 Okay, so before starting off today's lecture, I'd like to review what we've covered over last time. And last time we ended up, one of the things we did was derive the Bloch-Scholes equation for the price of a call option. And this involves recalling something that we did well before, and that was that under risk-neutral valuation, in order to avoid arbitrage, option prices are always represented as the discounted expectation under the risk-neutral measure of their payoff. And under that measure, asset prices that are traded, and any asset price that's traded, have to grow at the risk-free rate. And in particular, if we looked at the CRR model and took the continuous time limit of that, we found that the asset price actually looks like a log-normal distribution, at least at a fixed point in time it's log-normally distributed when we view it as a random variable. And then on when we view it as a stochastic process, we'll see that we will still have this equality in distribution, but it won't be true path-wise, and I'll explain what that means later on. What we ended up doing was computing this expectation explicitly just by writing it down in terms of the density of the standard normal that shows up, and carrying out that integral, took a few lines of work, but in the end we were able to derive this nice compact formula, which I will never actually ask you to memorize. You will, if a question about block shows, ever shows up on the test or an exam, it'll always be derived some result like this. It will never be memorized this result, okay? Then once we had that nice formula, and there's an analogous one for the put option, we discussed various aspects about the behavior of the price as various parameters in the model changes. And here were a few of them, so we looked at what happens as maturity increases or as you approach maturity. The option prices become equivalent to their payoffs. We looked at what happens when interest rates increase, and when interest rates increase, we saw that for call options the prices increase, but for put options the prices decreased. We saw that what happens with the volatility, as you increase the volatility, the price of an option, both for puts and calls, increased. And then we went on and discussed a couple of other sort of random sets of payoff functions, and we looked at a few examples, this was a straddle and strangle, and these are built out of the basic calls, and we could also build them out of puts as well, actually, calls and puts in this case. And we were asking about the price behavior of those, and I suggested that you use an Excel spreadsheet, which I'm not sure if I uploaded it. It should have been there from last year, but I don't remember if I did upload it. This portfolio spreadsheet that I mentioned at one point, and I showed you on the screen, that would allow you to explore various aspects of the price behavior. So I urge you to do that with a collection of other random, whatever random collection of payoffs that you would like. So this is in this one second here. Oh, didn't come back on. There we go. That's this sheet here that I mentioned that I'm talking about. And I will double check to see whether I uploaded it or not, but at the end of the day, you'll be able to play with the sheet and see what happens as you change things like volatility and interest rates and all of that, and you'll get a sense of what the payoffs do. So I urge you to use this sheet and get some intuition. Then the next thing that we discussed, and this was the last one, was American-style contingent claims, in particular American options. And one of the main principles behind American options is that they allow you, the holder, to either exercise the option at any point, not just at maturity, unlike the European claim. Europeans, you have to wait for maturity. American, you can exercise at any time. And the key feature there, when you're doing the analysis for the option evaluation, as you go back through the tree, is to simply look at the holding value, which is the analog of holding a European for one small time step, or the immediate exercise. So you have to compare these two values, and whichever is better, that's your right to choose. So you would, of course, choose whichever is maximum of these two. And that is what gives you the option price at one step in the tree, or at one node in the tree. And if you iterate this process throughout the tree, I mentioned that there's always going to be sort of two regions, well, for standard options. There's going to be two separate regions. For non-standard options, you may find sort of closed off and disconnected regions. But for the ones you'll be looking at, there are two separate regions, one in which you will always hold the option, and the other in which you would have always exercised the option. And the interface of where these two regions meet, this is called the optimal exercise curve, and it is always optimal to exercise once you touch that curve. And I showed you through some MATLAB experiments that this exercise curve for a put option, the diagram on the left there, tends to look like this. It hits K, and then it decays, and it's sort of nice and convex, and decreasing. And there are these two separate regions. And as I said, if you look at a path of the asset, the instant that it touches this, that's when you should exercise. You should not wait any longer. Even though if you do exercise at a lower value, it looks like this is more valuable than that. Exercising at this point seems to be more valuable. Exercising, if I compare those two points, let's call that one A and that one B, exercising at A seems to give you less value than exercising at B. Do you agree? Because it's a put, so if I'm lower, if my asset price is lower, I get more money. Imagine that this asset price corresponding to here, suppose that was $50, and this is $60, and the strike is $100. If I exercise at A, I'm only going to get $40. And if I exercise at B, I would get $50. So it seems the immediate exercise value is, in fact, more valuable to wait. It appears to be, but why is that wrong? OK, there is some elapsed time, but that amount of time, even if it was, can you imagine of an investment that's going to give you $10 in two months on $60? Probably typically no, right? Certainly not risk-free. Yeah? Yeah, I've drawn a single sample path. From here, the asset could have done that as well. There's no guarantee that once you touch the barrier, you're going to go deeper into the money. And the whole point of doing this backward induction where you solve for the holding value and look at the maximum between the holding and the exercise, the whole point of that is it takes into account the fact that there is a potential of the asset going up and going down, and what is the optimal thing to do. So you've already accounted for all of those optionalities that are there. And it's told you, and after doing all of that, the methodology is telling you it is, in fact, optimal to exercise at A in terms of value. Now, in terms of any one scenario, it may happen that a scenario gives you more money. That's true. Any one scenario, it may not be optimal to exercise along the optimal exercise boundary. But that's under a single scenario, while option pricing is protecting you against all possible scenarios. Because when you think about even without American options, you're thinking about matching the valuation of the option both in the upstate and in the downstate, not only one. You're not just looking for what happens in one situation. You're looking for what's happening in both. And here, when you do this backwards recursion throughout the entire regime, throughout the entire time frame, you're solving the problem looking basically at all possible paths of the asset. Yeah. Sorry, it gives you a big dividend. Oh, if there are dividends. Yeah, so I was gonna talk about dividends today a little bit. Well, it turns out that whenever there is a dividend payment, there's like a little kink. There's a kink in the surface, in the optimal exercise curve. And if you think about the call option problem, in fact, your quiz question last week, which I'm sorry to say about it, unfortunately, it was not done overall very well. And you see the solution. I gave it to you at the end of the class so you can take a look at it again. And if there are any questions you have about it, let me know. Why did I bring that up? Because when there is no dividends, yeah, we saw that it's not optimal to exercise if you have the options exercise between choice between now and the fixed maturity. Okay, that's not an American option. That's what's called a Bermudan option because you only have a fixed number of points. It's not optimal for a call, but if on that day, today, there is a dividend payment, then it may actually be optimal to exercise. And the sole reason is simply that the asset's price has to drop by the fixed dividend amount on the dividend day. And so this convexity argument fails right at that point. Okay? And yeah, now I'll talk a little bit more about that later on. Okay, so we went ahead and did a little bit of experiments. I showed you how this works in the MATLAB example. And this here was your little quiz question demonstrating that it's not optimal to exercise if you have the right to exercise now or at maturity, okay, just using Jensen's inequality. And that's where we ended up. Okay, so before continuing on, are there any conceptual issues that I can clarify from last lecture? Okay, so what I wanted to do today were a couple of things and I'm not sure how far we'll get. It depends on how fast we get through this and whether there are questions. I want to talk about trinomial trees in a little more detail and some exotic options. And if there's time, start discussing interest rate models. Okay? So these are the three things that I'd like to discuss. So why don't we start with the trinomial tree, first of all? So far in all of the dynamics that we've been investigating, at least taking a continuous time limit of it, is this simple two-asset case, right? You have an asset starts off at a zero, goes to AU or to AD. And in the CRR model, we took the particular situation that the asset or the particular model where the asset grew by factor of sigma square root delta t