WEBVTT 00:00:00.060 --> 00:00:02.500 The following content is provided under a Creative 00:00:02.500 --> 00:00:04.019 Commons license. 00:00:04.019 --> 00:00:06.360 Your support will help MIT OpenCourseWare 00:00:06.360 --> 00:00:10.730 continue to offer high quality educational resources for free. 00:00:10.730 --> 00:00:13.340 To make a donation or view additional materials 00:00:13.340 --> 00:00:17.217 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.217 --> 00:00:17.842 at ocw.mit.edu. 00:00:21.860 --> 00:00:23.570 PROFESSOR: So far we had a function, 00:00:23.570 --> 00:00:27.840 and then we differentiated to get an equation of this type. 00:00:27.840 --> 00:00:29.810 But now, we're given this equation. 00:00:29.810 --> 00:00:31.320 And we have to go backwards, want 00:00:31.320 --> 00:00:35.706 to find the stochastic process that satisfies this equation. 00:00:35.706 --> 00:00:45.700 So goal is to find a stochastic process 00:00:45.700 --> 00:00:48.295 X(t) satisfying this equation. 00:00:55.910 --> 00:01:06.130 In other words, we want X of t to be the integral of mu dS 00:01:06.130 --> 00:01:07.325 plus sigma... 00:01:17.460 --> 00:01:18.740 The goal is clear. 00:01:18.740 --> 00:01:19.340 We want that. 00:01:22.520 --> 00:01:26.505 And so these type of equations are 00:01:26.505 --> 00:01:30.630 called differential equations, I hope you already know that. 00:01:30.630 --> 00:01:33.220 Also for PDE, partial differential equations, 00:01:33.220 --> 00:01:35.590 so even when it's not stochastic, 00:01:35.590 --> 00:01:38.310 these are not easy problems. 00:01:38.310 --> 00:01:40.150 At least not easy in the sense that, 00:01:40.150 --> 00:01:42.310 if you're given an equation, typically you 00:01:42.310 --> 00:01:45.290 don't expect to have a closed form solution. 00:01:45.290 --> 00:01:48.430 So even if you find this X, most of the time, 00:01:48.430 --> 00:01:50.910 it's not in a very good form. 00:01:50.910 --> 00:01:55.130 Still, a very important result that you should first 00:01:55.130 --> 00:01:58.790 note before trying to solve any of the differential equation 00:01:58.790 --> 00:02:03.690 is that, as long as mu and sigma are reasonable functions, 00:02:03.690 --> 00:02:05.740 there does exist a solution. 00:02:05.740 --> 00:02:07.730 And it's unique. 00:02:07.730 --> 00:02:10.630 So we have the same correspondence with this PDE. 00:02:10.630 --> 00:02:13.080 You're given a PDE, or given a differential equation, not 00:02:13.080 --> 00:02:14.970 a stochastic differential equation, 00:02:14.970 --> 00:02:18.060 you know that, if you're given a reasonable differential 00:02:18.060 --> 00:02:21.550 equation, then a solution exists. 00:02:21.550 --> 00:02:22.730 And it's unique. 00:02:22.730 --> 00:02:26.416 So the same principle holds in stochastic world. 00:02:26.416 --> 00:02:28.360 Now, let me state it formally. 00:02:33.070 --> 00:02:47.040 This stochastic equation star has a solution that is unique, 00:02:47.040 --> 00:02:51.920 of course with boundary condition-- has a solution. 00:02:51.920 --> 00:03:18.344 And given the initial points-- so 00:03:18.344 --> 00:03:20.760 if you're given the initial point of a stochastic process, 00:03:20.760 --> 00:03:23.520 then a solution is unique. 00:03:30.840 --> 00:03:43.210 Just check, yes-- as long as mu and sigma are reasonable. 00:03:51.310 --> 00:03:55.180 One way it can be reasonable, if it satisfies this conditions. 00:04:27.800 --> 00:04:29.300 These are very technical conditions. 00:04:29.300 --> 00:04:31.258 But at least let me parse to you what they are. 00:04:31.258 --> 00:04:36.730 They say, if you fix a time coordinate and you change x, 00:04:36.730 --> 00:04:39.630 you look at the difference between the values of mu 00:04:39.630 --> 00:04:44.650 when you change the second variable and the sigma. 