and a couple of very brief comments about the proofs that are involved in this high-dimensional formula. One issue we have here, as I told you before, is that because this is a mismatched setting, there are things that are lost, some analytical tools that cannot be used in this setting. For example, this kind of easy relationship between the limits of the log evidence, the scaled log evidence, or the scaled log partition function free energy, whatever is your favorite name, is not really related to the MSC as it was in the non-mismatched case. So here, to work around a solution to the lack of this relationship, which is the IMMC formula, you have to consider a more general setting, which is not exactly the one we were saying before, but a setting where the noise you consider to have is the real structure noise, which is the z, plus a little Gaussian noise. And the thing you have to do is to compute the free energy, the limiting free energy, for this perturbed setting. Because now, this little Gaussian perturbation allows you to access the limiting value of the overlaps, which appeared in the MSC formula I proved before. And the good thing is that when you consider the setting where the prior is uniform on the sphere, you can compute the limit of the free energy of this log partition function that I wrote many slides before over here. So you can compute the limit of this quantity divided by n in the mismatched setting when you have a uniform prior on the sphere by using results on the asymptotic value of rank 1 spherical integrals and the spectral distribution of rank 1 perturbations of a structure matrices. With respect to the epsilon, yeah, it's convex. No, but still you have the usual relationship with the derivatives, and you can work it out. And using this limit for the free energy, you can access the magnetization and the overlap. The magnetization will be related to derivatives with respect to the real signal-to-nose ratio, lambda star, and the overlap with respect to the parameter epsilon that was introduced by this perturbation. Of course, because this perturbation was artificial, it's not really in the model we are analyzing. Then we have to take a limit of this perturbation parameter going to 0, but all the limits work out, and then you have your mean square error formula in the end. And the second main result we have is the characterization of the Gaussian wrong AMP, wrong in the sense I was saying before that the entire corrections are the ones corresponding to a Gaussian model and not to the real structured noise that the data matrix has. And the kind of characterization we have for this AMP is the usual kind of characterization. We have a characterization of all the iterates until time t in terms of asymptotic random variables that are described by some state evolution recursion. So this is essentially what is written here. And all this is in some specific metric that is the Wasserstein metric. Well, and a brief description of the proof techniques here, the idea is to write an auxiliary AMP, which has a correct denoiser that, in a way, corrects for the wrong Onsager term that is implemented by the Gaussian AMP. And the idea is that because the denoiser is chosen to correct this, but it still falls in the theory of orthogonal invariant AMP that can be found, for example, in this work. We go back to its theory to characterize the asymptotic value of this auxiliary AMP, which falls within that theory. And then using this characterization of this auxiliary AMP, we prove that the auxiliary AMP is really close to this Gaussian incorrect AMP. And we prove a kind of asymptotic bounce in L2. And we see that if we can track the auxiliary AMP, essentially we're also tracking the wrong Gaussian AMP. OK, so we have this asymptotic characterization for a Gaussian AMP and for the mismatched base estimator. And here what we do is to use these two theoretical characterizations that we rigorously built in the work. And we compare two characterizations of the other estimators we were considering that I described before. And we have a whole variety of these nice pictures, which shows the asymptotic behavior of all these estimators. We'll mention maybe just a couple of them if you want to really see the simulations and have a more detailed comparison, you can go back to the work. For example, in here we consider a setting in which the noise is not really structured. It's really big, but there is a mismatch only in the signal to noise ratio. So this is a milder mismatch than the general mismatch considered in the model we analyzed. And we're taking the assumed signal to noise ratio, which is lambda, to be two times the real signal to noise ratio. So in a way, what is the resulting model? The resulting model is a model which is correcting the kind of distribution it implements, but it's over-optimistic. It is like assuming that you have much more signal in your data than you really have. And this shows a kind of nice behavior in, for example, the Bayesian estimator, which is the dashed green line over here. I don't know how the colors are seen, if it's good enough. You see that now the limit of the MSC is not monotonous in the real signal to noise ratio. It's not monotonous in the real signal to noise ratio. And there is an intuitive interpretation of this. If you're really optimistic and you think you have much more signal intensity in your data, you will be over-estimating the norm that you should give to your estimator. Like if you think that there is much more signal in your data, you will make the estimator you have to have a much larger norm. But then there is not so much signal. So you put a large norm in your estimator, but you should really put a smaller norm. So in a way, you're making your MSC larger than it should be. And that's why you have these little peaks over here. Another thing that comes from the theoretical results is that there is an interesting relationship between Gaussian PCA and the mismatch Bayesian estimator. In this