All right, so I'm actually going to turn this. There we go, just a better light. So today's lecture, we're going to focus on multi-period models for asset prices. And we're going to look at how the continuous time limit for multi-period models work out. And by the end of the lecture, we'll understand how to calculate the so-called Black-Scholes formula within this continuous time limit study. So before we go on and cover the new material, as always, I'd like to review what we've done last time. And in the last lecture, what we looked at was the following situation. We were considering the most simplistic version of a discrete time model in which there are two tradable assets. And in that case, there was a risky asset and a risk-free asset. And we said that there were a couple of conditions that we looked at. One was the so-called no arbitrage condition. And the no arbitrage condition says that there exists no arbitrage in the market, if and only if. So maybe I'll just write that out explicitly. If and only if there exists some probabilities, q, such that the initial value of the asset is equal to the discounted expected value of the asset. And this, in fact, has to be true for any tradable asset in the market, not just asset A and that money market account, but any potential tradable asset. So let me change A here. And let's just call it some general C. So for any asset, any traded asset, that's basically the most important result that we found. And this result came about through this idea of replication. So one thing that we did before even going to replication, actually let me remind you of the other point. The other major point was, if we take this statement here and we actually choose A to be the asset C, this implies a no arbitrage condition on AU and AD and R. We can just take this expression here. And we can see that one is also equivalent to, the statement one is also equivalent to saying that the asset price that's tradable are ordered in this way. It's equivalent to that. In other words, this statement is equivalent to the so-called risk-neutral probabilities. And these things are called risk-neutral probabilities. These risk-neutral probabilities are, in fact, probabilities. They're within this range 0 and 1. Those were our key results. And the way that we basically showed this, first of all, this requirement on AU and AD was we constructed arbitrary portfolios which were potentially arbitrages and then looked at the cases where looked at the values of AU and AD such that we could avoid arbitrage. And we realized that if AU, AD, and R are in this range, then we can avoid arbitrage. And I asked you to think about the implication in the other direction to show that no arbitrage implies this. So there's really an if and only a statement. In order for us to see that expression that for any contingent claim, or sorry, let's say for any asset, we have this expectation result. The way that we found this result was via replication. And let me remind you of the basic idea. So the basic idea is you take alpha units of that contingent claim, beta units of the second contingent claim. And if you combine that portfolio together, then you will end up with a linear system for the two values in the states of the world. And what we do is we ensure that this is equal to the contingent claim or the secondary asset, the one that you're trying to value in both states of the world. And the implication, the no arbitrage implication, is that the initial value of this portfolio must in fact be equal to, so this sum, must be equal to the initial value of the contingent claim. Otherwise, there is an arbitrage. And in fact, in your little quiz session, you did exactly this exercise. You worked this out for a couple of numbers. And you constructed an explicit arbitrage when C0 was not equal to the value of the replicating portfolio. OK, so that's one of the key things that we ended up showing in the last lecture. And it's something that some of you may have seen already as well. What I'd like to do today is go on to this continuous time thing, but I also wanted to revisit the example that we didn't complete at the end of last lecture. So at the end of the last lecture, we were thinking about how do we apply this in the situation where it's not just two outcomes at every time step, but multiple outcomes. And I was doing one explicit example. So I'm going to do another explicit example. These are not exactly the same numbers as the end of the class, but that's OK. We'll reconstruct from the beginning. So let's take an asset here that has these three values. It can take on those three values in one time step. And another asset can take on these three values. So this is a risk-free one. And the first asset is a risky asset. What we'd like to do, what I'd like to show you again, is go through this whole analysis and see, ask the question, first of all, does there exist an arbitrage in this market? That's the question. Is the market arbitrage free? Or does there exist an arbitrage? How would I answer that here? So there are two approaches we can take. One is the direct approach. That means try to construct an arbitrage and then see whether we find conditions under which we can or cannot. The other thing is simply impose the risk neutrality criteria. Try to find a probability measure under which the asset prices are equal to their future-valued expected discounted. And if you can find such a queue, then there is no arbitrage. And if you can't, then there is an arbitrage. We have those two approaches. So let's do both approaches. Let's go through the direct approach again. So the direct approach just says, OK, if I want v0 to v0, and the direct approach will choose alpha units of asset A and beta units of asset B, if I want that, then it's clear that beta has to be negative 100 alpha. You must short 100 times the amount of the risky asset that you took. And if we do that, then this portfolio will start off at 0. And the three states, so if we're in the upstate, we'll get 110 times alpha minus 100 alpha, so that's 10 alpha. If we're in the middle state, we would end up with nothing because 100 times alpha minus 100 times alpha is 0. And in the downstate, we would end up with negative 10 times alpha. So the logic goes as follows. If I want to find an arbitrage, I must have, so we can say, for an arbitrage, we must have the probability that v1 is greater than or equal to 0. This must be 1, so I must never lose. I cannot lose. So we can ask, what values of alpha allow me to construct such a portfolio? Well, the only value of alpha that can do that is if alpha equals 0, because I have some states in positive and some states are negative. And if alpha is a positive number, then the up branch there is a positive number, which is fine. The middle branch is 0, which is also fine. The bottom branch would be negative, and that violates this condition. So that would not be an arbitrage. And if alpha was negative, then I'd have exactly the opposite. The top branch would be negative, which already violates this condition. So we already know that we can't have a negative alpha. We won't satisfy this. And we can't have a positive alpha. The only alpha that satisfies this is alpha equals 0. But then the probability that v1 is strictly positive is also 0, because you end up with 0 in all states. And in order for us to be an arbitrage, I think you can say, but an ARB also has probability that v1 is greater than 0 has to be strictly positive. So therefore, there are no arbitrages. That's our direct approach here. But it requires this very sort of strict logical argument. You have to start with let's look at all potential arbitrages. They have to start at 0. Let's make sure that we satisfy the condition 1 of an arbitrage. That puts a criteria on the portfolios again. It puts another restriction. So that's our second restriction. Those two restrictions already force enough constraints on our system that we pin down to one portfolio. And that one portfolio violates our third condition for arbitrage. Therefore, there are not. So we can therefore safely conclude for there are no arbitrages in this market. Now what about the other approach? So the other approach is using the risk-neutral probability approach. So for the other approach, we must have the condition that a0 must be equal to 1 over 1 plus r times the expected value under q of a1. We have to be able to find such a q, find such a probability. Now in our situation, that means 100 equals 1 because the discount rates are 1 here. Look at our branching probabilities. In fact, let me just make this small for a few seconds so you can see it all at once.  That's the condition that we need to solve. I'm going to make it bigger again now. So we need to see, can we find qs? And the condition, of course, that these qs, our probabilities implies that this sums to 1 and all of these are strictly positive. That's sufficient and necessary. Alternatively, I could say each q is between 0 and 1 as well and that they sum to 1. But if they're all positive and they sum to 1, they have to be probabilities. So we're seeking such qs. OK, so let's just go ahead and plug it into our system. And we can easily see here that if we put, for example, qm equals 1 minus qu minus qd, that satisfies my first constraint, then the 100 would cancel on both sides. You'd have 100 equals, and then you see you get 110 qu and then you subtract 100 qu, so that's 10 qu. This is just algebra. And then the next one is negative 100 qd plus 90 qd, so that's minus 10 qd. And the conclusion here is that actually qu must be qd. So let's call that little q. This is just a parameter, some parameter. OK, and then we go ahead and see if we can satisfy our constraints. Our constraints are that all the qs have to be positive. We've already satisfied that they sum to 1 by making the replacement of qm. So the last thing is all of them have to be positive. While qu be in positive and qm be in positive, these both imply the same constraint, that little q is positive. Now qd be in positive, on the other hand. That implies 1 minus qu minus qd is positive. But qu and qd are both equal to little q, so that's 1 minus 2 little q is positive. And that tells me that little q has to be less than 1 half. So we have two constraints that we've seen come out of this. We have this constraint here and this constraint there. But clearly, a q exists. A probability exists. As long as if q is between 1 half and 0, then there exists no arbitrage. Because if little q is between this range, then the actual branching probabilities, qu, qm, and qd, are in fact probabilities. They're in the range 0 to 1. They sum to 1. So a risk-neutral measure does exist. Therefore, there is no arbitrage. That's always the criteria that you seek when you use the risk-neutral approach, is you try to find a risk-neutral measure. As long as you can find one, there may be many. And this example here, you see there are many. And what I want to do is explore this idea that there are actually many possible qs. What does that mean for the valuation of another new asset that we throw into the mix? It means something a little bit interesting. But before going there, let me make sure does anyone have questions about this logic for the second approach or for the first approach for that matter? We're good? OK. So what I'd like to do now is I'm going to continue with this example in a minute. But before doing that, let me tell you what I'm going to do to continue this example. I want to continue this