00:04:44.650 --> 00:04:51.460 Then the change in your function is bounded by the distance 00:04:51.460 --> 00:04:56.570 between the two points by some constant K. 00:04:56.570 --> 00:05:00.730 So mu and sigma cannot change too much when you change 00:05:00.730 --> 00:05:04.660 your space variable a little bit. 00:05:04.660 --> 00:05:07.130 It can only change up to the distance of how much you 00:05:07.130 --> 00:05:08.540 change the coordinate. 00:05:08.540 --> 00:05:09.970 So that's the first condition. 00:05:09.970 --> 00:05:16.730 Second condition says, when you grow your x, 00:05:16.730 --> 00:05:18.382 very similar condition. 00:05:18.382 --> 00:05:20.340 Essentially it says it cannot blow up too fast, 00:05:20.340 --> 00:05:21.690 the whole thing. 00:05:21.690 --> 00:05:25.690 This one is something about the difference between two values. 00:05:25.690 --> 00:05:31.309 This one is about how it expands as your space variable grows. 00:05:31.309 --> 00:05:32.600 These are technical conditions. 00:05:32.600 --> 00:05:37.620 And many cases, they will hold. 00:05:37.620 --> 00:05:41.450 So don't worry too much about the technical conditions. 00:05:41.450 --> 00:05:45.160 Important thing here is that, given a differential equation, 00:05:45.160 --> 00:05:48.940 you don't expect to have a good closed form. 00:05:48.940 --> 00:05:52.880 But you do expect to have a solution of some form. 00:05:57.260 --> 00:06:01.060 OK, let's work out some examples. 00:06:05.260 --> 00:06:08.240 Here is one of the few stochastic differential 00:06:08.240 --> 00:06:09.590 equations that can be solved. 00:06:21.090 --> 00:06:22.360 This one can be solved. 00:06:22.360 --> 00:06:26.376 And you already know what X is. 00:06:26.376 --> 00:06:29.700 But let's pretend you don't know what X is. 00:06:29.700 --> 00:06:31.316 And let's try to solve it. 00:06:34.490 --> 00:06:36.530 I will show you an approach, which 00:06:36.530 --> 00:06:40.280 can solve some differential equations, some SDEs. 00:06:40.280 --> 00:06:43.390 But this really won't happen that much. 00:06:43.390 --> 00:06:48.890 Still, it's like a starting point. 00:06:48.890 --> 00:06:50.280 There's that. 00:06:50.280 --> 00:06:55.500 And assume X(0) [INAUDIBLE]. 00:06:55.500 --> 00:06:56.855 And mu, sigma are constants. 00:07:05.130 --> 00:07:07.080 Just like when solving differential equations, 00:07:07.080 --> 00:07:10.430 first thing you'll do is just guess, suppose, guess. 00:07:23.870 --> 00:07:45.252 If you want this to happen, then d of X of t is-- OK. 00:07:48.210 --> 00:07:51.290 x is just the second variable. 00:07:51.290 --> 00:07:52.872 And then these two have to match. 00:07:52.872 --> 00:07:54.080 That has to be equal to that. 00:07:54.080 --> 00:07:55.830 That has to be equal to that. 00:07:55.830 --> 00:08:10.320 So we know that del f over del t, mu of X of t 00:08:10.320 --> 00:08:11.516 is equal to mu of f. 00:08:20.240 --> 00:08:24.260 So we assumed that this is a solution then differentiated 00:08:24.260 --> 00:08:24.990 that. 00:08:24.990 --> 00:08:28.470 And then, if that's a solution, these all have to match. 00:08:28.470 --> 00:08:30.210 You get this equation. 00:08:30.210 --> 00:08:32.669 If you look at that, that tells you 00:08:32.669 --> 00:08:36.429 that f is an exponential function in the x variable. 00:08:36.429 --> 00:08:40.640 So it's e to the sigma times x. 00:08:40.640 --> 00:08:44.750 And then we have a constant term, plus some a times 00:08:44.750 --> 00:08:45.730 a function of t. 00:08:50.680 --> 00:08:54.770 The only way it can happen is if it's in this form. 00:08:54.770 --> 00:08:57.000 So it's exponential function in x. 00:08:57.000 --> 00:09:02.130 And in the time variable, it's just some constant. 