case, for example, Gaussian PCA, that's something which is similar. It has a similar phase transition in which the MSC starts to be decreasing again. You see both of them start increasing at some point, and then they become decreasing again. And these things happen at the same time. And this comes from the fact that they are both assuming the same kind of noisy solution. But they don't have exactly the same values. But we will see that in a lot of settings of structure noise, Gaussian PCA and this mismatch Bayesian estimator, they do essentially the same. There is another interesting behavior, which is that this Gaussian AMP is not doing the same as the mismatch Gaussian estimator. So even though this Gaussian AMP is implementing the same model as the posterior you have, the mismatch posterior, they are doing different things. And you can see that in all the simulations we have, always this Gaussian AMP is having a worse performance. And as you can see, in this setting in which you have the Wigner noise and uniform prior, the AMP that implements the correct on-side corrections and the optimal PCA, they are doing the same, and they are the best estimators. And that is kind of natural because these are the estimators that have really the true noise distribution. Here, this is a more mismatch setting because we are considering a noise matrix that has some uniform spectrum between these two values. These two values are only fixed so that the eigenvalues in the spectrum of the noise have variance 1. So this is comparable to a standard Wigner matrix. And as you can see, in this setting, Gaussian PCA and base PCA, they match exactly, which was not the case before. So there is a kind of interesting and complex relationship between this Gaussian PCA and the mismatch Bayesian estimator. And again, we observe that the Gaussian AMP is doing worse than with respect to all the other estimators. And the estimator is different from the mismatched Bayesian estimator. And the similar behavior is seen also if we study the problem in which now the real data generating mechanism has some additive rotational invariant noise with some Marchenko-Pastore spectrum of a spec ratio equal to 1. In this case, the picture is very similar to the situation in which the mismatch was only in the signal to noise ratio, but the relationship between the Gaussian PCA and Bayesian estimator now inverts. Now here, you can see that the Bayesian estimator is doing worse than the Gaussian estimator and the places of the phase transitions. Now they don't exactly match. You can see that the mismatch Bayesian estimator starts making worse predictions before Gaussian PCA. And another intuition that we had from the theoretical results is that if you plot the overlap of all these estimators in the case in which the prior is uniform on the sphere, you can see that the behavior of the overlaps, the asymptotic behavior of the overlaps as a function of the real signal to noise ratio for all of these estimators is the same. So with this intuition, what does it mean that all these estimators have the same overlap? It means that the angle between the estimator and the real signal you want to infer in all these cases is the same. So in a way, when the prior is uniform on the sphere, all these estimators, in a way, behave like if they were PCA, but scaled with some constant, which is different in every case. One important thing that I don't think I mentioned before is that while the characterization of the mismatch Bayesian estimator was restricted exclusively to the uniform prior on the sphere, the characterization of this wrong Gaussian AMP does not require that. So you can assume some general factorized prior or spherical prior. It's OK. But in the special case in which the prior is uniform on the sphere, you have that all these estimators have the same overlap, which tells you that they are essentially PCA with different adaptive norms. OK. So to close a little bit the talk, we have all these asymptotic characterizations on all these family of estimators in this heavily mismatched setting in which the statistician completely forgets, completely disregards the structure of the noise. And we see that this heavy mismatch in the model brings this complex phenomena into the picture. It makes all the relationship between all these estimators that in the situation in which there is no mismatch, they all coincide, makes all these interesting behaviors in different settings. In particular, I think one of the take home messages is that in this setting we have that the Gaussian AMP and the Bayesian estimator, although they are implementing the same model, the estimator in the end, they do not match. So the mismatch makes these two things to differ. That's it. Thank you for your attention. If you have any questions. Many questions. So if no one wants to start, I'll kick in. So you told us what is the price of ignorance. But I was curious whether you ask if you have any insight about the opposite question, like the price of knowledge. So if I had some underlying model, which is just like the standard one with Vigner noise, but for some reason, I read your paper, so I know like let me use just orthogonally invariant AMP. Would you think that I would pay a price for that as well, or somehow that would adapt to? You mean the situation in which you're trying to kill a fly with a bazooka? Exactly. I wouldn't really know. But that's an interesting question, like if you overdo it instead of underdo it. So I statistician, I don't know what the model of the data, but let me use the most powerful spectra estimator that I have out there. Do they pay a price if the model is simpler than? I guess in that case, I mean, if you really want to use this kind of base optimal algorithm, the first thing you would need to do is to have information about the spectrum. Because if not, you would always be in a mismatch setting, which wouldn't be kind of optimal. But it's still interesting to do. OK, thanks. Any other questions, otherwise? I have other ones. Thank you, Manuel. It was really nice. Please, can you go back to slide 16? 