example. And I want to see, can we find the value of a claim that, say, does, I don't know, this. It pays that, 5, 6, 10. Any three numbers, doesn't really matter. How do we value it? Well, we know from the risk-neutral probability rules, we must have the value of this new claim have to satisfy that same discounted expectation result, where the expectation is the q expectation. So we know, in fact, let me pull this and I'm going to put it down below. How to value that thing? So if we call that as b, according to the risk-neutral expectation rules, since there is no arbitrage, and to avoid arbitrage further, we must make sure that b0 has to be equal to the expected value of b1. But we have a problem. The risk-neutral measure is not unique here. We found, in going through this approach, approach number two, we found a whole range of risk-neutral probabilities. And that's kind of interesting already, from something that you may have not already seen when you did this in your previous course on binomial models. You may have only ever seen that assets have one price, compute discounted expectation. But here's a clear example where that is not true. And what would, you know what, I'm going to simplify the payoff just to make the question really, really, let's make it, let's do this. This is 0, 10, 0, 10, 0. And just so our payoff is really nice and simple. So according to this risk-neutral expectation, we've seen from above that the branching probabilities have to be, so there's a one there, first of all, we'll get 0 times q, little q, which was our parameter. We'll get 10 times qm, but qm was 1 minus 2q. And we'll get 0 times little q again, which was our other parameter. And we see that here the answer is 10 minus, 10 times 1 minus 2q. So the range of b, b actually can take on a number of values. And that range of values, we can see what does it range over. Well, that's a linear function of little q. 10 times 1 minus 2q is a linear function, so it takes on a maximum in. And it takes on the maximum in at the two end points.  If little q equals 0, the maximum value is what, 10. And what's the minimum value? 0. So this is a little weird at first when you first see this. It tells us that if we have a market in which there are three possible outcomes in one time step from now, and there were two assets that were traded in that market, and we avoid arbitrage. So there is no arbitrage with just those two assets. If we throw a third asset into that mix, its price is not unique. Anything in the range 0 to $10 is perfectly viable. It's a perfectly viable no arbitrage price. So all of these values, any such B0, is a viable no arbitrage price. That says viable. So allowed, OK? Any of those are no arbitrage prices. And I urge you, suppose you now take any number in that range. B0 equals 5, for example. And try to use method 1, which is the explicit construction of arbitrages. Try to use method 1 and see to convince yourself, actually, there is no way for you to construct an ARB. Take any number in that range. So just try it and use method 1. We know from method 2, whatever B0 you choose there will give you a unique Q. So what I mean by that is, suppose somebody went ahead and told you that, actually, the price is B0 star. It's some B0 star. They tell you 3, 2, 8.9. That's the market's price. From that, you can find little Q, can't you? Right?  Right? So no matter what price someone tells you, if it's in that range, this little Q will still be in the range 0, 1. And it's in that range by construction, right? Because we kind of just did a little closed loop. We said, with asset A in the money market, to avoid arbitrage, or I shouldn't say to avoid arbitrage, with asset A and the money market there, the implied Q, the implied risk-neutral probabilities do exist. And they imply that this little Q has to be in the range 0 and 1 half.  Well, if little Q is in the range 0 and 1 half, the price of asset B is bounded above by 10 and below by 0. So if I take any value between 0 and 10 and I work backwards, I have to get a little Q between 0 and 1 half. It's just trivial arithmetic. And that will avoid arbitrage. So what I'm asking you to do as a check of sanity on your own is to use method 1. Take a B in that range, use method 1, and convince yourself you actually cannot construct an arbitrage. So what's the intuition then? So now let's step back. That's what the math is telling us. The math is telling us there is no unique value for this asset. So what is the intuition as to why there is no unique value? And why was there a unique value when we had that other example that we looked at last lecture? Anyone want to venture an answer? There's a very good financial reason for it. Well, the specific B that I would choose in that range would depend perhaps on my personal risk aversion because now there's a choice. I can choose anywhere between 0 and 10. And if you look at the payoff, 0 is the minimum that you can get and 10 is the maximum that I can get. So it seems that, OK, well, someone who's very risk-averse will pay almost nothing. And if they were selling it, they would sell it for almost all. That's true. But what I'm asking is a little bit different than that. I'm not asking how would I pick one of these prices. I'm saying, why does there exist many possible prices? Nope, there's no bid and ask spread here. There are unique prices. For the traded assets in the market, A is uniquely trading at $100. And the money market is uniquely trading at $1. Those are unique prices. What I've done is I've also put in another thing. This is a unique set of potential outcomes, 0, 10, 0. But why? OK, there's another. Very good. So the idea is that, how did we come up with this concept of no arbitrage that had to satisfy the risk-neutral expectation result? What did we do? I reminded you just now at the beginning of the class. We took two traded assets. We took linear combinations of them. And we replicated the third. We traded the two things. Let me go back one slide. We took the asset A, we took the money market account, and we made a linear combination of those things such that the value at time 1 was always equal to that third asset, no matter what the outcome was, whether it was the upstate or the downstate. And that's the replication idea. This is the so-called replication. And when I replicated in this way, I found a unique portfolio that allowed me to replicate. Now imagine doing this replication, attempting to do this replication in your current situation, where you have three outcomes. There are two traded assets. So let's say try replication. So we've got 100, 100, 110, 90, alpha, 1111. This is beta. And we can see that if we combine these two things together, you get 110 alpha plus beta, 100 alpha plus beta, 90 alpha plus beta. And if we're attempting to do replication, we will want this to be equal to be in the upstate, be in the middle state, and be in the downstate. That's what we like. But there are only two degrees of freedom, alpha and beta. So this linear system is over-determined, right? Or yeah, we have two unknowns, three equations. We may be able to satisfy. In fact, you can satisfy. Take any two of the ones. Take any two of these equalities. Take any two. And you would find a unique alpha and beta. But then the third one, you will not. There will be no unique alpha, beta. Or sorry, I shouldn't say there will be no. You will violate the third equation. So if I took equation one and equation two, the first, the up and the middle, if I solve for alpha, beta there, I would definitely get Bu and Vm. But then I will not get Bd. So you cannot use replication arguments to construct the value. And because there is no replication to construct the value of that extra claim, what it means is that that extra claim is actually like a really fundamentally new asset. Unlike when there are only two outcomes, the extra claim, the extra asset that you throw into the mix, it's not fundamentally a new asset because you can build it out of the other two. In the case when there are only two outcomes, you can take a combination of those two to make that third thing. So these two assets are really the same thing as that third one. Or a linear combination of these two are the same as that third asset. Third asset doesn't add value, does not increase the amount of outcomes, does not do anything for you. Now the third asset does something new. And because it's doing something new, it is allowing, you have now a new degree of freedom to put a price on it. Let's think about this pictorially or geometrically because there's an interesting geometric interpretation as well. Let's suppose, let's go back to the case of just the two and just for the sake of having numbers in front of us, let's go back to this case, all right? And we're trying to replicate some assets, okay? If I interpret, I'm gonna draw axes here. Whoops, I draw straight lines, I need this tool. What I wanna do is I'm going to interpret these two axes. What they represent are the value in the upstate and the value in the downstate, okay? U and D. This is D, this is U. If we take asset A, asset A is somewhere 90 and 110, so it's kind of, it's somewhere like that. It's kind of a vector, right? This is a point, 90 comma 110. This is representing asset A. Asset A in the downstate will take on the value 90 and in the upstate will take on the value 110. And if I take the money market account, the money market account is just, it's a straight line with slope one, that's one, one, right? If I think of them as vectors, these two assets, and I think of a linear combination of those two vectors, what does a linear combination of two vectors do for me? It's gonna span a plane, right? And it's gonna span exactly this plane just this two-dimensional plane. You can get to any point in this plane. No matter where I am, I can get to that point by taking some linear combination of this vector and some linear combination of this vector and I will be able to produce something that points to any particular place and let's call that BUBD, okay? That's my asset that I want to find. So in this two-dimensional case, in this case where there are two outcomes, I can get to any point on that plane by trading in those two assets unless what? There's one situation where I won't be able to do that, even if it's two-dimensional. What if these two arrows are pointing in the same direction? They're pointing, they're collinear, right? So if they were basically pointing in the same direction with the same or opposite directions, right, collinear, then I couldn't span anywhere. Then I would just be able to go along the line. But if they're not collinear, then I can get to any point that I want. What's the interpretation in terms of this three asset or three outcome situation? Well, now I have, now we've got, instead of a point in 2D space, we're gonna have to describe what happens in the upstate, what happens in the downstate, and what happens in the middle state. Well, that's close enough to the center. So that's like U, M, D.  