00:09:02.130 --> 00:09:04.540 When you fix a t, it's a constant. 00:09:04.540 --> 00:09:06.310 And when you fix a t and change x, 00:09:06.310 --> 00:09:08.590 it has to look like an exponential function. 00:09:08.590 --> 00:09:11.064 It has to be in this form, just by the second equation. 00:09:13.730 --> 00:09:16.590 Now, go back to this equation. 00:09:16.590 --> 00:09:25.520 What you get is partial f over partial t is now a times g 00:09:25.520 --> 00:09:30.350 prime of t, f... 00:09:43.580 --> 00:09:44.580 AUDIENCE: Excuse me. 00:09:44.580 --> 00:09:46.200 PROFESSOR: Mm-hm. 00:09:46.200 --> 00:09:49.420 AUDIENCE: That one in second to last line, yeah. 00:09:49.420 --> 00:09:52.790 So why is it minus mu there at the end? 00:09:52.790 --> 00:09:55.364 PROFESSOR: It's equal. 00:09:55.364 --> 00:09:56.364 AUDIENCE: Oh, all right. 00:09:56.364 --> 00:09:57.030 PROFESSOR: Yeah. 00:10:03.860 --> 00:10:06.260 OK, and then let's plug it in. 00:10:06.260 --> 00:10:13.270 So we have a of g prime t, f, plus 1/2 of sigma 00:10:13.270 --> 00:10:17.094 square f equals mu of f. 00:10:17.094 --> 00:10:25.525 In other words, a of g prime of t is mu minus... 00:10:25.525 --> 00:10:26.025 square. 00:10:29.590 --> 00:10:36.502 g(t) is mu times some constant c_1 plus c_2. 00:10:44.380 --> 00:10:48.980 OK, and then what we got is original function f of t of x 00:10:48.980 --> 00:10:54.980 is e to the sigma*x plus mu minus 1 over 2 sigma square t 00:10:54.980 --> 00:10:57.590 plus some constant. 00:10:57.590 --> 00:11:01.520 And that constant can be chosen because we 00:11:01.520 --> 00:11:09.270 have the initial condition, x of 0 equals 0. 00:11:09.270 --> 00:11:14.203 That means if t is equal to 0, f(0, 0) is equal to e to the c. 00:11:14.203 --> 00:11:16.510 That has to be x_0. 00:11:16.510 --> 00:11:21.701 In different words, this is just x_0 times e to the sigma*x plus 00:11:21.701 --> 00:11:25.550 mu minus 1 over 2 sigma square of t. 00:11:25.550 --> 00:11:27.802 Just as we expected, we got this. 00:11:33.210 --> 00:11:35.940 So some stochastic differential equations 00:11:35.940 --> 00:11:39.920 can be solved by analyzing it. 00:11:39.920 --> 00:11:43.010 But I'm not necessarily saying that this is a better way 00:11:43.010 --> 00:11:45.020 to do it than just guessing. 00:11:45.020 --> 00:11:46.830 Just looking at it, and, you know, 00:11:46.830 --> 00:11:50.027 OK, it has to be exponential function. 00:11:50.027 --> 00:11:50.610 You know what? 00:11:50.610 --> 00:11:54.430 I'll just figure out what these are. 00:11:54.430 --> 00:11:56.030 And I'll come up with this formula 00:11:56.030 --> 00:11:59.110 without going through all those analysis. 00:11:59.110 --> 00:12:01.435 I'm not saying that's a worse way than actually going 00:12:01.435 --> 00:12:02.310 through the analysis. 00:12:02.310 --> 00:12:05.250 Because, in fact, what we did is we kind of already 00:12:05.250 --> 00:12:08.780 knew the answer and are fitting into that answer. 00:12:12.940 --> 00:12:14.790 Still, it can be one approach where 00:12:14.790 --> 00:12:17.590 you don't have a reasonable guess of what the X_t has 00:12:17.590 --> 00:12:21.772 to be, maybe try to break it down into pieces like this 00:12:21.772 --> 00:12:24.410 and backtrack to figure out the function. 00:12:24.410 --> 00:12:26.260 Let me give you one more example where we 00:12:26.260 --> 00:12:28.580 do have an explicit solution. 00:12:28.580 --> 00:12:30.700 And then I'll move on and show you 00:12:30.700 --> 00:12:33.367 how to do when there is no explicit solution 00:12:33.367 --> 00:12:35.700 or when you don't know how to find an explicit solution. 