and minus sigma square delta t. And we just recurse this through the entire tree. And this allowed us to get a recombining tree, which we then used for valuation of whatever kind of options were out there. And the limiting distribution of this is what? What happens when we take, if we repeat this and we take many, many time steps and we're under the P measure? So it's a lot of normal, right? And the drift, the exponent that shows up here is mu minus a half sigma squared when we calibrate two in the way that we normally do it. And here we would have sigma square root t, Z. And Z is normal zero one. And what about in terms of the risk-neutral probabilities? What is the distributional property? So it's also log normal, but the drift instead of mu, you have R. So I should put a little subscript here, P and under here, Q. So we have these two results. Now, what I wanna do is discuss, is actually, first of all, let's think about taking two steps of this binomial tree. And I think we, did we discuss this before? The fact that if I take two steps of this tree, this effectively, if I only look at the end point, if I only look at the end point, so I hand up with those outcomes. And so I imagine just those yellow branches, and I forget about the middle branch, I forget about the t equals one step, the one step situation, I ignore it entirely. That actually gives me a trinomial tree, does it not? And if instead of taking step size of delta t, I took step size of delta t over two, then I would have this situation. And I can imagine taking single steps in a giant tree, where I go up and down by this factor, square root, so that the divide by two is underneath the square root here, and there's a minus sign missing there. So this is one way to sort of motivate the idea of a trinomial tree, is it's really just multiple steps of a binomial tree. And what might be the advantage of using a trinomial tree? If I took a binomial tree and I took step sizes of delta t, what's the step size here? This is also delta t, right? Because I took two steps of delta t over two, so the total step size is delta t, versus just, versus binomial model, which would have that. Clearly I'm introducing one extra state, right? That's one thing that I'm doing. I have one extra state in the middle, so I'm able to, I can get to that central limit theorem more quickly, right, to the limit and the central limit theorem. What do you do? You have a sum of independent random variables and this will eventually become normal. And if I include one more state, I'm able to get to that limit more quickly. As well, I'm able to capture the dynamics of the asset a little bit more realistically. It's still not that much more realistic. I've added only one more node. Rather than two, I now have a third one in the middle. But still, it actually improves things a lot. Do you remember an issue that showed up in the American option valuation? Let me pull it up for you. In the American option valuation, when we implemented the binomial tree and we recursively solved backwards and found the optimal exercise curves, there was a feature that showed up that was definitely undesirable. Okay, let me pull this around. What does it give me? Okay, time. Okay, this is the result of that optimal exercise curve from that American option. And what do you see? Hopefully you all see this kind of jagged motion here, this up and down ticking. So it seems a little bit weird that you get that, that you exercise here and then the next exercise point is below and what's the reason for that? Well, if you overlaid the actual underlying tree there, there's a little binomial tree in here which one step goes down this way and then the next step is up. So this up and down tick is actually just due to the discretization of the tree like this. And if I take two steps again, let me just draw that or maybe three steps. So we saw from that diagram there, there was an optimal exercise point there, there and then there, right? This gives us our up-down problem, our up-down, up-down, up-down, up-down. And that's what we're finding, that's what we would see here. That's just the next two nearest nodes. So if the exercise was somewhere in between, we would run into this problem. That every other node, it goes, one time it goes up and the next time it goes down. One way to avoid that is to use this trinomial tree. So that's the motivation, one of the motivations for it. So there's a few of them, right? One is you get to the central limit there, limit faster, you approach a normal more quickly. You converge to it faster. Second, you're expanding the state space a bit more. Instead of just having two outcomes every little step, you now have three outcomes. And third, you're going to avoid some of these kind of numerical oddities, such as this up-and-down ticking that's showing up in the American option. Okay, so there was a question. No matter how much you blow it up, you'll always see that little up-and-down tick. Yeah, but look at how many steps. I'm already taking a hundred, right? This is a hundred steps. Let me increase it for you. Let's go to a thousand. And yeah, it will clearly get reduced. It's still there, okay? And to really get it, to really remove it, I'll probably have to go to, I don't know how long 5,000 is going to take, probably 20 seconds. Let's let it run in the background and come back, okay? So you can kind of get rid of it by taking more and more steps. But that's another point, is that this issue of taking multiple steps of the binomial model where it can kind of capture that by taking a single or half the number of steps, and in fact, usually much less than that, of the trinomial tree. Because you're spanning the state space more densely. No, no, no, I'm saying, okay, if I take, if I have a model where I take one step and in one step I do a binomial tree, versus another model, I take one step, same amount of time span, but I have three outcomes, okay? Because I want to take the same number of steps, but I want my model at every step to be more accurate. And this is why we're able to achieve the limit better, more quickly. Okay, so is it done? Yeah, it's done. And now you kind of get something that, you know, you can still see the, this is not a thin line, right? It's still ticking up and down. And in fact, it's kind of, yeah, that's our, that's the visuals that we get. So I'll show you what happens when we implement this trinomial tree. But before doing that, let's actually work out. What should the probabilities be? The risk neutral one. How would we figure it out? So we know, we already know from the general theory, that if I start with a model that has three outcomes, and I'm going to replace the factor of two divided by square root two, I'm just going to call it an arbitrary constant lambda. And we will let lambda be a parameter. Okay, if we start with a model like this, and we have our interest rates, all right, or our money market account, this is growing at the risk-free rate, okay? If we start with this model, and I try to, and I ask you, okay, what is the risk-neutral measure? Is the model arbitrage free? Oh, there's a minus sign missing again. I'm constantly forgetting my minus signs today. So there's my question to you. Does there exist a risk-neutral measure? Yeah, exactly. So, answer would be yes, as long as lambda and sigma are non-zero, okay? Then we know that we have, in one state of the world, we're doing better than the money market account. In the other state of the world, we're doing worse. So there's no way to make an arbitrage here, just with these two assets. But it isn't going to be unique. Why? There are two traded assets, and there are three states. I told you this little folklore theorem a while ago. If you have more states than you have traded assets, then the risk-neutral measure is not going to be unique. Okay, so that's a problem here. Seems to be a little bit of a problem. So we can resolve that, though, quite simply, just by saying, okay, instead of imagining deriving the risk-neutral measure based on this model, let's do the following. Let's say, I know that in the continuous time limit, I have to have the log-normal distribution with that appropriate mean, and that appropriate variance. We know that I would like to have this property here under the measure Q. I want that one. So why don't we force that over the small time step? Because it has to be true also over any time step. Capital T here is any time, so in particular, it could also be delta T. So force this condition over delta T, at least, and whatever we mean by force it. What we're gonna do is we're gonna force two moments to be accurate. Let's force the first and second moment under the Q measure to match this property. Rather than deriving the Q measure and then having to deal with the non-uniqueness, this is a way of actually picking a particular Q, picking a particular risk-neutral measure. We know that Q, so let me write the statement down here, Q itself is not unique, but we can pick one. And later on, we'll see that this particular measure comes about, it's basically a measure that's induced by discretization of a partial differential equation which the pricing function has to satisfy. And later on, we'll see where that connection is made. But for now, you can simply view it as, you have a collection of risk-neutral measures. We can certainly enumerate what they are. They're parameterized by one parameter family. So we're gonna pin ourselves down in that one parameter family. And we're picking it such that the expectation under Q of the asset price over the time step delta T, over, let's call it any little T plus delta T. This has to be equal to its current asset price grown at the risk-free rate. We know that we have to enforce that anyway. We've already enforced some kind of symmetry here. We have forced the fact that the up tick and the down tick are in log space equally separated. One is going up by lambda sigma squared delta T, the other is going down by lambda sigma squared delta T. So there's some kind of symmetry there and that's gonna help us reduce our parameter set or our variables. And the next thing we'll do is match the variance, but just like what we did when we were matching the P properties, we're gonna match the variance of the log ratio rather than the variance