And then there's the money market account, which is just, again, a straight, that's just pointing off in the direction one, one, one. And if I take a linear combination of those two things, which would be my general portfolio, what am I gonna do? I'm only gonna line a plane, right? This will not span the entire three-dimensional space. Instead, I'm going to land on some sort of plane, something like that, right? If I take a linear combination of those two vectors, I will be in some sort of tilted plane that goes through the origin, but some sort of plane. And if asset B, in general, asset B is gonna be somewhere there. It's gonna be BU, BM, BD. It will not necessarily land on that plane. And so there is no linear combination that will get me there, right? There isn't one. So there are certain choices. There are certain Bs, there are certain claims, there are certain assets, which would lie on that plane. And those assets, I could get to. And so you can think, if you think ahead really quickly, that means for certain assets, there is a unique price, but not for every asset, right? For any asset that lands on that plane, there will be a unique price. But if you're off, there won't. So given that there is no unique price, what do you think is a reasonable price? What's another, so we could try to incorporate risk aversion and go through the similar kind of analysis of this indifference valuation stuff that we did in the first class, but now with three assets in the mix. And I said it's a little bit, it's possible to do it, it's just a little bit annoying and long detailed algebra. But if we think about intuitively, and particularly this picture in front of you, is there some point, is there some way that you could think graphically of getting a unique price that makes sense for the general asset? Remember, you can't, before unique prices were, we found because we were able to exactly replicate the claim. Now we know we cannot do it in general. But what's the next best thing? What about minimize the least squared error? Something like that. Minimize, take all possible portfolios that you can make and find the one which is as close as possible in the L2 sense to your claim. Does that kind of make sense? And what does that mean geometrically? Well, it simply means take the plane, put up a perpendicular, find the perpendicular that goes through that point, goes through the asset's point. So that would be there, right? That's a, whoops, that's a, ugh, damn it. Trying to draw that little perpendicular sign that you usually do when you see a triangle. And it's a 90 degree triangle. Is that picture reasonably clear to you guys? You see what I've done, I've just projected onto the plane in a perpendicular way. And now that point that touches the plane, this point that's on that plane here, that one, I can get to that. That's the point that I can trade using my asset A and my money market and I can get there. So if I do that, I would get the replication of the payoff that is as close as possible in the L2 sense, right? In least squared error, in sort of squared error sense from the claim that I'm actually trying to value. And that would be a unique point. It would be a unique portfolio, right? It's two, you have two vectors, you're trying to get to a point. So those two vectors, there's only a unique combination that will get you there. Right, there's a unique value that, so this gives me a unique alpha and data which would imply a unique quote unquote price. Now, it's not really a price of B. It's a price of the claim that is as close to B as possible. Right, price of asset. And maybe we can call that thing, that asset there, let's call it B hat. So what we've done is we've created a new asset, B1, or B hat, such that B1 minus B1 hat when we square it and we compute the expected value. And this is actually the expected value, here's the interesting question, is do I do that expectation under P or under Q? Under the real probabilities or under the risk neutral one? So the question's still open. No one's given me an answer yet. P or Q, how many people think that we should use the P measure there? The real world probability measure. Just a couple, okay? How many think we should use the risk neutral probability measure there? More than a couple, okay? So the most of you are wrong. If P measure is the correct thing, because why, why is it the correct thing? Because you are trying to minimize the error in your replication. If you think about it, you're trying to minimize the error of the actual outcome. You're not trying to minimize the error of the fictitious outcome. Cues are not real probabilities. They do not actually happen. They are simply there to tell you whether arbitrage is in the market or arbitrage is not in the market. And it so happens that asset price is given by an expectation in terms of those cues. But that Q itself does not tell me the probability of the real outcome. The real probabilities are the P. So you should be minimizing the expectation under the real world probability of the squared error. And then you find such a B1, okay? You find that, and then you price this thing. So this is, I'm just writing below what I mean by that diagram, okay? So this perpendicular thing, it's orthogonality but not orthogonality in Euclidean space, right? Because they're not equal probabilities. It's orthogonality that are perturbed by your probabilities, yeah. If it's perfectly perpendicular, well, then the price is zero, right? That's the closest point. If I have a plane, okay? Take, rotate into the frame of the plane. So the plane is flat. You can always rotate things around so your coordinate system is flat. And you're saying what if B is right here? Well, what's the closest point to B? Zero. The value of that claim would be zero. I shouldn't say that's not the value of the claim. That's what this principle would tell you to value the claim at. This principle would say the claim cannot be value or the claim's value is worth nothing. Why? Because there's nothing you can do to hedge it, right? The best you can do is put zero in everything. That's it. So that's the value that you would associate with it. Whether you can actually find somebody willing to buy or sell in that way is another question, all together, right? This does not tell me that the market will allow that price. This just tells me this is a rational approach to giving a value to something for which I don't know its unique value. All I know is that there's a range of prices given by the risk-neutral expectation, just like what we have here, right? There's a range, oh sorry, I need to slide up more. We found this range and we're trying to understand what is a reasonable range, okay? Now you notice if B is always positive, if it always makes a positive payment, B would not be perpendicular to my claim. So there will always be some positive portion to it. Do you realize that? If B had all positive components to it, or at least non-negative, then B is hanging out somewhere in that quadrant in this direction, right? Somewhere in this quadrant. So it's clearly not perpendicular to the plane, which is somewhere sort of slanted in that direction. The point that will be closest to B is very typically in the way I've drawn it. And it wouldn't be at zero. So very typically you'd have a non-zero price. Now the interesting question to ask at this point is, does such a pricing mechanism lead to no arbitrage prices? It's not obvious, is it? We started off by saying, okay, there is no arbitrage with A and money market. There is none. And if we want to value B, B's got to be in the range zero to 10, according to the example we did. And now I've sort of thrown that all out and I said, okay, there is no arbitrage. I don't have a unique price. How am I going to think up of a unique price? What's a reasonable principle? And here's a reasonable principle. Find a portfolio that's as close as possible that I can get to by trading in my traded assets. Find a portfolio that gets me there in the closest possible way in the L2 cent, in the square cent. And value that claim and say the value of B actually is the value of B hat, right? That's a principle. I didn't use anything about arbitrage. I didn't use anything sort of foundational there. But an interesting question is, does it lead to no arbitrage prices? And I'll leave that open for you to think about. So if you want to think about it, do it. This is not a trivial question. It's not very hard, it's not trivial. So let's say, so B1 hat is the, or so what is argmin? It means the thing which attains the minimizer there. So you find the B1 hat that actually gives you the minimum. And then you say that the value of B is equal to the value of B1 hat, which is one over one plus R expected value. Or sorry, not one over one plus R. It would be the alpha times A0 plus beta, okay? Which replicated it, all right? So then the question is, is this arbitrage free? It's an interesting, more slightly theoretical question to work on. This strategy, by the way, this is called the risk minimum variance hedge. Minimum variance hedge price, okay? So what is hedge here? Hedge means hedging in the sense of removing risk. You're doing the best you possibly can to remove risk, right? Because you found the B1 that minimizes this error as much as possible. Now remember again, this B1, these hats are only those portfolios which you can get to by trading an asset A in the money market. Okay, they're only those ones. They're not any arbitrary ones. They're only the ones that, so if you like, you can think of this as alpha A1 plus beta in fact, right? That's the most general portfolio from trading in the asset A and the money market. Yeah, yeah, but some will be negative. Some will be negative and some will be positive in order for that to happen. Yeah, so, but that's because if some, so if all of them are negative, you won't have this situation either. You'll only have it, you'll only have the projection being at the origin if some are negative and some are positive and it's okay for an asset which has in some states of the world positive outcomes and in some states of the world negative outcomes to be valued at zero. Perfectly viable asset or perfectly viable price, seems like. And you can actually prove that yes, these are in fact no arbitrage price. So I'll let you know ahead of time. The answer is affirmative and you can try to show it. Okay, any questions about this discussion? Nope, all right. Okay, so I think I'm gonna stop the discussion. Let me see, was there anything else I wanted to say about this case? No, so now I'll stop the discussion about this non-uniqueness in the market. Sometimes that's also, this is called non-unique or incomplete market. So let's put that little title here. So it's incomplete because what does incomplete mean here? Incomplete just means that when you throw the new asset into the mix, you cannot replicate it from the other two. Okay, you can't. So that's what it means to be incomplete. When there are more outcomes than there are traded assets. That's what it means in the discrete setting. Okay, let's, before going for a break, one example just to show you that there are cases which it is unique. Let's think up of the following claim. I just had to do the quick calculation in my head. So what about a claim that does this? Zero, 10, 20. Okay, this is my claim. Same, so we're doing the same, continuing that same example. Okay, all the same numbers are flowing through. So what is our B zero supposed to be equal to?  