00:12:38.304 --> 00:12:39.710 Maybe let's keep that there. 00:12:52.240 --> 00:12:55.781 Second equation is called this. 00:13:16.655 --> 00:13:18.080 What's the difference? 00:13:18.080 --> 00:13:22.830 The only difference is that you don't have an X here. 00:13:22.830 --> 00:13:26.560 So previously, our main drift term 00:13:26.560 --> 00:13:30.330 also was proportional to the current value. 00:13:30.330 --> 00:13:33.980 And the error was also dependent on the current value 00:13:33.980 --> 00:13:36.350 or is proportional to the current value. 00:13:36.350 --> 00:13:38.580 But here now the drift term is something 00:13:38.580 --> 00:13:42.610 like an exponential-- minus exponential. 00:13:42.610 --> 00:13:45.460 But still, it's proportional to the current value. 00:13:45.460 --> 00:13:49.120 But the error term is just some noise. 00:13:49.120 --> 00:13:51.690 Irrelevant of what the value is, this has 00:13:51.690 --> 00:13:54.070 the same variance as the error. 00:13:54.070 --> 00:13:58.550 So it's a slightly different-- oh, what is it? 00:13:58.550 --> 00:14:00.770 It's a slightly different process. 00:14:00.770 --> 00:14:09.100 And it's known as Ornstein-Uhlenbeck process. 00:14:18.530 --> 00:14:23.410 And this is used to model a mean-reverting stochastic 00:14:23.410 --> 00:14:24.980 process. 00:14:24.980 --> 00:14:30.230 For alpha greater than 0, notice that if X deviates from 0, 00:14:30.230 --> 00:14:33.940 this gives a force that drives the stochastic process back 00:14:33.940 --> 00:14:36.940 to 0, that's negatively proportional 00:14:36.940 --> 00:14:37.902 to your current value. 00:14:40.740 --> 00:14:44.330 So yeah, this is used to model some mean-reverting stochastic 00:14:44.330 --> 00:14:45.130 processes. 00:14:45.130 --> 00:14:50.785 And they first used it to study the behavior of gases. 00:14:57.230 --> 00:14:58.330 I don't exactly see why. 00:14:58.330 --> 00:15:01.810 But that's what they say. 00:15:01.810 --> 00:15:03.710 Anyway, so this is another thing that 00:15:03.710 --> 00:15:05.510 can be solved by doing similar analysis. 00:15:10.490 --> 00:15:14.590 But if you try the same method, it will fail. 00:15:14.590 --> 00:15:34.550 So as a test function, your guess, initial guess, will be-- 00:15:34.550 --> 00:15:36.576 or a(0) is equal to 1. 00:15:40.878 --> 00:15:44.070 Now, honestly, I don't know how to come up with this guess. 00:15:49.137 --> 00:15:50.720 Probably, if you're really experienced 00:15:50.720 --> 00:15:52.870 with stochastic differential equations, 00:15:52.870 --> 00:15:56.510 you'll see some form, like you'll 00:15:56.510 --> 00:16:00.340 have some feeling on how this process will look like. 00:16:00.340 --> 00:16:03.330 And then try this, try that, and eventually something 00:16:03.330 --> 00:16:05.266 might succeed. 00:16:05.266 --> 00:16:06.890 That's the best explanation I can give. 00:16:06.890 --> 00:16:08.264 I can't really give you intuition 00:16:08.264 --> 00:16:10.590 why that's the right guess. 00:16:10.590 --> 00:16:13.570 Given some stochastic differential equation, 00:16:13.570 --> 00:16:15.356 I don't know how to say that you should 00:16:15.356 --> 00:16:16.730 start with this kind of function, 00:16:16.730 --> 00:16:18.722 this kind of function. 00:16:18.722 --> 00:16:20.430 And it was the same when, if you remember 00:16:20.430 --> 00:16:22.980 how we solved ordinary differential equations 00:16:22.980 --> 00:16:25.680 or partial differential equations, most of the time 00:16:25.680 --> 00:16:27.206 there is no good guess. 00:16:27.206 --> 00:16:31.690 It's only when your given formula has some specific form 00:16:31.690 --> 00:16:33.334 such a thing happens. 