itself. It's also possible to do that. So you could try to retry this exercise where the second condition that I'm writing down here is a condition on the variance, not of the log of the ratio of A's, but just of A itself. You can try that. And what should this be equal to? Sigma squared delta T, right? If we just look at this expression here, I take the log, I have sigma square root T, the T is delta T here now times the standard normal so the variance is just sigma squared delta T. Okay, so if I force this, I will in the continuous time limit have the same distribution. So now we've got two equations and they're only two unknown. We have three probabilities, but they have to sum to one, that's our third equation. So this will uniquely provide us with the risk-neutral probabilities. So let's solve that, right? It looks like I did, but I don't know exactly where it is. So let me leave this as an exercise for you rather than wasting time. And I'll post up the correct version of it. Okay, that's your exercise, fix this. Because it's really algebra at that point. There's no finance in here at all. Okay, so let me go to a spreadsheet, not a spreadsheet, but some MATLAB code that actually basically implements this idea and show you what the difference is. Okay, so here is, actually, let me step through this code for you with a bit. Here's the contract parameters. These are the model parameters as before. These are the number of steps that you're taking. Okay, it's only a hundred in this case. I'm actually just setting lambda to one in this case. And, aha, yeah, that's why. That simplifies a lot when lambda's one. Right, that's the step I forgot to do. I was supposed to do that for you guys in class. Just choose lambda equals one, so that it goes through simply. But you can do it with any lambda. And those are the probabilities that you end up finding. This is gonna store the value in the tree. And, in fact, I do need to say a little bit more, but I think I can, about what does V top and V bottom are. Here is the value of the asset along the tree. And here I'm just doing the risk neutral expectation as before. Okay, you have the Q up, Q middle, Q down, times the various nodes in my tree. And this is doing an American option valuation. So you check to see whether it's optimal to exercise or not, and record the boundary. And what's plotted here, this is our exercise boundary. Remember what it looked like before. Let's go back to this American tree. Actually, I don't need to do that. I can just open up a new figure. There's a hundred steps. And, there we go. So that's the trinomial. Okay, hopefully that convinces you that they give you the same results. But, this is basically giving you a more stable boundary. Now, there was one little caveat in this code that I said that I wanted to tell you about. And I don't want to tell you about it just because it's something with respect to programming. There's actually some interesting finance in there. Suppose you wanted to plot. Actually, here's a plot. Suppose you wanted to make this kind of plot. On the horizontal axis is a spot price. On the vertical axis is the option price. And you were using a tree to do this. Okay? How would you calculate, how would you be able to plot this curve? If I wanted the spot price at, say, if I wanted the value at 60, what are you gonna do? You're gonna take your tree, you're gonna start it at 60. I think this is what probably 90% of you in the class would do. You'd start your tree off at 60. And even if it was a trinomial tree and you calculated the right probabilities, you then, you know, you build your tree. You go ahead, you solve back. And that would give you the value of the option where the option value equals to 60. Right? That's one point. So you would have been able to find that value there. Right? Now, what if you want all of those values though? All at once. Doesn't make sense to now go build another tree. That's starting at 65 and doing the same thing. That's the naive approach, right? And probably if I just assigned it as a quick question, you'd probably all do that. But there's a much more efficient way to do it, particularly when we have a trinomial tree. It doesn't work with a binomial tree so well. But with trinomial it works very well. And that is you extend the tree to look like this. So what you do is you build a lattice, actually. It's not a tree anymore. It's really building an entire lattice. And this extends out to however far out the maturity date is, okay? And you work backwards just as before, but there's a little bit of an issue. All of these points along here, which are the edge of this lattice. Do you understand how you would use this lattice to build your option price? So you know the values, you know the terminal values. Just as before, suppose it was a European option. You know those terminal values all along here. Discounted expectation, how do I get that value? Well, it's a discounted expectation of those three nodes. How do I get the next value? Discounted expectation of the next three nodes, et cetera. And then I'm able to get the value all along here. And then I repeat. I get the value all along here. And I repeat. I get the value all along there. I repeat, and so on. And eventually you'll get the value all along there. And once you have that, you now actually have the option price not just at one spot, current level of the spot, but for all current levels of the spot that are in your grid. But when you do this discounted expectation, there is one little caveat, right? This issue of the edge of the tree that I mentioned just now. These three nodes. How do I compute that value? Well, in maturity it's okay because you know the maturity value. You know that one and you know that one. So these guys are okay. Let me draw one more step. Then the problem becomes apparent. Okay, I'm not gonna fill in all the rest there. Let's just go to the edge. So at this point, we have all of those nodes highlighted in green. The vertical ones there, okay? At one step prior to maturity. We've got all of them. Got all of those. How do I compute this discounted expectation? Do I know the value here? You don't. We haven't actually been able to calculate it using discounted expectations because it doesn't lie in our, there is no node connecting to that node. And that's the edge of my tree. So yes, I could put one more layer there but then I'll run into the same problem again. So how do you solve that? Anyone wanna suggest a solution? How would you obtain the value at this edge of the tree? Or the edge of this lattice, I should say. No suggestion? All right, yeah. You wanna discount this value here? Well, it's not quite that because you do know that there has to be some connection to, it's not just gonna be the discounted value because if I use that rule of thumb, then I would have all the nodes in the interior also being kind of just discounted values. But they're not. They're discounted expectations, right? So what you're effectively doing is you're thinking there's only one connection between there and there and then nothing else, right? That's kind of what you're thinking. But that would lead to instability. It will not lead to a good solution. Any other suggestion? How do you think, yeah? Yeah? Hold on, you don't know the distribution of prices of the option. You know the distribution of the prices of the underlying asset, right? You know the asset's distribution but not the option. In fact, that's what you're trying to find. Well, you're trying to find its price and then you could use that to tell you something about distribution. Any other suggestion? What about the behavior of the option prices? Are they very badly behaved functions? Option prices? Don't they look kind of smooth generally? When we plot price versus spot, we usually have some nice smooth things, right? What's the asymptotic value for a call? Do you remember? Just minus K discounted, right? Okay, so in fact, this is quite smooth asymptotically. It's a straight line. Well, we don't know that some generic option will be a straight line, but what if I took a derivative of that? I know I get one. I know a generic option won't have a slope of one. If I took a second derivative, now I know that I would get something having a second derivative of zero. And if I leave it just generic, that the curvature vanishes. This would be valid actually for a call and for a put. Because a put is doing this, and the curvature would vanish out of zero. The call, as I approach zero, is also going to zero with a zero slope. The put approaches its price here with a straight line as well, K minus F, basically. Discounted K minus F. So in fact, these conditions at the edge, one way to put some reasonable conditions, it's not always true that this works, but one reasonable set of conditions is to put as a boundary, that the second derivative of the option price is zero. Same thing at the bottom, okay? These are sort of standard kinds of conditions that you see if you've ever taken courses in PDEs. You often put vanishing while they're either Neumann or Dirichlet conditions. In this case, we're putting conditions on the second derivative vanishing. And how would that be implemented in terms of a tree? How would I make the second derivative be zero at the edge of the tree? Well, since you have a tree, you only have an approximation for the second derivative. You don't actually have the second derivative. How would you approximate the second derivative along that line, central differencing? Have you seen that before? Okay, I can say that second derivative function is approximately the function evaluated at S plus delta S minus twice the function at S plus the function at S minus delta S all over delta S squared. Have you ever seen that? Who has seen that approximation before? I think, really, just a few of you. Okay, so how would you prove that result then? How would I prove that that's actually true? Since you haven't seen it before, now we have to convince you that it's correct. Okay, those who've seen it before, how do you show that's correct? What's the step? Come on, exchange students. What's the first thing you do? You want to expand these two guys for small delta S, okay? And use a Taylor expansion. So this is going to be F S plus delta S times F prime or times