According to this example. And here you can see, you see I've chosen something very judiciously. The 20 Q cancels out two Q times 10 and you're left with exactly 10. So even though this market was in general incomplete, we found a unique price for that specific asset, 20, 10, zero. It's actually valued at $10, it's unique. And this is going to the point that I said before that not all assets lie, so if we slide back up to our diagram, the generic asset does not lie on our plane, does not lie within the span of our asset, but certain assets do land in the span. And those assets will have a unique price, okay? This is an example of one. So you can try to think up of some others, right? Probably, what would happen if I put the 20 in the downstate and zero in the up? Same thing, right? It's actually valued exactly the same. If instead I had 20, 10, zero, this is also going to be valued at $10. You can easily check that. And then the other, of course, there's another natural exercise here. And that is, what if B zero equals 11 in the market? There's an arbitrage, right? There has to be an arbitrage. How would you find it? So you can think about that as a good exercise. Find the ARB. Okay, so maybe that's about it for this topic for now. Would you like to take a break before we start the multi-step? Okay, let's take a break. All right, so we're going to start something new now. We're going to be looking at the continuous time limits and continuous time versions of these discrete models. And we're going to do this in the most simplest setting back to the situation where there are only two outcomes at every step. And the reason is basically just for simplicity. You can also formulate this analysis when there are multiple outcomes at every step. So, so far what we've been doing when we talk about these traded assets is I've been telling you effectively what is AU and what is AD and what is R and also the initial price A zero. Either value through indifference or I just give it to you, right? We've been saying here's a picture. I've given you these numbers. Go ahead, value a third claim. But if we want to think about having this model say something about the way that assets are actually behaving, we have to go and figure out what a reasonable model for AU and for AD itself. How should we write those down in order to replicate behaviors that we actually observe? And in order for us to understand that, we need to talk about a multi-period setting and before going to the multi-period setting and taking this continuous limit, I just want to say something about two steps. How would we use what we've developed so far in two steps for valuation? And then come to the question of how should I write this in terms of observable, okay? How to write in terms of things that we can actually observe, okay? Because you can't observe what the two future outcomes are, it's not something you actually have access to. So far we've assumed that. We said, suppose we can, then all of the stuff that we've done makes sense. But in principle, we actually have to have a way of figuring out what are those two possible outcomes? And in general, what are the multiple step outcomes? So let's discuss just two steps of the model, first of all, and just stylistically, I'm not gonna write down a lot of details here. So one way to think of a two step of the model is to imagine the situation where once you get to the second step, there are two further outcomes. So I start off at some value, here, I get to either I move up or I move down. And if I move up or I'm down, then again I can take on two other values. And if I'm in this state, I can take on these two values. And you could keep going on forever like this, right? You could make this into a multi-period setting. And this is a perfectly reasonable approach, but there's one main problem with it. The main problem is that the number of points, the number of nodes that you have after making a total of n steps is two to the n. It grows exponentially. And that's just the number of nodes at the end. So if you sum them all up, you of course get more. This is gonna grow basically, well certainly geometrically exponentially. So this is far too many. So we need to think of a way of modeling that reduces this complexity, this computational complexity. And the way that is often done is by designing what are called recombining trees. So this is just a binomial, a generic binomial tree, but there are other kind of trees that are called recombining. And the idea here is very simple. If I think of the two-step case, when I make, if I go up and then down, I want it to be equivalent to going down and then up. So I want this path to give me the same value as that path. So if I wrote down the values for say a single asset, we would have had AU, A0 going to AU, and then here it would have been AUU, and in this node it would have been AU and then D, and here it would have been AD and then U, and here ADD. So when we recombine, we make those two things equal. That's what implies recombination. And the advantage of recombining is, if you imagine going out to multiple steps here, it's purely a computational one actually. But you can imagine if I go out to multiple steps, how many nodes do you get? You always just add one more, right? It grows linearly, okay? So here you've got two, here you've got three, one, two, three, four, here you've got four nodes, and if you sum them up, it'll grow quadratically. So this is little o n squared, or big o n squared. It's of order n squared. Something like n times n minus one over two or something like that. While here this is of order exponential, clearly of very different nature. So there's a huge advantage to doing this. Okay, it makes sense? Why we're gonna use recombining? Okay, so now let's go forward and say, okay, let's suppose we use a recombining tree to model two assets. And we ask the same set of questions that we asked in our last lecture. We asked about, does the model have no arbitrage? What are those conditions? If we take those conditions and then we add it and we assume there's no arbitrage, we throw in a third asset, how do we value that? So we'll readdress all of those questions very, very quickly, because it's actually extremely simple. Once we've understood the one period case. So here's our asset A, and over here let's put our money market account. I'll just call it M. And notice I've only put a subscript on it, and it's the same, oh, that should be two, M sub two, M sub two. Okay, so these are equal to one plus R. One plus R all squared. All right, just grows at the risk-free rate. Okay, just for simplicity. And let's say I give you these values. I gave you some numbers there. How would you decide whether this system admits an arbitrage or not? What would you do? Well, you could use the definition, the fundamental definition is, you set up a trading strategy, and you check to see whether you can find a trading strategy which starts with zero value, and at some point in time, has the criteria that you never lose, and you sometimes win. Those are the three things. But now, you have to rebalance the portfolio. So if I started, if I originally started with some position in alpha, let's call it alpha zero, I'm gonna, darn it, I'll draw it on a separate tree here. Alpha zero, beta zero. This represents my portfolio. Positions in asset A and the position in the money market account. Once I get to the next time step, I'm allowed to change that in principle. I don't have to have the same alpha and beta anymore. I can change it because I'm allowed to trade, and I would have some new value, alpha one up, beta one up, and down here, once I get there, again, I could in principle change it to alpha one down, beta one down. And then the story's over, right? So I only get to change it once. But is there a condition that I have to impose on these strategies? Can you think of a condition that I have to impose on the new alphas relative to the old ones? I'll think about it financially for a second. You put $10 in one asset, $5 in another asset. You've waited one period in time, and you're now gonna change it. You're now gonna change the number of units, so let's restate it. I put $10 worth of my money in one asset, and that may have cost me one unit of asset. I put two units, and I bought two units of the other asset, so one, two, that's my portfolio. I hold that to the next period. I'm sitting here now. I now know whether the asset went up or it went down. I'm now going to change my portfolio from one asset, one unit of asset one, and two units of asset two to something else. What constraints do I have? Yeah, yeah, right. Remember this concept that I mentioned briefly last time about self-financing? I have to be able to use, I can only buy and sell whatever I get from the strategy. So one unit of asset one and two units of asset two are currently worth some amount of money, currently worth a fixed amount, and all I can do is buy assets worth that fixed amount, that's self-financing. So let's write that down in equations. I must have that once I go from, so if I bought this many originally, how much is that worth at this point in time? It's worth alpha zero, a one, whatever a one happens to be, plus beta zero, m one, whatever m one happens to be, and here m one is always one plus r. That's how much money this system is now worth. And after rebalancing, I have to ensure that I hold this same amount of money. So if I'm in this state of the world, I have to make sure that alpha one times u, a u, the new value of my asset, oh, sorry, yeah, that's right, yeah. Alpha one, so this is my new amount of alpha one of a, plus beta one u of m one, that has to be equal to this amount of money in the upstate. So that has to be alpha zero, a u, plus beta zero. And if you want, you could have put this as mu here. It doesn't really matter, right? The money market account is the same in mu and md. Does that make sense to you, this equation? This is self financing. I have this amount of money in my hand available because I had alpha zero units of the asset, it's not worth this. I had beta zero units of the money market account, it's not worth that. And in the upstate, that's the bottom line. And I rebalance now, so I have some new position, alpha one u and beta one u. But those are degrees of freedom that I get to choose, but I'm constrained. I can only choose it so that I self finance, so that I don't throw money into the system or pull money out of the system. And if you're in the downstate of the world, you of course have a similar, basically the same equation, right? Oh, this is alpha a1d and that has to be equal to again. You could also write that, you can also write that same system in general here. Alpha one times a1 plus beta one times m1 has to be equal to alpha zero a1 plus beta zero m1. Both of those equations can be encapsulated like this. My new position times a new asset value plus my new position times a new asset value must be equal to the old position times a new asset value plus the old position times a new asset value. And if you rewrite this, you can see that you can write it like this. The change in my position times a zero plus the change in the position times m zero must be zero, times m1, sorry, must be zero. In other words, any money that you put in one has to come from the other, that's it. Does it make sense? So this is a self-financing condition. Okay, so if we were asking again, going back to that question that I recently asked, can I find arbitrages? Well, it will be a little bit difficult in this setting, right, because I've got to find, I've got to check putting a portfolio at time zero that has zero value, go one time step, make sure I'm satisfied with self-financing condition, try to find a portfolio again, which now has the conditions that I never lose and I sometimes win. That's what I need to do. But is there another approach that you could take that may be able to avoid this? Well, today's class, we talked about two approaches for just the one period