00:16:33.334 --> 00:16:37.250 So let's see what happens here. 00:16:37.250 --> 00:16:39.310 That was given. 00:16:39.310 --> 00:16:42.990 Now, let's do exactly the same as before. 00:16:42.990 --> 00:16:46.940 Differentiate it, and let me go slow. 00:16:46.940 --> 00:16:49.240 So we have a prime of t. 00:16:49.240 --> 00:16:54.440 By chain rule, a prime of t and that value, 00:16:54.440 --> 00:16:59.240 that part will be equal to X(t) over a(t). 00:16:59.240 --> 00:17:00.440 This is chain rule. 00:17:00.440 --> 00:17:03.390 So I differentiate that to get a prime t. 00:17:03.390 --> 00:17:05.580 That stays just as it was. 00:17:05.580 --> 00:17:09.609 But that can be rewritten as X(t) divided by a(t). 00:17:09.609 --> 00:17:15.329 And then plus a(t) times the differential of that one. 00:17:15.329 --> 00:17:17.109 And that is just b(t)*dB(t). 00:17:29.180 --> 00:17:31.870 You don't have to differentiate that once more, even 00:17:31.870 --> 00:17:34.630 though it's stochastic calculus, because that's 00:17:34.630 --> 00:17:35.500 a very subtle point. 00:17:35.500 --> 00:17:39.090 And there's also one exercise about it in your homework. 00:17:39.090 --> 00:17:42.300 But when you have a given stochastic process written 00:17:42.300 --> 00:17:46.190 already in this integral form, if we remember 00:17:46.190 --> 00:17:49.410 the definition of an integral, at least how I defined it, 00:17:49.410 --> 00:17:52.560 is that it was an inverse operation of a differential. 00:17:52.560 --> 00:17:55.660 So when you differentiate this, you just get that term. 00:17:55.660 --> 00:17:57.290 What I'm trying to say is, there is 00:17:57.290 --> 00:18:02.830 no term, no term where you have to differentiate this one more. 00:18:07.060 --> 00:18:11.780 Prime dt, something like that, we don't have this term. 00:18:15.180 --> 00:18:16.170 This can be confusing. 00:18:16.170 --> 00:18:17.360 But think about it. 00:18:22.580 --> 00:18:26.690 Now, we laid it out and just compare. 00:18:26.690 --> 00:18:32.110 So minus alpha of X of t is equal to a prime t 00:18:32.110 --> 00:18:36.260 over a(t) X(t). 00:18:36.260 --> 00:18:42.590 And your second term, sigma*dB(t) is equal to a(t) 00:18:42.590 --> 00:18:43.400 times b(t). 00:18:47.240 --> 00:18:48.660 But these two cancel. 00:18:51.240 --> 00:18:57.746 And we see that a(t) has to be e to the minus alpha t. 00:19:06.510 --> 00:19:09.785 This says that is an exponential. 00:19:09.785 --> 00:19:10.660 Now, plug it in here. 00:19:10.660 --> 00:19:11.201 You get b(t). 00:19:17.772 --> 00:19:18.940 And that's it. 00:19:18.940 --> 00:19:21.096 So plug it back in. 00:19:21.096 --> 00:19:32.085 X of t is e to the minus alpha*t of x of 0 plus 0 to t sigma e 00:19:32.085 --> 00:19:33.067 to the alpha*s. 00:19:56.640 --> 00:20:02.670 So this is a variance, expectation is 0, 00:20:02.670 --> 00:20:05.210 because that's a Brownian motion. 00:20:05.210 --> 00:20:08.420 This term, as we expected, as time passes, 00:20:08.420 --> 00:20:12.776 goes to 0, exponential decay. 00:20:12.776 --> 00:20:18.576 And that is kind of hinted by this fact, the mean reversion. 00:20:18.576 --> 00:20:21.610 So if you start from some value, at least the drift term 00:20:21.610 --> 00:20:25.380 will go to 0 quite quickly. 00:20:25.380 --> 00:20:29.120 And then the important term will be the noise term 00:20:29.120 --> 00:20:30.200 or the variance term. 00:20:35.237 --> 00:20:35.820 Any questions? 00:20:45.750 --> 00:20:49.991 And I'm really emphasizing a lot of times today, 00:20:49.991 --> 00:20:51.490 but really you can forget about what 00:20:51.490 --> 00:20:53.370 I did in the past two boards, this board 00:20:53.370 --> 00:20:54.789 and the previous board. 