partial S evaluated at S plus one half delta S squared. Tudor is F S, right? Plus little o delta S all squared. So something that goes to zero faster than delta S squared. And then the second one, this term here, that would be F S minus delta S partial derivative plus one half delta S all squared. Second derivative evaluated at S, again, plus little o delta S squared. And I hope now you see how it works, right? You get a cancellation when you add those two terms together. These two terms will cancel one another. These two terms will cancel one another. Oh, sorry, these two terms will add up to two. Two times F S, but that will cancel this one. So I'll circle, let me highlight for you. So you can, those three terms cancel each other. Those two terms cancel each other. And then you're just left with the sum of the last two terms here, which is delta S squared times the second derivative. But you're dividing by delta S squared. So the delta S squared is gone and you end up with just the second derivative, okay? So this gives you the proof, quote unquote. Okay, so if I have that approximation of the second derivative, and of course delta S will be my, will be the little window size that I'm using here. What is this equation tell me then? What does that boundary condition tell me at that edge? So it tells me if that equals zero, the boundary condition is this equals zero. So it tells me that this difference, F at S plus delta S, which is the function value just outside my grid. Okay, let me write this down here. So if I have this equals zero, let's say at the upper edge, off edge. This implies that F S plus delta S has to be equal to twice F S minus F S minus delta S. Okay, and this is a point just outside the grid. And these are inside. So those are known. Okay, so on in terms of this diagram, that circle would be F S. That circle would be F S minus delta S. And the, maybe I should do this in a different color. I know it's starting to look messy. Let me draw a diagram below it instead. Okay, so we worked one step backwards. We have found all of those values by working one step backwards. We need to find this point. And that's being given by, that point is this outside. And these two points here are those two inside points. At S plus delta S, one point outside has to be twice the edge minus the point one step in. This gives you the smoothness of the price function. This allows them to have enough convexity. Well, zero convexity in fact, but enough slope. And if you wanted, you could enforce the third derivative as zero if you wanted. Or any other kind of boundary condition that was reasonable for the type of option that you are valuing. And that would require some other analysis to know what that boundary condition is. For these kinds of options, these American calls and puts and so on, and even European calls and puts, permutant calls and puts, the second derivative vanishing is a valid boundary condition. Okay? And you can imagine, of course, on the flip side, if you look at the down, the lower edge, what would the result be? You would have known those points, and you're trying to find that one. And the way that you would find it is by saying, you basically, you still have the same approximation here. Second derivative is still approximately that. But now what you're interested in is F S minus delta S. Right, that's the point just outside the grid. It's the lower point. And you use this approximation to give me that point there. And those two are used for this combination. So in this code, that's where this, where is it? Here we go. That's where this V top and V bottom come in. These are the values of the outside edges, and they're computed via this extrapolation. Okay? All right. Any questions about this trinomial tree idea here? Yeah? No, well, it's the financial reason is motivated by what I drew here. The fact that asymptotically for calls, you end up with the call approaching the straight line, S minus K. And that has zero curvature. Second derivative is zero. And for the put, asymptotically as you probe zero, it's K discounted minus S. And that, again, is asymptotically zero derivative. Second zero second derivative. And on the other two edges, the outside ones, you know that the curvature is going down to zero, right? The slope is going to zero. The curvature is going down to zero as you go out to infinity. And for the put as you go to zero. So that's the finance part that comes from few lectures ago, those limits. But now it comes into this to tell us that the second derivative should be zero. So these kind of lattices are much more efficient than solving the tree many, many times, right? It's kind of ridiculous to do that. You wouldn't build a tree every time you wanted a new spot price. Instead, you build a lattice and you get them all at once. Now, what I have here is a comparison of the lattice result to the black-should result. What's plotted here? Okay, this is for a put option, it's American put option, and then the black-shoulds equation for a put option in blue. So you notice that they don't quite line up, right? Why? And the answer is not because of discretized. So again, the sorry, the blue line is the American put option price and the X's are the black-shoulds prices here for a put, yeah, that's right. The blue is the price of the American option that has to be larger than the price of the black-shoulds put. Because the black-shoulds formula for a put is for Europeans. So it has to be above. If I repeat this analysis, and I offer the tree and I implement the European version of things, so let me just, this is the one line where I replace the value with the maximum of the exercise and the holding value. So if I just ignore that line, my code will produce the price for a regular put. If I run that, that exercise boundary means nothing, it's junk. So here's our comparison now from that trinomial tree and the black-shoulds formula. And you can see they lie right one on top of another. There's almost no error. I mean, yeah, if you probably zoom in a lot and you look at the difference, whoops. Uh-oh, I lost track of where the point is now. They do deviate very, very, very small deviation, okay? All the way down into fourth or fifth significant figures. Okay, great. Any other questions about this little trinomial implementation? I urge you to try it yourself, do it in Excel. If you don't know how to program in MATLAB or R or any other programming language, just do it in Excel. You can do the same thing in Excel. Okay, before we go for a break, I wanna do one more example of an exotic option because we haven't done any yet. Okay, who's heard of barrier options? Okay, so what's a barrier option? What are they? Give me an example of one. Do you have one in mind or no? No, okay, I'll say. So they come in two main flavors, knock-in and knock-out. Okay, there are two types. And within those types, you have many subtypes. Okay, so let's start off with the most simplest version. Knock-in European call, okay? In fact, let's say an up and in European call specifically. So I'll say knock-in European call. So what is that? This is an option and in particular, I'm thinking of an up and in. And I'll explain that terminology for you in a second. So European call, you understand what that is. We all know what a Euro call is. Knock-in, what that refers to is the fact that when you buy this option, when you buy this barrier option, you do not have the European option. Instead, the barrier option is a right, basically. In fact, it's not even a right. It is an obligation for you to exchange that option for a European option at the time in which the asset rises above a certain level. Okay, that's a mouthful. So we can draw a diagram and then I can write down an equation that represents what that payoff is. So there's a barrier here, it's an upper barrier, and that's what the up and in refers to because it's up, once you get up to that level, then the option gets knocked in. So the idea is that there's an upper level there, there's your standard maturity. This is asset price on the vertical axis here. This is time. If the asset starts off at a certain level and it moves around and it hits that level, what you receive right then, the instant that you hit it, hit that level, you get a European call. And you only get it if it does touch that level. So if there was another path where the asset did whatever it does and it ends up there, even if the strike, let's say the strike was set to that level, that's our K. So both of those paths, in terms of the European counterparts, they would have ended up, oh, actually I didn't finish the green path. And let's say the green path continues on and does whatever, both of those paths would have ended up in the money, right? I would have received a payment if I had the European option. But if I had the knock in option, I will only receive the value under the green path and not the red path. So in other words, the payoff for the European up and in call it's equal to the indicator that the event tau occurs prior, well, from the time at which you sign until maturity date of the call option price. So this is my payoff at capital T and tau is the first hitting time and usually it's greater than or equal to. But since we're dealing with models that diffuse, that is, they have continuous paths, it will hit it, it will actually attain the value, the level U. Okay, so tau here is the first time at which the asset hits the level U. So if tau happens to be between now and maturity, then you get this option payoff. Does that make sense? Now you can imagine what a down and in option would be. The barrier is just on the lower side. So I would have tau be the first time that F becomes less than or equal to L where L is a lower barrier. There can be different versions. There can be knock-in, a double knock-in option, such as an option which gets knocked in. So you can have two barriers here, L and U and then there's a maturity date again. And the option gets knocked in the instant that you hit either here or there. The first time that you enter in, and in general that's how the knock-ins work, is you define some sort of region and you say the first time you enter the region, you receive the other option. And in this case, the other option is the European one, the embedded option, the embedded