case. We said the explicit direct construction by using the definition of no arbitrage or inducing the risk-neutral probabilities and seeing whether or not the risk-neutral probabilities really are probabilities, right? So let's talk about how can we adapt that idea here? Well, the approach is actually quite simple once you do one very simple thing, circle this. Now, what do I mean by circling that? Well, think of the following situation. You are a time one and you are in the state you. If you are in that state, there are only two possible outcomes, AUU and AUD. Those are the only two things that could happen. And M could either, and M grows to one plus R squared. M just multiplies as it always does. So if I look at those two circles, and then I think through the logic that we had before, I'd realize that if I start there, I have to avoid arbitrage, or sorry, avoiding arbitrage is equivalent to the existence of queues for that branch. It implies that if I can find risk-neutral probabilities just for that portion of the tree, then there is no arbitrage in that portion of the tree. Then you go to the other portion of the tree and you say, okay, let's look at this portion of the tree now. Compute queues there. If those queues are also between zero and one, then there is no arbitrage from there going forward. Then finally, you focus on this portion of the tree. And then you say, well, there's no arbitrage in the future steps. And if there's no arbitrage in this step, then there is no arbitrage anywhere. There can't be any. So that's all it is. It's exactly the technology that you've already developed. There's nothing new to add. All you need to do, I guess the only new twist, let's say if we wanna call it a twist, is that each of these branches, they have their own queue. So there's a QUU associated with that branch. There's a QUD associated with that branch. And there's a QU associated with that branch. And these don't have to be the same. Those queues don't have to be the same. They could in principle be all different. And if all of them are between zero and one, then there does exist a risk-neutral measure on the whole tree. And if there does exist a risk-neutral measure on the whole tree, then there is no arbitrage. So you see, this approach doesn't require you to try to impose a self-financing condition and do all of those little intricate computations. It simply boils down to a linear algebra system for three parts of the tree. And just to be very clear, let's say, what is the linear algebra system that I'm solving for the, I don't know what color that is, this portion of our tree? What would it be? How am I going to solve for the QUD? What equations would I write down? Well, I'd write down AD equals one over one plus R times QUD AUD plus one minus QUD times ADD. And I solve for QUD. Right? That's it. There's nothing more to it. And then similar kind of system for the yellow circle and then for the purple circle. So it's a little bit of annoying calculation, but it's straightforward. Three linear systems. Make sense? Questions about that? Because I'm not going to do an example, right? You can cook up your own example and try it out. Or look up in the notes. Look up in the book, sorry. Okay, so now, so at this point we can answer, so let me write down a statement here by the way. So if I want to put that in brackets because that's sort of not the approach we're going to take. We're going to say this is the direct approach that's in the brackets. Instead of talking about the Q approaches, if there exists a risk neutral measure Q such that any A at some T equals one over one plus R expectation under Q of A at T plus one, then there's no arbitrage. And in particular, this is really an if and only a statement. So we can modify this, we could say, there exists a Q such that if and only if there is no arbitrage. So find QU, QUD, QDU, and check they're all in the interval zero one. And that will be, then there won't be arbitrage, if that's true. The algorithm Claire, and it just follows from what we did last time. Okay, so now let's answer the next question that we usually ask, or that we have asked in the past. Suppose we've done that exercise and we've found there is no arbitrage. Okay, and you can guess what the no arbitrage conditions are gonna be, right? Right off the bat from here. You know, remember that ordering of the asset? A, D is smaller than one plus R, A, zero is smaller than A, U. Well, basically that has to hold for these three different notes. That's basically it, nothing more. So the next question that we usually asked was, throw in a third asset into this mix, put in an asset B. How are we gonna value the asset B? And what information are we given? So let's draw that. So I'm gonna repeat the diagram I have up there. This time I'm not gonna put the values in the notes, I'm just gonna draw it like this. Okay, this is my asset A and my money market. We throw in asset B, and what if I tell you the value of asset B here? How do we figure out its price? We have two approaches. One is risk-neutral expectations, the other is replication. If we already did the exercise in part, one, checking for no arbitrage, if you've already done that, then you already know the cues in this tree. Right, remember, this tree here is not a different tree. It's the exact same tree, we've just split them up into three. So this tree, the branch, the probability of going here is still the same Q U U, this is still Q U and this is still Q U, the same one. The risk-neutral probabilities are the same one. So what we can do is the same kind of exercise. I can focus in on this part of the tree and I can use the Q risk-neutral probability and I can find B U equals one over one plus R times the expectation under Q, which would be Q U U, B U U, one minus Q U U, B U D, okay? That's just the straightforward approach. Alternatively, if I wanted to replicate, what I would do is look at this part of this tree, look at this part of the A tree, take a linear combination and make the linear combination, only in that portion of the tree replicate this portion of the tree for B. And that would give me a unique portfolio at that node. All right, so if I did this, I would find an alpha one up and a beta one up and the replicating value would be equivalent to this. All right, it would have to be the same. And then you can just repeat the exercise, right? Go to this node, there and there. And at that point, you would have computed the value of B, you would have found, sorry, you would have found B U and you would have found B D, right, by doing this procedure, either the replication or using the discounted risk-neutral expectations. And once you calculate those two, then you focus in on the last portion of the tree. Again, the system is identical. So you've reduced this whole problem of computing multiple time steps into a whole sequence of little binomial models, right?  They're just a sequence of little binomial models.  Like that. All you do at every single point in this tree is the same exercise, okay? You just recursively, you just focus in on some little node here, compared it to there, there and then it's the same old binomial story. Nothing new, okay? Good. Are we happy with that explanation? And I think some of you have already seen that before, so it shouldn't be completely new. All right, so now what I'd like to do is go on to discuss what's an appropriate model itself, because now we have a way of taking a model. If I've given you these values, A, A in all of those nodes, we now know how to check to see if there's, if there's free of arbitrage, and then we also know how to go ahead and value a new claim, a new asset. But now we ask the question, what's the appropriate model? Okay, that's what we're going on to now. So we've got time, so I'm gonna go with that. What's the appropriate model? So let me reflect it back to you. What is it about the price or the dynamics of the asset price that you would like to capture using this model? What's the most basic things that you can capture? You see asset prices moving around, right? That's what they do. They move around, and in general there's some sort of, there may be some positive return, and there's some fluctuation. So you wanna at least capture the return, expected return, and the variability of that return. That's what you want to at least capture from based on your historical observation. If I can have a discrete binomial model that somehow reflects that same characteristic, then I can be confident that at least I'm getting the first two moments correct in the behavior of the asset's dynamic. And then you can do much more complicated things, but for now that's all we'll do. So let's talk about how we would go about and do that. Well, imagine having a set of observations of the asset. So you have a set of observations in history, or in fact, let's put these at a number of historical times, and equally spaced, okay? C2, dot, dot, dot. You have this whole sequence of historical observations. These are observed, and what we'd like to do is at least compute the return of this historical observation, and we'd like our model to replicate that behavior. So let's look at how would you calculate returns? There are two ways to do it. If we look at the return over any time period N, I could compute just the sort of relative return, which is this, agree? There's also another type of return that's a little bit more useful for our purposes, and that's the logarithmic return. So you take the logarithm of the ratio of the asset prices, and you compute these things historically, and then you might say, well, why one versus the other? And my one answer to you would be, well, actually they're not that different. It looks like they're completely different, but they're not that different. Here's why. Tracked one, add one. Okay, sorry, let me not do it directly in there. Let's do it here. So we've got this ratio, call that ratio alpha, okay? Subtract one from alpha, add one. That's a log of one plus or one minus one minus alpha, agree? Yeah, okay. Then I would say alpha is the relative price, right? Alpha is this ratio. It's this ratio that's showing up there. Is that ratio generally very big? Is that ratio generally close to some number, close to zero, close to three, close to 3.1 for one, five, nine, two, six, five, three, five, eight, nine, seven, nine, three, two, eight, four, one? No, it's close to one, right? Normally. So what I've underlined there is actually close to zero, generally. So I can do a Taylor expansion. I have log of one plus something small, well, one minus something small. And what is that approximately? You're gonna need your Taylor expansions in a minute. Quite a, you're gonna need quite a few of them in a minute. So do you remember this? Log of one or one plus x is approximately what? X, isn't it? Minus x squared over two plus cube over three, dot a dot. Okay. So this is approximately alpha minus one. One plus x is approximately x. So you have one minus x, so just flip the sign. I guess I could have kept this as plus alpha minus one. You get alpha minus one, plus dot dot dot, higher order stuff. And what is alpha minus one? Well, that's a tn over a tn minus one, minus one. And you can see that this is actually more or less the same thing, okay? So this logarithmic return is approximately the same as this relative return. But there is an advantage to working with the logarithmic return. So here's the story for the advantage. Suppose I told you that we were going to assume that