00:20:54.789 --> 00:20:56.705 Because most of the times, it will be useless. 00:20:59.220 --> 00:21:01.970 And so now I will describe what you'll 00:21:01.970 --> 00:21:05.090 do if you're given a stochastic differential equation, 00:21:05.090 --> 00:21:06.991 and you have a computer in front of you. 00:21:15.860 --> 00:21:18.038 What if such method fails? 00:21:30.030 --> 00:21:31.570 And it will fail most of the time. 00:21:36.210 --> 00:21:38.100 That's when we use these techniques called 00:21:38.100 --> 00:21:53.840 finite difference method, Monte Carlo simulation, or tree 00:21:53.840 --> 00:21:54.340 method. 00:22:04.130 --> 00:22:06.250 The finite difference method, you probably 00:22:06.250 --> 00:22:09.610 already saw it, if you took a differential equation course. 00:22:09.610 --> 00:22:10.740 But let me review it. 00:22:19.520 --> 00:22:24.230 This is for PDEs or ODEs, for ODE, PDE, not 00:22:24.230 --> 00:22:26.636 stochastic differential equations. 00:22:26.636 --> 00:22:29.010 But it can be adapted to work for stochastic differential 00:22:29.010 --> 00:22:30.640 equations. 00:22:30.640 --> 00:22:32.540 So I'll work with an example. 00:22:36.090 --> 00:22:46.445 Let u of t be u prime of t plus-- u prime of t 00:22:46.445 --> 00:22:53.220 be u(t) plus 2, u_0 is equal to 0. 00:22:53.220 --> 00:22:54.630 Now, this has an exact solution. 00:22:54.630 --> 00:22:57.095 But let's pretend that there is no exact solution. 00:22:57.095 --> 00:23:00.890 And if you want to do it numerically, 00:23:00.890 --> 00:23:04.180 you want to find the value of u equals-- u(1) numerically. 00:23:07.155 --> 00:23:08.660 And here's what you're going to do. 00:23:08.660 --> 00:23:10.660 You're going to chop up the interval from 0 to 1 00:23:10.660 --> 00:23:12.590 into very fine pieces. 00:23:12.590 --> 00:23:17.050 So from 0 to 1, chop it down into tiny pieces. 00:23:20.460 --> 00:23:22.100 And since I'm in front of a blackboard, 00:23:22.100 --> 00:23:25.190 my tiniest piece will be 1 over 2 and 1. 00:23:25.190 --> 00:23:27.550 I'll just take two steps. 00:23:27.550 --> 00:23:29.400 But you should think of it as really 00:23:29.400 --> 00:23:33.090 repeating this a lot of times. 00:23:33.090 --> 00:23:35.480 I'll call my step to be h. 00:23:35.480 --> 00:23:40.800 So in my case, I'm increasing my steps by 1 over 2 at each time. 00:23:40.800 --> 00:23:44.710 So what is u of 1 over 2? 00:23:44.710 --> 00:23:47.135 Approximately, by Taylor's formula, 00:23:47.135 --> 00:23:53.844 it's u(0) plus 1/2 times u prime of 0. 00:23:53.844 --> 00:23:55.010 That's Taylor approximation. 00:23:57.590 --> 00:23:59.910 OK, u(0) we already know. 00:23:59.910 --> 00:24:02.650 It's given to be equal to 0. 00:24:02.650 --> 00:24:05.700 u prime of 0, on the other hand, is given by this differential 00:24:05.700 --> 00:24:07.520 equation. 00:24:07.520 --> 00:24:12.830 So it's 1 over 2 times 5 times u(0) plus 2. 00:24:12.830 --> 00:24:13.930 u(0) is 0. 00:24:13.930 --> 00:24:15.040 So we get equal to 1. 00:24:17.940 --> 00:24:20.932 Like we have this value equal to 1, approximately. 00:24:20.932 --> 00:24:22.015 I don't know what happens. 00:24:22.015 --> 00:24:24.410 But it will be close to 1. 00:24:24.410 --> 00:24:28.170 And then for the next thing, u1. 00:24:28.170 --> 00:24:32.300 This one is, again by Taylor approximation, is u of 1 over 2 00:24:32.300 --> 00:24:37.720 plus 1 over 2 u prime of 1 over 2. 00:24:37.720 --> 00:24:40.520 And now you know the value of u of 1 over 2, approximate value, 00:24:40.520 --> 00:24:41.277 by this. 00:24:41.277 --> 00:24:41.860 So plug it in. 00:24:41.860 --> 00:24:48.552 You have 1 plus 1 over 2 and, again, 5 times u(1/2) plus 2. 00:24:48.552 --> 00:24:50.000 If you want to do the computation, 00:24:50.000 --> 00:24:51.510 it should give 9 over 2. 