European call. So the payoff is identical to this one. It's still the indicator of tau in zero T, ST minus K plus, but tau now represents the first time that ST is not in the interval L, U. The first time it exits that region. Now you can imagine that this leads to, it's a very interesting probability questions, right? The first exit time of asset prices and stochastic processes from regions. These first hitting time problems are notoriously difficult in general. But under specific modeling assumptions, they can become tractable. Very tractable, very easy to deal with. And what I would like to talk about when we come back from our break is how would you value an option like this based on the technology that you already have? Let's not do any, you don't need to do anything really sophisticated to actually value these kinds of options. You can value them based on the knowledge you have. And before that even, I want you to think about during the break, why would an investor want such an option in the first place? What's the purpose of them? Okay, so let's meet back in 10 minutes. All right, so now the task is we're gonna try to figure out how to value these kinds of contingent claims. Just knock in contingent claim. And let's deal with it in the very, very simplified setting of just a binomial tree, and then we can extrapolate to what you do in this trinomial lattice. So I'm gonna draw a few steps of the tree here and let's suppose that we're doing the valuation of the knock in option, okay? The up and in call. And the knock in level is there. Is that level? Okay, this is my U. So how do I go about valuing this? Well, the way to think about it is to imagine what happens once you're knocked in, once the option has been knocked in. Well, once it's knocked in, the option is simply a call. Is it not? It's a European call. And that means that if I look at this term, this value at this node here, the value at that node there has to be the same as the European call option. It has no other choice. It has to be equal to a European call option. So what I should do is I should draw a separate lattice over here, and in this case, corresponding to that node. So I'm drawing a lattice corresponding specifically to that branch of the tree, or drawing a tree corresponding to that branch in the tree. And we know what the option payoff would be. It's whatever C up, C down, okay? Doesn't matter, whatever it is. Yeah, and that's it. And then I calculate the C zero as I normally would just by calculating the standard European option price. And that C zero feeds into this node here. So that gives me one value in the tree, only one so far. And then I have to do the following change of, you have to change your thought for a second. And not, and think about an option, a barrier option which never breached the barrier as yet. Okay, so what do I mean by that? If we're looking at this tree and we're thinking about what to do at a particular time step, this last time step, for example. And I want to put option values in that terminal time step. What I have to do is I have to ask myself the question, what is the value of this barrier knock-in barrier option conditional on the barrier not having breached? So if the barrier had not breached, if I didn't get into this exercise, into this knock-in region, what's the value of the barrier option? It's zero. So that tells me I get zeros in these nodes. All I have to do is value the option below that barrier, right? I don't need to do anything above the barrier because I know that once I enter into that region, it's a European option. So I'm only really interested in anything below that barrier. So this is, and this is how you should view this tree, is it's the value of a barrier option conditional on the barrier not being breached? Conditional on, here's a way to write it simply. Conditional on tau is bigger than t. So whatever time step you're at, you assume that the barrier is only gonna breach further in time, not now. It has never done it in the past and it can only do it in the future. Seems a little bit strange. And why does that seem strange? Because it seems as if, well there's not enough paths here but let me scroll up to the diagram. It, take a look at this green path. There's certain points where it went up and then it came down below the barrier and then it goes back up again, right? Goes up, down, up, it can do any of those things and cross in and out of that barrier region. But when you're valuing the option, you should only look at the barrier breach forward in time. And the reason for that is once you do, once you introduce the barrier option that has only breach that looks for breach only forward in time, then that option does not depend on the past. Do you agree? If I look at, if I define a new option that's a knock in option and it can only be knocked in from now forward, then I don't care what happened in the past. And I use that option as my payoff structure to go one step back. And once I'm one step back, I view that now again as an option which is a barrier option and I only am concerned about what happens one step forward. I don't care about what happened in the past. And then I go back again. And again, I'm considered interested in a barrier option which is looking at breach only from that point forward. I don't care about what happened in the past. But once I do this process and I get to the start of the tree, I now actually have the price of a barrier option for a breach in the entire future. Do you see how you got, how that works? You don't, you forget about the past dependence in the past. You only look about the past dependence, the breaching of the barrier, toward the future. So you condition on tau never, tau having not occurred. Okay? Or tau being bigger than little t. So that's why these nodes get zeros. And then you go ahead and do discounted expectation. And maybe there are too few nodes in here to really get a good sense of what's happening. So let's draw in a few more. Okay? So you'd also have a zero there. Okay. So now if I do the discounted expectations, in this particular tree, I'm still gonna, I'm gonna get zeros there, zeros there, zeros there. But now all of a sudden I'm gonna get a value that's not zero at this point. Because I'm saying, okay, I don't care about what's happened in the past, but currently the barrier option is gonna be worth either zero or the value that I calculated already here, the C zero number. So if I focus in on that portion of my tree now, just that portion, I got C zero zero. And then I can calculate, I don't know, that'll be C zero times Q. Let's suppose interest rates are zero, just to make things really simple here. I got Q times C zero there. So this gives me a non-zero value at that point. And then that non-zero value is gonna feed into a non-zero value for this node. All right, once I calculate that portion of the tree. As well, what's the value at that node that I just circled here? How would I compute it? So we already figured out over here is Q C zero. Okay. Sorry, it looks a little bit messy there, I know. Whoops. What should be the value at this node here? Do I use Q C zero at all? The answer is you don't use Q C zero. That value below the barrier is irrelevant because once you touch the barrier, the barrier option portion ceases to exist. And the option changes into a European one. So in fact, what I should have is let me draw, run another colors here very quickly. Let me draw tree on the side here. Okay, so C zero was there. And the node that we're interested in is that node there. We're interested in this node now. That's the corresponding node into my European tree. So what you should have done is on the side, you do a side calculation for the European option. You have this tree stored. You do that computation first. And then whenever you breach the barrier, you put the corresponding node into this tree. And, oh, sorry. I circled the wrong node that corresponds, didn't I? Oh, okay, that's fine. It should be at the same level of C zero. It's that one. It's the same height. Do you realize that? It's at the exact same height. So we want this to be the same height as C zero. So that node gets fed into this tree. And then you work backwards. So the rule of thumb for the knock-in option, always look at the barriers. Look at where the barrier is. Calculate the price of a European option along that barrier. So that's the computation we do first. Then come back and fill in the remainder of the tree by putting in zero for the boundary condition and then recursively going back with discounted expectations. Okay? So why don't we do an example of this? Yeah, that's the maturity. Those are actually zeros. Because remember, the tree that we're drawing there is a tree not, it's not the tree for the actual barrier option price. Because if I got the maturity and I hit the barrier before, then I would get a different price. I wouldn't have a zero price. I'd have something else. But what's drawn on the tree is the value of the barrier, of a barrier option conditional on barrier breach occurring in the future. And since you're at maturity, there is no future. So you have to have zero. Okay? You can kind of think of it. There's another way to think of this. And it might look a little bit strange. But let me draw the tree up there kind of, I'm gonna try to draw it in a 3D manner, sort of speak. So imagine this is the barrier tree there. It's sitting, it's a plane here. Okay? The barrier tree is in a plane like this, up here. And then you have the European tree in a plane below it. So we have these two trees, one in one plane and another parallel plane. The barrier breach, let's suppose that, let's suppose that this line here corresponds to the barrier breach. Okay? So that's this line here corresponding to barrier breach. What happens is as the asset moves, so I'm drawing now a sample path of the asset, as the asset moves along in here in the barrier tree, once it hits there, you drop down to this level. You drop down to the corresponding point in the European tree. So this is a Euro tree. And this is the barrier. And now the asset continues moving along in the European tree. And if the asset took a different path, well, it may take this path, in which case you never dropped below. You never dropped down to the European