these returns are normally distributed. First order approximation, right? We're thinking of perhaps central limit theorem type thoughts going through our mind. The returns should be independent of one another. Presumably they have finite variance. So presumably they're going to at least in the long run look like some sort of long normally distributed random variable, or sorry, normally distributed random variable. So we thought the returns were normal and we used the first version of things. What could potentially happen to the asset price? It could become negative. So if we use the first version of things, assuming R is normal is not actually going to be accurate and could lead to theoretical problems. If I assume that the second type of return is normal, now I'm perfectly fine. Because the log of something positive could be negative for sure. Lesson one, it's negative. So that's the reason why we're going to deal with the logarithmic returns. I'm going to use this. Okay, so I've told the story, said that we have this historical asset prices. We compute this historical return and we're going to compute an estimate of its mean and its variance. So we get from the data some estimate of the mean and the variance. And how would I, so if I view these as, R is actually a random variable, right? And I want to compute the expected, the real world expected return, or sorry, expected value of this return. And let's say that from observation, we find that it's equal to something called mu star. That's just the observation. And how did I compute that from observation? What's my estimate? You're all actuaries, right? I have a random variable, I have a bunch of observations. You compute an estimate of its mean, sample mean, right? Nothing miraculous here. Just compute the sample mean. Now, because of something that's going to happen later, okay, actually, sorry, I won't tell that story yet. I'll come back to it in a second. Now what about the sample variance? Or the estimate of the variance? We can call that sigma star squared, okay? And again, I can compute that. Well, we can make it the unbiased version. Let's not bother. So we have some estimate of the variance. We've observed it. Now, because of two things, we're going to modify these expressions here. First point, why we're going to modify it. Suppose I did these observations daily. And I told you the return was 0.1%, okay? Then that really means what annualized amount. So it depends on if you continuously compound it, et cetera, right? But if we just did simple interest, we would say that we'd have to scale it by how many days in a year. So when you talk about return in finance, you usually talk about annualized versions of things. So we're going to annualize this thing, first of all. And variances, as well, will be annualized. Variances, if the random variables are independent, then the variance of the sum is just the number of terms, is just sum of the variances, right? Variance of sum equals sum of variance if the random variables are independent. So we would simply scale it. And this would tell me that I'm measuring them in an annualized quantity. So the stars, mu star and sigma star, annualized things. There's one more correction that happened. And this is going to look a little weird why I put this in now. It comes in because later on, it leads to a cancellation that allows you to interpret mu star in a very particular way. So what we're going to do is we're going to say, okay, we've got data, we calculated the period return. We estimate the mean of that random variable and the variance of the random variable. And we simply call that estimated mean and that estimated variance in this way. We just define it. So mu star and sigma star are equivalent to estimations of the mean and variance modified to be annualized and modified to correct for what is called a convex decorrection. And you'll see why it comes in later on, why I put it in there. I didn't have to put it in there. And all of theoretical work that I've derived does not require me to put this there. But for me to interpret mu star in a particular way, I do require it. Okay, it's only for the interpretation that I need it. For the mathematics, it's not necessary. I could have called everything in the round bracket there mu star instead. Okay, is this clear at this point? You've gone through, got historical data, estimated the mean, estimated the variance. This is what our data says. Now we want our model to behave this way. That's what we want to do. We want our model to replicate this behavior. So here's what we will do. We'll say, let's take, we're gonna make our model, because we know that returns should be independent of one another, these logarithmic returns should be independent of one another, then our model should more or less look the same through time. It should be stationary. So it shouldn't depend on where I am in time. And for that reason, I'm just gonna draw generically one node in my giant tree. So if we take, you know, we have in general some giant tree here, and I'm only gonna take this little node in the middle there, and I'm blowing it up, and we're gonna say that at that node, our asset price is whatever it is, and it's gonna take on two values. It's gonna go to either, grow, remember we took logs before when we talked about return. So that's why I'm putting an exponential factor here, because when I take logs, then the exponential will go away. So it's gonna grow by some factor C, or decay by some factor minus C. Why do I make this particular choice? At this point, it's a modeling choice. I'm just simply thinking of it as the asset goes up or down. I didn't have to make it symmetric, but if I didn't make it symmetric in log space, it's symmetric in log space, not when you remove the log, but in log space, it's symmetric. If I didn't make it symmetric in log space, I'd find I have too many degrees of freedom. We only have two things to match, mean and variance. And what's the other parameter that I have to play with here? So C is one parameter, what else? The branching probabilities, P, the actual probability of the outcome. I have two parameters, P and C, and I've got two things, two observables that I need to match, mean and variance. So that's enough. I can't throw in a third. I can't put E to the C and E to the minus D. I'll have over-specification. I mean, I could, but then I'd have over-specification. And I'd have to use some other thing to decide on what should D be, okay? So I'm trying to give you a rationale as to why this form. Make sense? Okay, so what we're gonna do now is we're simply gonna find what is the appropriate P and what is the appropriate P, which allows my model to behave like the data. That's the goal. Because now we'll have a sound way of stating how an asset should be modeled, as opposed to before where I simply said the asset for sure has these two values in one time step. Now I'm saying go to the data, estimate these things, build a model that behaves like the data. Okay? All right, so it's actually quite simple now. It's now just a matter of some simple algebra. Okay, from the model, let's compute the expected return. Well, what is R actually in this case? It's log of A, E, to the one way that I can rewrite this asset, actually a time one. Let me put one more little note here. One way that I can write this asset to time one is to say that asset of time one, time step, or any time step n, in fact equals its value before times E to the C times little xn. And what is little xn? All of these little x's, x1, x2, dot, dot, dot, these are iid, Bernoulli random variable, okay? But they're not standard Bernoulli. They're not zero one. Instead they're plus minus one.  Okay, so if I compute the return, you can see that the return at any, the returns they're all identically distributed, they're log of A n divided by A n minus one, and that's just C times xn, right? The return over any nth period. So what I need is I need the mean of this return. I want it to be equal to mu star minus a half sigma star squared. Times delta t, and that has to be equal to C times the expected value under p of xn, and that's equal to what? What's the expected value of this Bernoulli? Two p minus one, right? p times one plus negative one times one minus p. Maybe I'll do that in two steps for you. One times p plus minus one times one minus p, right? X takes on those, those are the probabilities, plus or minus one, so it's just that. So that's C times two p minus one. So in fact, you immediately get the branching probability, the real branching probability, in terms of C, if we knew what C was, and the observable. So we can just check immediately, this is just one half, one plus mu star minus a half sigma star squared all over C times delta t, okay? The next thing that we have to do is match the variance. Well, from the data, this was sigma star squared delta t, whatever that is, right? We just get that from the data, and we've got the variance of Bernoulli. Well, the variance of R is the variance of C times x, so that's C squared times the variance of x, and what's the variance of x? Well, that's the mean of x minus the square that's mean, the mean of x is one, right? Plus minus one, so if you square it, it's always one. Maybe I'll write that in two stead stages for you. And we've just computed this thing, it's two p minus one. We got two p minus one all squared, okay? And now we can just replace, so from the top line there, we know that two p minus one is equal to this divided by C, so we can use that to replace this two p minus one squared, okay? I just underlined it in purple, I don't know if you've seen it, let me highlight it for you. I'm gonna use this equals that to replace that. So we get C squared one, sorry, one minus mu star minus a half sigma star squared there's a squared there, there's a squared there, and there's a squared there. And therefore, so you can see the C squared cancel in this denominator here, and all I have to do is take the sigma star delta T and add it to that. Or alternatively we can write this in the following way, I can find that C equals, can C, again C could be positive or negative, so let's just take the positive root, okay? And I'm gonna factor out sigma star delta T, and I'll get one plus the square root of one plus mu squared minus one half sigma star squared all over sigma star, and this would be squared, and that would be delta T, and there would be a big square root on this thing. Okay, so I'm sorry I'm going through this slightly annoying algebra just because I need to show you what the correct order of magnitude things are at. So these are our two equations, they're not our final ones yet, this one and this one, we've now found P and we've found C, and in principle that's what they have to be equal to, right, they have to be, so you take this expression for C and you plug it into that expression for P, but before doing that, let's just think a little bit ahead. What we would like is we wanna look at, we don't wanna have these time steps being very large, we want them to be relatively small. So if we want them to be relatively small, we should really be only concerned with the largest contribution to P and to C. So what terms contribute the most? Well, delta T is small, so square root of one plus something times delta T, this is gonna be a small contribution and there's already a delta T out here. So we can in fact do an approximation. When delta T is small, we could say that this is just approximately sigma star delta T plus things that are gonna go to zero faster than delta T, so this is little o delta T. So everyone know the notation little o delta T? So let me remind you. This means that the limit of whatever is there divided by delta T equals zero, okay? So just means something that goes to zero faster than delta T, such as delta T squared, okay? Or delta T to any power bigger than one, okay? But there are a lot of other things that can go to zero faster, but in this case, sufficient to know that. So if I have that approximation for C, then I also know that there's a nice, simple approximation for P. Look, P is way up at the top of the screen there, so if I substitute this expression I have for C into P, you get a nice, simple expression.  