16? Yeah. Yeah. Yeah, one thing that wasn't clear to me is how does this z is related to the ground truth noise z? This is the real ground truth noise. So what we did is to really analyze a problem in which it's like an auxiliary problem that you need to analyze theoretically to be able to access this quantity Q in the overlap, which is needed to characterize the MSC. And in this case, you have to analyze a kind of more generic problem in which you have the real noise z plus a little component of an extra noise, which is Gaussian. But the z is the same as in the model you want to characterize. Thanks. Welcome. Thank you. Please, can you go to the slide where you have a PCA, OPT, and this one? No, no, no. This is where you have where you rank which one is the best. OK, the graphs. For example, yeah, all of these settings, which one? Yeah, you said, if I'm not mistaken, you said the PCA, OPT is the best. Are you saying that, like for example, in future selection, in a like data set, for example, we used to in a like data set, for example, we used to use PCA to like a dimension reduction of a data set before we use it for machine learning. In this case, are you with your own research? Are you saying with your observation or with what you have done, is this the best we can use? The best is PCA, but with a scaling constant. Remember that when we define the PCA, we said there are three variants of PCA. There are like real PCA that I'm going whatever please. Now the jet lag is starting to kick in. Luckily, it started to happen in the end of the talk. Here, you see we have three variants of PCA. One which is just PCA, so you just construct the rank one matrix coming from the eigenvector from with the largest eigenvalue. And the optimal one is what we call optimal PCA, which has a constant in front. And this constant is a function of the signal to noise ratio that has all the information about the spectral distribution of the noise matrix. So this is like a really smart PCA in a way. And that smart PCA is the one that is optimal in achieving the minimum MSc, which is this one. Any other? Yeah? If you don't have rotation in variants, but you still have a formula for the free energy, are you able to do a similar analysis with the AMP in those settings? If you have? If you do not have rotation in variants, but so instead of using the formulas for the spherical integrals, you have another way to compute the free energy. Are you still able to do the AMP analysis or is this rotation invariance important when you do the PCA analysis as well? I'm a bit confused. You mean for the Bayesian estimator? So you don't have a uniform prior on the sphere? So that's right. But let's say you have some other way to basically compute all of these quantities. Are you still able to do some AMP in the other settings? Yeah, the results for AMP that I forgot to say at the moment, you don't require for rotation in variants of the signal. The signal can be whatever prior that is factorized or Gaussian. It's much more general, this. Thanks. Can I just ask a quick follow-up? So this result here, the state evolution for this is not Gaussian. Is that correct? Or are these Gaussian? It's a state evolution coming from this auxiliary problem that has the state evolution of the orthogonal ensemble AMP. So it involves all the cumulants and it's a complex expression. So these variables here, these x1. I didn't write everything because it's a mess, you know. OK, but I just wanted to clarify. So even the x's here are not Gaussian variables in the limit. The x's, you mean these x's? Yeah, yeah, yeah. Because they have all these extra terms, they are not. So none of this, OK. Can I ask another more basic question? So in the very last slide that you have on page 22, I think it was? 22. This one? Right, so you have this comparison saying that all of them achieve the same overlap. I was wondering, do you also have a simulation of this in a non-sphorically invariant signal case? And then what does that comparison look like? I'm just curious. I think we did some simulations for AMP for the overlap. But because we didn't have a comparison with a Bayesian estimator because the result for the Bayesian estimator is only on a uniform prior on the sphere. We didn't plot them. I don't remember. You should ask Marco if you want to. He was the one running the simulations. Maybe I have my last question. Is putting my physicist's hat, if I understand correctly, like all your results are done by computing explicitly these spherical integrals. So we know that when we have mismatch, usually we have RSB. So from the paper of Fabry to Antenuch and so on, when we mismatch the likelihood, do we know here what kind of structure of the measure we have, like which level of RSB or if it is RS? Because we didn't use any of these kind of techniques. We went directly to spherical integral. It's not clear the structure of the replica symmetry breaking or not. But we had some discussion, John, about this. You remember? And we had this kind of heuristic argument, which might be wrong, that was pointing to the fact that this was not replica symmetry. We never knew for sure. Maybe we can discuss this at some point if you want. From the paper of Fabry to, at least in the case where you just mismatch the SNR, like your first, first, first plot, we know that I think two is not in the 80 line. But probably in general, I think, would not be RS. We thought about it for a while. We had this heuristic argument, which we were not completely sure it was correct, which kind of pointed out that there was a kind of replica symmetry breaking, but that had a really simple structure, like not kind of Schrodington-Kerr-Patrick model replica symmetry breaking. But maybe we can discuss it. I think it's interesting. OK, thank you. There is not other questions. Let's thank Manuel again.


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