tree. You just stayed in the barrier tree. And if it took, if it just went up immediately, so it drops right down here, and that could be a path. Okay? So this is another way to imagine it. And sometimes people find this a little bit easier. You have one tree representing the barrier option price, another one representing the European, and you switch from the one tree to the next once you touch that barrier. Okay? And so you imagine now why you have to do this European first? Because those values feed into the barrier. All right, along that red line, the barrier option equals the European option price. Because once you hit it, the contract says, okay, the barrier contract is ripped up, you now get a European. Okay? Do you want to see an example of a calculation? Okay, let's do it quick. I'll do this one in Excel so that you can see the development of the tree rather than just the final result. And we'll just do it with a binomial tree for simplicity. Let's say it starts at one. Our volatility is 20%, risk-free rate is 2%, and let's take one month time steps, why not? So we're using the CRR model, so it's going up by of all times square root DT and DT, our risk-neutral probabilities. Okay, so this is the asset tree that I'll draw here. And I'm not going to tilt it this time because it's just so that you can see the analog with these diagrams. I won't bother tilting the tree, so I will kind of try to implement it like this so that you see the actual steps. So this is a zero. This is equal to this times up factor. That's enough steps, I think. And this equals this times down. Whoops, divided by the up factor. It's a little bit tedious. Okay, so that's our asset tree. And let's value, let's put the barrier level at, just so that there's some interesting structure, let's put the barrier level at 1.13. Okay, so 1.13 means that all of these are, are going to be barrier breaches, okay? So U equals 1.13. So, oh, and what's the strike of the European? Let's make the European strike be 1.2. Why not? Actually, yeah. No, 1.1, okay? 1.1 is my European strike. So that strike, okay. So I'm going to build my European tree here. And what do we do? We start at maturity, and that equals maximum of this minus its European call. So minus strike in zero. How many do I need here? One, two, three, need five of them. Six, sorry, two, four, six of them. One, two, three, four, five, six. Okay, good. Now, you can see that at the end of the day, you can see I actually don't need to hold European tree. Do you realize that? I'll do the whole European tree, but we don't actually need it. So it equals X negative RF times DT times Q, Q up times this value there. Plus one minus Q up times that value there. And that's it, I didn't define Q up. Okay, did I miss, did I miss a step somehow? Two, four, six, I thought I had six here. Two, four, six. Where am I missing? Does anyone notice that I'm missing something somewhere here? Sorry? D27. What's about D27? Oh, jeez, thank you. Yes. Okay, good. I don't know why, is it looks, it looked like, I was just stunned. Okay, so there we go. So the price of the European that's embedding this is about two cents, it's nothing. But anyhow, the underlying asset is a dollar, so that's expected. All right, so now what about the barrier option? So what we should do for the barrier is, again, we're gonna start off at maturity, but we say conditional barrier breach happening only into the future. So, wow, actually it should be including now and the future. So since we're at maturity here, we're in fact gonna get just when we're above the barrier in this two nodes, those two nodes, we will get the European value. In fact, I wanna lower the strike because I want you to see, let's use a dollar, do I get values? Good, yeah. And maybe it would be more convenient for me to make this tree next to this one and make my font a little smaller. Can you still read that? Is that, it's reasonable, I know it's not optimal, but so what I'm doing is, here, for this terminal boundary condition, we're putting in zeros when we're below the barrier, okay? We're below the barrier, so the barrier is worth nothing. We're above the barrier at maturity, we should get the same price as the European. And let's highlight the European cells which correspond to breach, those ones. And when I'm at this node, I'm still in the breach region, so this is actually just equal to European value. And this node, I'm still in the breach region, so that is equal to the European value. Now, there are no more points that are in the breached region, in the knocked-in region, so I simply stop and now all I do is apply discounted expectations, as before. So I can use the same, copy that same formula over here, discounted expectations going up and down, and just apply that everywhere here. Okay, good. Now you see there's definitely a difference between the barrier option, which is 4.2 cents, and the European, which is 5.8 cents. The barrier's a little bit cheaper. And that makes sense because there are certain states in the barrier option, which are paying nothing, while in the European option, they pay something. One of those nodes is the one where the cursor is right now. Here, the European option pays you 6 cents. Here, the barrier pays you nothing. And now how would an actual asset price process work? Well, the asset or the pricing for this contingent claim, how would it actually work? I would be starting here, moving through this tree, or let's say I went up, up, and as soon as I hit this node, I drop to this tree. Okay, and if I went down, that's the value. If I went down again, that's the value. So the value, the path of the value of the barrier option, if I went up, up, went up, up, up, down, down, it's not zero. Okay, if I went up, up, up, down, down, up, that's up, that's the second up, that's the third up, down, and then down, you might think it's zero, but it's not. Because once you hit this breached region, you drop down to the European tree, and then you go down, down, and it's actually worth something. Okay, that's the slight tricky thing with barrier options, is that it's not really one option, it's an embedded option. It's like a compound option. Okay, can you imagine what you do for a double knock-in barrier? It's a barrier option in which you have restrictions on both sides. Well, all you would do is you'd paste the other boundary condition on the lower side as well. So, let's explore that case here on the same worksheet. Let's suppose there was also a lower barrier, which was at the level 0.84, and the strike of this call option, let's make it really low, let's make it 0.5. Okay, just so that there's some interesting price behavior in there. So now, actually 0.83. For the lower one. Okay, so now, those are also breached points. Right, these are also breaching points. So I need to go here and replace the corresponding nodes in my tree, which somehow it's offset by, am I offset by one somehow here? No, it's this row, okay, good. Yeah, I need to replace those with the European counterparts. Ah, crud. 85, yeah. Thanks. Yeah, 85. Okay, so now this is our region and then we're gonna just replace this by the corresponding nodes down here. So that's that one. Good, those are replaced. And then we still have zeros here because this is within the knock-in region. It's, sorry, it didn't enter the knock-in region, so we still get nothing at maturity. And then again, you just apply the discounted expectation for all of these interior nodes, okay? Procedures, just that straightforward. And if there was a sample path that went down, down, down, up, up, it's not worth zero, rather it's worth down, down, down, down, then we move over to this tree here and we go up, up, and it would be worth 0.4439. Okay? So here's an interesting question and this will lead into your little quiz. Suppose you have an option which did the following. You got $1 at maturity, but you had to do two things. First, the option has to go above one level, call that level U. And once it goes above that level U, then it has to go below some level L. And only then do you get a dollar and you get it at maturity, okay? Let me write that down. Sometimes these are called onion barrier options because you have to peel away one layer before the other one gets activated, okay? So you're gonna get $1 and this is happening at the maturity date, but only if S first arises above two and then it drops below one. If it hits one first, sorry, if it just hits one, you don't get anything. You have to hit two first, then you have to hit one. Then you get a dollar, okay? So you have to get knocked around basically. I'd have to bounce. So a path which does this, da-da-da, that gets you nothing. You get zero for this path. Let's suppose that's maturity here. This also gets you nothing because you did hit two, but you never dropped below one. What about that path? That one, you do get one because you hit two first, then you dropped below one. So as long as you hit two now, you all of a sudden you have to drop below one. So let's do this in a very, very, very simple model. I'm gonna tell you, suppose the risk-neutral probability is in this model or one-half. These numbers here are telling you the value at the value of the asset. So this level is one and a half at the middle. One step up is one and three quarters and then another step up gets you to two, okay? And one step down gets you to one and a quarter and one more step down gets you to one. Okay, maturity is there. It's out one, two, three, four, five, six steps. That's your maturity date, okay? You're out six steps. And yeah, I want you to tell me what's the, R is zero. R is zero. Suppose Q is a half. Tell me what this option price is worth. What this option is worth, sorry. So the hints, of course, is very much related to what we just did, right? Think of it as two separate trees. I'll walk around and give you a hand. Okay, so good. It looks like pretty much you all recognize the fact that there's really just these two paths that lead to an option payoff. Everything else has an option payoff of zero. And so that means if I'm computing the expectation under risk-neutral measure of the discounted payoff function, phi, then there are only two paths that do this. And the probability that those two paths occur are, well, as long as I get to here, to