Square root delta T. And again, it's gonna be in this case plus little o square root delta T, something that goes to zero faster than square root of delta T. So these are our two final results. So let me write that. This is our key new result for today. So let's recap, okay? If you start with the observation of the data and you compute at the mean of those returns, the log return, to make our binomial model match that behavior, all we need to do is choose this probability and choose the asset to grow at every node like this. A goes to E to the sigma square root delta T. Sigma star, square root delta T times A. The minus sigma star, square root delta T times A. That's the result. Then this replicates the behavior that we observe in the market. Questions? Okay, I think it's about time for a break. So let's take a little 10 minute break. Okay? Okay, so I just like to point out a small typo that I made here. This should be square root delta T in these three places. It's sort of obvious you're taking a square root of something and I missed the square root, okay? So square root, square root, square root. I drew it correctly in the diagram. And what that tells you is there should also be a square root there and here. There we go. So in all of these things here, in fact, they're square roots. So it's a little o square root delta T. Stuff that goes to zero faster than square root delta T. Little o square delta T divided by square root delta T is zero. Square root delta T, little o square root delta T. You can fill that in on your own. Okay, so what I'd like to do in this last part, before giving you your little quiz, is to talk about, now that we have a discrete model that matches the behavior of the returns over the daily return, we can ask a couple of interesting questions. One is, what is the continuous time limit of this model? So what happens when delta T goes to zero? What type of distribution does the asset A have? Then the next question is, what is the behavior of asset A when we change to risk-neutral probability? Because at this point, what we've done is we've figured out the real world evolution and we've made the asset behave as it does in the real world. But we know when we wanna do derivative valuation, we have to go ahead and calculate risk-neutral probability and maybe the behavior under risk-neutral measure is quite different than real world. Okay, so those two questions will answer and then we'll end. Okay, so first of all, the continuous time limit. Well, imagine, so the idea is that we've got some big time frame here, zero capital T and we break this up into pieces and of course, each one of these is delta T and delta T is big T over N and we're gonna keep big T finite and we're gonna take N going to plus infinity. So in other words, delta T going down to zero. So we'd like to make this mesh become extremely fine and in terms of the one-period steps, we can see there's an interesting simple behavior or at least an interesting simple interpretation of the behavior. At every step, we go up or down by a small amount, by a small percentage amount, that's the sigma star squared delta T and we can see as delta T gets smaller and smaller, they become closer and closer together. The sizes of the steps get closer and closer but the number of them get larger. So it's not obvious whether or not this actually converges to anything reasonable at the start, it's not obvious. Something's getting small, something's getting big, maybe they cancel in the right way, maybe they don't. So we'd like to answer the question, what is the asset price distribution at capital T? And to answer that, we can simply just use this recursion relation that we have and we know that this becomes sigma star square root delta T we'll get x one plus x two plus dot dot dot all the way up to x n, right? Every time we step, we go up or down and the up or down is decided upon by x by this little Bernoulli random variable x. So what we'll do is focus on what's up in the exponential. Call that random variable, let's call that random variable capital X. And in fact, why not capital X subscript T, big T? And then we can ask questions about that random variable. Well, it's a sum of independent random variables. In fact, those are just Bernoulli's. So there are some of independent random variables that have finite variance. So what do you know about such type of random variables? In distribution, this becomes normal with some mean and some variance, right? We don't know what, what mean and what variance, we have to figure it out. So the arrow with the D above it means in distribution, limit in distribution, and this is by CLT. So all we have to do is compute the mean and the variance of this for any finite capital N and then take the limit as n goes to infinity and see, do we get a finite mean? Do we get a finite variance? If we do, then things have converged and we know what it would converge to. So the expectation under P of X capital T, well, that's pretty simple. It's sigma star delta T times N, there's equal number of them, times the expected value of any one of these, right? What's the expected value of any one? 2P minus one, remember? Do you agree? What was 2P minus one equal to? Well, if we just flip back, we can see, okay, well, here's P, okay? So if you multiply it by two and subtract the one, you just get this ratio, okay?  So we're gonna take delta T going to zero. So we can ignore the plus little o delta T in the end. We don't need it. But if you want for completeness, you can put that there. And now we can see delta T goes down to zero, same thing as N going to infinity. What happens? Well, let's look at all of the factors that we've got in terms of big N. This here is capital T over square root over N. This here is capital T over N. Multiply those two together. I'll get T over N, but I also have a factor of N there. So I'll just get T times this constant plus something that goes to zero. So in the limit, I will simply get this constant times T and then there's a sigma star out front so the sigma star is canceled. And I'm going a little bit fast here, but this is sort of simple, it's algebra. Check it on your own if you get lost. But at the end of the day, it's just computing these expectations and doing a little bit of algebra. So that's one, we've got that. And the variance of this random variable, well again, we've got a sum of, I'm just scrolling up to show you what X is again. So sum of independent things that are identically distributed and that we've got a sigma star square delta T there. So if we take the variance, we'll get sigma star squared delta T times N times the variance of any one. And we already know that we fixed the variance to be sigma star squared. Actually, sorry, we fixed the total variance to be that. So the variance of this guy is one minus two P minus one all squared, okay? And I'll remind you again what's this two P minus one all squared is going to be this. It will be that. And here we see something that's going to zero being added or subtracted from one. This is not going to zero. This is big T over N, so the Ns will cancel and this just gives me big T. So in the limit, as N goes to infinity, this is very simple again. For delta T going down to zero, this becomes sigma star squared times big T. So what we found is that the limiting distribution is normal with mean mu star minus the half sigma star squared times T. It scales linearly with time, which is nice. And the variance scales linearly with time, which is also nice. And we can write the asset price equals its initial asset price times e to the xt. So such kind of random variables, which are written as the exponential of a normal, these are called log normal random variables. Oh, and I should be careful here. This is the limiting distribution under the measure P. Equals in distribution equals dot and let's say, trying to think where I wanna put the P in here. I'll just put it there, sort of at the end of the arrow. This is a log normal distribution, log normal distributed random variable. You could write this in terms of a standard normal. Sometimes people write it like this, where z is just a standard normal random variable. So all I've done is I've introduced something which is in distribution the same as x capital T. So this is its mean, and if you compute the variance of this last term, since this is standard normal, the variance of this is the same as the variance of x. So they're equal in mean, equal in variance, they're normal distributed, so they're the same in distribution. So you sometimes see it written that way as well. Okay, now we can actually understand why I put that minus a half sigma squared in the definition of the return. Remember back, way back here, I did this sort of odd thing. I said, we have the observation, we compute the mean of those observations and we look at the, and we get an estimate of that. And that estimate, I said, suppose we write it as mu star minus a half sigma star squared delta T and we were saying, why sigma star, why minus a half sigma star squared? And I said, for now, take it at fate. Later on, it'll mean something. It's called a convexity adjustment, convexity correction. Well, let's see, what does it mean? Why did that come in useful? Well, suppose I asked you to calculate the expected value of the asset a capital T. What is it equal to? Well, it's a zero as the expected value of e to the x t. And if you remember your moment generating function for standard normals, so first of all, just write this in terms of the z random variable here. Reminder, MGF of normal. You take the expected value of e to the uz, that's e to the one half u squared, if z is normal zero one. So I can use that result and you can see, well, the u there is sigma star squared t. If I square it, I'm gonna get a half sigma star squared times t and that kills that half. That kills that term. So you get just mu star t. And it only killed that term because that term was present. So if it was absent, I would find that the mean is not the initial value grown at the rate of mu star. So there's the interpretation. Mu star, when we write it in the way that I did, in terms of calibration, mu star becomes the expected continuously compounded return of the asset. Sometimes simply called the drift or the expected return. So this gives you the interpretation why I subtracted the half sigma squared in the beginning. Question? Thinking it's getting a little bit late. I don't know if I wanna do the other thing. Okay, let's start it and see where I get to in seven minutes. Okay, so the last point that I wanted to do was look at the risk-neutral measure, q. So how would I calculate the risk-neutral probability here? We need the money market account. Since we're in continuous time setting, it makes sense to think of the money market account growing in a continuous way. So over some small time step like this. The condition of risk neutrality to calculate the risk-neutral measure, q would be that A equals e to the minus r delta t, that's the discount factor, times the expected value of A one time step later, q times A e to the sigma star square root delta t plus one minus q times A e to the negative sigma star square root delta t. And we see one very nice thing, one very nice fact is that this A canceled everywhere and what does that tell me? That tells me that this risk-neutral probability doesn't matter where in the gigantic tree I am. No matter where in the tree you are, it's the same q. q is independent, so there's some giant tree and you're often some node there, q is always the same. Probability of going up, always the same. And in fact, that was an outcome of the previous calculation. The probability of p going up was always the same. It didn't mean that there wasn't obvious that because p is always the same, doesn't imply q is always the same, it doesn't have to be. But in this case it turns out to be. So here we'll get q equals, if you just solve the system, you'll get e to the minus r, e to the r, sorry, negative sigma, sigma star square root and negative star square root. So you get that expression. So this is another very useful result. It's not so obvious any interpretation of that result is, but if you do the following thing, look at small delta t. What does q look like for small delta t? So in the numerator we get one plus r delta t, all right, exponential e to the x, so here's in one of these Taylor expansion things again. It's approximately one plus x. Okay, so we get that plus a half x squared, okay, we're gonna need that term, plus dot, dot, dot, minus one minus sigma star square root delta t, plus one half sigma star square delta t, plus dot, dot. In the denominator you get one plus sigma star square root delta t, plus one half sigma star squared, there's no square root there because I squared it, minus the exact same thing, except there's a minus sign there. Eventually I'm gonna stop writing this star on the sigma. It's purely at this stage, it's a reminder of the fact that it comes from calibrating. Comes from calibrating the historical data, it's the historical estimates. Okay, and if you look at these expressions, there's a series of cancellations. Let me highlight them. This cancels with that. There's a cancellation of this term with this term, this one, and this one also cancel. So everything that I've highlighted goes away.  And in the denominator the only thing that's left is two sigma star square root delta t, plus higher order stuff, things that go away. And now we just simplify that, that's one half one plus, that's q. Now remember, what was p? There's kind of a miracle happening here. Okay, that's what p was. The only difference between q and p, for at least small delta t, is what I just highlighted. Instead of having mu, you get r. The return of the asset, or that should be mu star I guess, instead of mu star, you get r. So the risk-neutral probabilities are actually exactly the same as the real world ones, except you have risk-free rate in place of the drift of the asset, instead of the expected return of the asset. And that makes a lot of sense. So if you think quickly about it, if you think, suppose I repeated the calculation we did before for the asset's distribution, what would A capital, we know that we would be able to write A capital T to be equal to this, and XT under the q measure should be what? Normal? What? It's mean would be this, and its variance would be the same. That's the surprising fact. So you can check that calculation for you. Go ahead and check that this is true for under the measure q. So there's two surprising facts. The variance under the p measure and the q measure are identical. Oh, that's not delta T, that's full T. That's full T. The variance under the p measure and the q measure are identical, but the drifts are modified. The return of the asset under p is mu, and the return of the asset under q is r. As expected, we kind of expected that. That's the definition of risk neutral. Expected returns are equal to the risk-free rate. So you can try to flush in this detail for yourself, this last part. It's just some more algebra, more or less the same as what we just went through for p. Okay, so let's, I'm gonna pause the,


All right, if we're going to make any progress describing the physical and chemical properties of solutions, mixtures of two different or more different compounds in the liquid phase, we're going to need to do the type of thing that we often do in physical chemistry, which is to define a simple model that describes the system as simply as we can. And kind of like we've done for the ideal gas, when we talked about gases, we describe the system as a gas with as simple a model as we can and see what the repercussions of that model are, even though it's not an absolutely correct version of describing a real world physical gas. It gives us a starting place in some things we can correct when talking about real gases. So we're going to do the same thing for solutions and define what we call the ideal solution model. And again, it'll be the simplest possible model we can construct of a solution. It may not be right. It may not perfectly describe real solutions, but it gives us a starting point. So we'll start with a sketch of what a solution looks like. So I've got a solution, and we're going to zoom down to the microscopic level and look at individual, at least a cartoon of individual molecules. Let's say we've got some two different species, mole fraction of A and mole fraction of B in this solution. And I'll draw my A molecules as one type of circle and my B molecules as a different type of circle. So I'll just draw a whole bunch of molecules in this box and some of them will be A's and some of them will be B's. So that goes on in all directions. So I've drawn those in a grid like we would need to if we're going to study this system with the lattice model of the type that we got used to when we first started talking about the probabilistic character or the entropy and the multiplicity, for example, of systems at the microscopic level. So if I think about a lattice model for a solution, I just put one molecule in each one of these boxes. None of the boxes are left empty. So the question becomes, what do we assume about this simple model to make it a good model for a solution? We can't do what we did for the ideal gas. We can't start by saying the molecules don't interact with each other at all. That was our assumption for the ideal gas. In fact, for a solution, they have to interact with each other. They have to interact strongly enough that they've condensed out of the gas phase. When we talk about Van der Waals gas model, for example, having that interaction between molecules is key to making the molecules condense and form the liquid phase. So we have to have some interaction between A and B molecules or between A molecules and themselves. But what we can assume at the macroscopic thermodynamic level, I can put the molecules down in the arrangement that I've just drawn them. That's, of course, only one of many different configurations I could have drawn. What I'm going to assume is that whichever configuration I draw, it's going to have the same energy. So if I rearrange this and make this molecule an A molecule instead of a B and put a B molecule there instead, if I rearrange the molecules in some way, then the energy is no different as long as I haven't changed the mole fractions of the solution. So every different rearrangement doesn't lead to a change in the internal energy or equivalently the enthalpy for that system. That would be a macroscopic way of understanding the system. If I want to think about it more at the microscopic level, the individual atomistic or molecular level, an equivalent statement to that requirement would be to say the total energy of the system, the total potential energy of the system is identical for every microstate. Every way I have of placing those molecules down on that lattice model, they all give me the same energy. In order for that to be true, so that's the assumption I'm making. One thing I could require in order to guarantee that this requirement is met, an equivalent statement would be what if I have the potential energy of interaction between an A molecule and an A molecule, this AA interaction? What if I require that that is exactly the same as a BB interaction and is exactly the same as an AB interaction? If a molecule doesn't care who its neighbors are, it has the same potential of the energy of interaction if it's surrounded by A molecules or by B molecules, then if I reshuffle those molecules, if I replace its neighbors by a different type of neighbors, then that's not going to end up changing its energy. This is certainly a good enough, a sufficient condition to guarantee that the energy doesn't change when I rearrange the molecules. It turns out that's actually a little bit stricter than I need. I won't go into the details too much, but it's actually good enough to require the heterogeneous interaction, the interaction between an A molecule and a B molecule. As long as that's equal to the average of the two different homogeneous interactions, as long as that's equal to the average of the AA and the BB interactions, that's also a good enough condition to guarantee that the energy won't change. Essentially the reason behind that is if I pick up a molecule and I exchange it with another one, I'm going to gain as many of one type of interactions as I lose of others. It turns out that this is also just as good a requirement. Whatever one of these we think about, the purpose of making this assumption at the microscopic level is to guarantee that the energy doesn't change for each one of my microstates. All those microstates are degenerate, the same energy as one another. Whichever one of these you want to think about, these are all equivalent assumptions for an ideal solution. What we would normally do at this point, or at least what we've done in the past, once we've defined a model, I've defined some energy of interaction for the system. We could write down a partition function for the system and use some thermodynamic connection formulas to derive entropies and energies, internal energies, for example, free energies and so on. And we'd be on the path toward calculating the thermodynamic properties of an ideal solution that obeys these assumptions. We've actually already done some of that already. We have written in particular for a lattice model involving fluids. We've already done the work of driving what is the entropy of mixing when I take a pure A liquid and a pure B liquid and I mix them together in a lattice model. We've seen already that that is equal to, for this two-component solution, so negative Boltzmann's constant times mole fraction, log mole fraction, summed up over each of the two different components in that solution. So like I said, we've already made a little bit of progress up toward the thermodynamic level and it's much easier to use thermodynamic relations to connect the different properties of a solution than it is to use the microscopic statistical mechanical equations. So we're going to start with this entropy of mixing at the thermodynamic level to describe this ideal solution and we'll see what we can learn about the thermodynamics of solutions from there and that's what we'll do next.