this node, then one of those other two branches will occur. So I just need to compute the probability along there. And that's just one half to the sixth times my payoff, which is one plus one minus one half to the sixth times zero in every other outcome I get zero. And the discount factor is one, so that's my answer. Done. There's nothing to do other than that. But suppose this was a particularly simple situation, right? And it's kind of unfortunate. So what I wanna do is show you quickly what would happen if we did this in the, hey, you can't have that in after I just went over it. What I wanna do is extend this out to many more steps and show you what would have happened, okay? Because then we'll see something kind of interesting. So we had one, 1.25, no, sorry, what was it? 1.5, 1.75, and two, 1.25 and one, yeah. And that's it, that's my asset tree, one, one, two. How many steps do I have there? Got 12 steps out, okay. And suppose we wanted to value this kind of knock, this sort of ratcheting option. So the idea is to first of all value the embedded option, which in this case, in the previous case, the embedded option was a European. Here the embedded option is really the second situation, right? You assume the first has happened, so you bar your breach, you've reached on number two. So what you're after is trying to find the value of an asset which pays you one if the asset price dropped below zero. That's what you need, okay? So how would I calculate that price? I need that first. So if the asset price dropped below zero, so here we're out to some very far time step. If the asset price has already went above two, and now I'm looking for the asset price dropping, sorry, below one, not zero, by mistake, I misspoke. We're looking for breach below this level. What would my maturity value be? So zero's here and it's one's here, is it not? Discounted expectation, all along here, this is going to be one, all of these will be one, everything below here, so you can see that there's going to be some sort of region here in which all these asset values will be one, one, and draw it a little more carefully. And it's only at this barrier level here that we have to be a little bit more careful, right? So we're going to get one and all of this in this entire region where I've drawn this orange line all below there, it's one, and then the question is, what do we do right there? What's the value here? Why? You think it's just a half times that expected value? What do you get upon barrier breach? When I touch this barrier, what do I receive? I receive $1 at maturity, but $1 at maturity, since interest rates are zero, is $1 now. So the value of this node is not a half, it's one. The value of this node is also one, because once I touch it, I receive one. So this barrier option, this second, this is the embedded barrier option, looks like that. It's one all along this barrier. Now, the region up here, so this region below here, this is the barrier breach region. In the European example I gave you, what you would have had in that shaded region would be the value of a European option. Now what you have instead is the value of receiving one at maturity, which in this case is one, because interest rates are zero. So everything in that region is one, not just below that sort of orange line or below the one layer higher. The reason is again, because once you touch the barrier, you get one, that's the contract specification. But in the internal region, this is where you get this discounted expectation working. In the region above the red line, this is the continuation value, this is the value that you just compute your discounted expectations for. And here, yes, at this node here, I would get, let me write in another color, I would get a half, because this would be zero. I need to draw in some other nodes here. How many spaces are there between? No, no, between the lowest, so this is one, one and a quarter, one and a half, one and three quarters, and that's two there, okay. So what we're after is we need to know the price along that line, right? The price along that line is what will feed into our actual option price that we wanna compute. This is just the embedded portion, this is replacing the European option. So what I want eventually is the value along that line. But I'm currently valuing only the option that says, you get a dollar if the asset drops below one, touches one. That's the option that I'm valuing currently. And we can see that that zero there, that zero there, that zero there. And here now, I have a half and zero, that's a quarter, and between one and a quarter, we have what's that? I can't think right now, average between one and a quarter, that's five over four divided by five over eight, okay. I think it's five over eight. It's five quarters, five over eight, et cetera. And then you fill in all of this and you'll get some values along this tree here. So let's do that on the Excel spreadsheet because it's a pain to do it here. So we're gonna put one all along here. This is a lower barrier. We have a zero, how many zeros do we have? One, two, three, one, two, three. And then this is just a half times, oops, this is just a half times this, plus a half times that. And we probably need one more. Okay, oops. This is level one, this is level 1.25, 1.5, 1.75, and that's two. So I skipped, sorry, I haven't. 1.25, 1.5, 1.75, and that's two. So I need one more line here. Somehow I'm getting extra nodes and I'm not sure why. 1.5, one, two, three, four, five. Oh, no one's telling me. I'm just skipping, I'm going way too, my spaces are mismatched here, right? Does this make sense here? I have asset level one, 1.25, 1.5, 1.75, and two. So this again here, this is the embedded option which is the one that is giving you one upon hitting the level one. So my barrier conditions are one all along here and zero all along here, and then I'm just doing discounted expectations, okay? It should be clear that from up there, I'm getting zero everywhere there, so all those discounted values are going to give me zero at this node. I don't need to correct anything there. And then what I wanna do is what we're interested in is the analog of the value at this point, that along the layer asset price equals two, right? This is where we wanna find as an input for our next layer, okay? So this is the input for the next layer, so I take those values, actually let's say equals, and I'll really only need this one here. And I know that if I'm below two, I get nothing, one, two, I only have two steps to do, that's it. And then I just use discounted expectation again here. So everything below there is all gonna be zeros, right? This whole portion of my tree, it's all gonna be zeros here, and I've inserted the value that I got from, I inserted the value that I have from the first option, the first barrier option, as my payoff for the second, and then I feed back and that's my value. So it's kind of a tricky problem in general when you have lots of steps, because you can't identify in general all of the steps that lead to the payoff. If you could enumerate them, then you would be fine. All right, I think that's enough on this funny option. Okay, let's do one last piece of work and then we'll be done for the day. Just a word of warning, so far the course has been fairly computational and kind of, it hasn't been terribly theoretical, it's gonna change gears coming from next week onwards more or less, we start to do some more analytical work. So the first few lectures were more so to take what you've already been known from binomial trees and do something more interesting with them. From next week onwards, we're gonna be doing some more mathematical work. Okay, so last little bit that I wanted to talk about today is another class of exotic options that we can value analytically. And these are called forward starting options. Okay, has anyone heard of forward starting options? Okay, so the idea behind forward starting is here's there are two dates embedded in the option. And let's specifically think of a forward starting call. Okay, the idea is that at one date, this is the date on which the strike of the call gets set and it's usually set as a percentage of what the prevailing spot price is. And here is the maturity of the option itself. So if you think of the payoff here for the forward starting call, let's call it FC or FSC forward starting call, it looks just like a regular call option except the strike is not known ahead of time. Okay, so sitting here today, you don't know what that strike is gonna be. It's something random. And the reason that someone might be interested in this is that they may have a view on what the future volatility or future behavior of the asset price is, but you're not sure what's happening in the short term. Between now and T zero, you don't know what's gonna happen, but you do want to get exposure from T zero to T one. Okay, that's the reason why you might want such an option. And to value it, we're basically gonna use, well, we're gonna use the same techniques that we've already got, that is discounted expectation under risk-neutral measures. We have this one tricky issue to deal with, the fact that the middle of time, it's stochastic. So any suggestions as to how you might deal with that? Just to get you started, you know that the option price is always computed as an expectation under the risk-neutral measure of the terminal payoff that's discounted from when you get the payment. So I need this expectation. Now, is asset time T one independent of asset time T zero? Those are definitely not independent. Can you write, so let me just draw something here. T zero, T one, so this would be asset T zero, this would be asset T one, and they're somehow dependent on one another. Now, is it correct to do the following thing? T zero, is this a correct statement? Okay, let you think about it for 10 seconds. Okay, let's see, who believes that this is incorrect? Just one person, and who believes that this is correct? Two people, one plus two is three. I think there's more than three here. So what do you think? You gotta believe either this is correct or incorrect, right? There's only two choices. Okay, what do you think is incorrect about it? Somebody said that it was incorrect, so why? Okay, it turns out this is actually correct. This is a correct statement. It is true, we've already proven the fact that asset prices under the risk-neutral measure in the CRR model, okay? So this is under a modeling assumption. In the CRR model, taking the steps going to zero, you find that the asset price at one point in time is normally, is log normally distributed, and it has that mean and it has that variance. But what is not true is if I want to jointly consider FT zero and FT one, I cannot consider them as both being equal almost surely, as opposed to in distribution, being given by these expressions. So that means if I want to generate a scenario for asset time T zero and asset time T one, I do not simply go ahead and pick one normal random variable and then take that sample from the normal random variable and plug it into this equation. That would be incorrect. Because this would be saying that the same normal random variable is generating the outcome for ST zero and for FT one. That's incorrect, okay? Distributionally, do you understand the difference between distributions versus almost sure? Yes or no, right? Almost sure means you take an event with a probability non-zero and you evaluate the outcomes on that event. And if those are equal, then they're equal almost surely. In distribution, it means the probability of the events are the same. Those are two very different things, right? Here's a classic example. X equals Z, Y equals Z, Z is normal zero one, okay? These are equal, not even almost surely. They're just equal, okay? They're equal quantities, but certainly equal almost surely. They also have the same distribution, agree? Because in fact they're generated by the same normal random variable. But suppose I said X equals Z one, Y equals Z one, Z two. Z one, Z two independent. Normal zero one. Is X equal Y almost surely? In fact, almost surely not equal. Two outcomes of a normal random variable are almost surely not equal. But they have equal distribution. X, the distribution of X, is identical to the distribution of Y. Distribution of X is normal zero one. Distribution of Y is normal zero one. Here they're equal in distribution, but they're not equal almost surely. They're equal in distribution and equal almost surely. Here, these distributional quantities are true, but it's not true that I can then just say ST zero, I cannot remove that equal sign, the D. They're not equal almost surely. They're only equal in distribution. Now, in order for me to compute an expectation of this kind, I need to know how ST zero is related to ST one. You need to know how they're related and how are they going to be related. Well, if you think about how this is generated as a limit of a CRR model, there's a whole bunch of little, you know, there's a tree inside of here, right? There's a really small tree sitting in there somewhere. And that we've taken a limit on. And to get ST zero, we've had to sum up all of the little X's that are corresponding to that tree, all of the little Bernoulli's that tell me whether to go up or down. From time zero, it's time T zero. And once I've gotten to a fixed point here, to get from there, to move from time T zero to T one, there's another set of Bernoulli's that tell me how to get there, right? That tell me how to move from ST zero to ST one, right? There's still the tree sitting in here, right? It's embedded under there. And so the Bernoulli's that I flipped are different Bernoulli's. It's another collection of them. Here's say one up to little N, and here's from little N plus one up to big N, for example. Right? And these Bernoulli's are independent of one another, agreed? That's by definition of how we make the tree. We decide whether to go up or down just by flipping a coin. It doesn't matter how I got there. So if you think of it like that, you can easily imagine that calculating the distributional property of ST one given that I start at a point ST zero at time T zero has to have the same distribution as what we had in the first interval, right? It's got to have the same distributional properties. So what I can do is you can write ST zero in distribution as ST zero e to the r minus the half sigma squared times T zero plus sigma squared with T zero Z one. And that's given me that normal random variable is generating where I get to at time T zero. And then where do I get to from time T one? That also equals in distribution starting at ST zero and then being pushed forward. But the amount of time I get pushed forward is only the difference of T one and T zero plus sigma T one minus T zero again, just the difference times another independent normal random variable Z two and under the risk-neutral measure normal zero one. So if I were to generate just these two outcomes I'd have to actually use two normal random variables one to tell me where to get to at time T zero and then another how to get from that point in time forward. Does that make sense? And in terms of the tree, all it means is that well when I'm deciding where to go in the tree I keep flipping a coin and the coin doesn't matter how I got there, right? The Bernoulli coin doesn't matter how I got there. So once I get to a point moving forward it's independence of its past, okay? So this is basically saying the same statement. And you can try to prove that little statement there just using the property of how you write S in terms of the sum of the little x's. I could suggest that you use, you know in terms of the little Bernoulli's you know that the S at any point in time is exponential sigma square root delta T times the sum of all the little x's from one up to n, whatever n is, however many steps I go to get from S zero to ST. So if you use this basic relationship you can easily show those two results. And I urge you try it, okay? Try that. That's a very big hint. So let's assume that we've done that work and we've got that result. Now when we want to compute this expectation how would you approach that? You have two random variables, right? Imagine putting in this expression here up into that expression there. And that expression for ST zero up into that expression. You have a two dimensional integral in principle to compute, agree? Because in fact you can take the expression for ST zero and put it in there. And then ST one would have Z one and Z two in it. And that's okay, you're allowed to have that and in fact you need it. That shows the co-dependence between the spot price of time one and the spot price of time zero or T zero, sorry. And if you took those expressions and plugged it into that expectation you can imagine you get a mess, don't you? It's big and horrible. And it's a two dimensional integral integrated over the density, the product of the densities for Z one and Z two. But there's a much slicker way to do it. I need a hint from the audience, come on. You guys are fourth year students who've done enough probability to tell me, give me one good suggestion here. Yeah, expectation of the conditional expectation. Yes, iterated expectations. You should be able to condition, condition on the asset price S at time T zero, S at time T one is independent of Z one, right? Conditional, conditionally. You have a conditional independence effectively. So use that fact. So let's compute this expectation by iterated expectations. So we have ST one minus alpha ST zero plus. So that's the expectation under Q of the expectation under Q of ST one minus alpha ST zero plus given the value of ST zero. Now, if you are given ST zero within this inner expectation, this is basically a constant strike. And if it's a constant strike, you know this expectation, this inner expectation is the block shows formula with that strike. Now the discount factor is missing, but we could put that in. Okay, that is basically the block shows formula with the strike equal to alpha ST zero. So let's just write that down. Block shows formula is going to give me ST zero phi D plus minus the K, the strike here which is ST zero, discounted from time T one to time T zero because it's been valued at time T zero, phi D minus. And since the discount factor wasn't there, we have to multiply by it, okay? Normally the block shows formula alone doesn't have this coefficient out front, but that's because block shows discounted expectation. Here we don't have the discount. Eventually we'll put it back in. Okay, and what are D plus and D minus? This is a log of the strike, which in this case is alpha ST zero divided by the spot which is ST zero plus R plus or minus one half sigma squared T zero minus T one. All over sigma, sorry, T one minus T zero. Now, what do you notice about D plus D minus? There's something special about it. Not just a, no relation between D plus and D minus, but both of them. Look, that cancels, doesn't it? D plus and D minus are constants. They're independent of S at time T zero, okay? They do not depend on S at time T zero and that's very important because if it doesn't depend on S at time T zero, I can further simplify this inner expectation. I can simply write it as, or sorry, let me, I'm gonna put an arrow there. That equals a constant times ST zero because ST zero appears only there and there. And what's that constant equal to? The constant, in fact, I'm going to keep the discount factor as part of the, because that will cancel eventually. What's that constant? The constant is just phi D plus minus alpha e to the minus r, big T minus that, phi D minus, it's that. That's a constant. Independent of ST zero. And so this gives me a way now of computing that expectation. So now continuing the main equal sign. That equals e to the r, T one minus T zero. Sorry, I realize it's just after five now. I'll be finished in less than two minutes. Times constant. The inner expectation is done. And what's this expectation equal to? S at time T zero. Is it not that? By definition of risk neutral. And so we have the final value. We just have to put in the discount factor and we can see if e to the minus r, T one, that'll cancel that e to the r, T one there. And you just get constant times that zero. So if you did it by brute force method, you would be in a lot of trouble to show that. You'd have these double integrals with some nasty functions. But if you condition, use conditional and iterated expectations, things simplify a lot. Okay, so we'll call it a day there. And.