00:24:55.740 --> 00:24:58.240 It's really simple. 00:24:58.240 --> 00:25:01.790 The key idea here is just u prime 00:25:01.790 --> 00:25:05.720 is given by an equation, this equation. 00:25:05.720 --> 00:25:09.267 So you can compute it once you know the value of u 00:25:09.267 --> 00:25:09.850 at that point. 00:25:12.610 --> 00:25:16.030 And basically, the method is saying take h to be very small, 00:25:16.030 --> 00:25:17.590 like 1 over 100. 00:25:17.590 --> 00:25:19.540 Then you just repeat it 1 over 100 times. 00:25:19.540 --> 00:25:23.760 So the equation is the i plus 1 step value 00:25:23.760 --> 00:25:30.232 can be approximated from the i-th value plus h times 00:25:30.232 --> 00:25:33.064 u prime of h. 00:25:33.064 --> 00:25:34.560 Now repeat it and repeat it. 00:25:34.560 --> 00:25:36.570 And you reach u of 1. 00:25:36.570 --> 00:25:38.250 And there is a theorem saying, again, 00:25:38.250 --> 00:25:42.060 if the differential equation is reasonable, then 00:25:42.060 --> 00:25:44.355 that will approach the true value as you take h 00:25:44.355 --> 00:25:45.520 to be smaller and smaller. 00:25:48.350 --> 00:25:50.360 That's called the finite difference method 00:25:50.360 --> 00:25:52.921 for differential equations. 00:25:52.921 --> 00:25:55.170 And you can do the exact same thing for two variables, 00:25:55.170 --> 00:25:55.669 let's say. 00:26:07.240 --> 00:26:10.050 And what we showed was for one variable, finite difference 00:26:10.050 --> 00:26:18.412 method, we want to find the value of u, function u of t. 00:26:18.412 --> 00:26:23.040 We took values at 0, h, 2h, 3h. 00:26:23.040 --> 00:26:26.680 Using that, we did some approximation, like that, 00:26:26.680 --> 00:26:29.670 and found the value. 00:26:29.670 --> 00:26:33.340 Now, suppose we want to find, similarly, 00:26:33.340 --> 00:26:38.920 a two-variable function, let's say v of t and x. 00:26:38.920 --> 00:26:42.340 And we want to find the value of v of 1, 1. 00:26:42.340 --> 00:26:44.292 Now the boundary conditions are these. 00:26:44.292 --> 00:26:45.666 We already know these boundaries. 00:26:49.650 --> 00:26:51.520 I won't really show you by example. 00:26:51.520 --> 00:26:55.560 But what we're going to do now is compute this value 00:26:55.560 --> 00:26:59.150 based on these two variables. 00:26:59.150 --> 00:27:00.300 So it's just the same. 00:27:00.300 --> 00:27:01.970 Taylor expansion for two variables 00:27:01.970 --> 00:27:05.610 will allow you to compute this value from these two values. 00:27:05.610 --> 00:27:09.072 Then compute this from these two, this from these two, 00:27:09.072 --> 00:27:11.230 and just fill out the whole grid like that, 00:27:11.230 --> 00:27:14.380 just fill out layer by layer. 00:27:14.380 --> 00:27:16.510 At some point, you're going to reach this. 00:27:16.510 --> 00:27:20.320 And then you'll have an approximate value of that. 00:27:20.320 --> 00:27:24.080 So you chop up your domain into fine pieces 00:27:24.080 --> 00:27:25.970 and then take the limit. 00:27:25.970 --> 00:27:27.500 And most cases, it will work. 00:27:31.160 --> 00:27:33.495 Why does it not work for stochastic differential 00:27:33.495 --> 00:27:33.995 equations? 00:27:38.690 --> 00:27:40.490 Kind of works, but the only problem 00:27:40.490 --> 00:27:44.490 is we don't know which value we're looking at, 00:27:44.490 --> 00:27:47.670 we're interested in. 00:27:47.670 --> 00:27:51.050 So let me phrase it a little bit differently. 00:27:54.770 --> 00:27:58.140 You're given a differential equation of the form dX 00:27:58.140 --> 00:28:12.930 equals mu dt plus t dB of t and time variable and space 00:28:12.930 --> 00:28:15.540 variable. 00:28:15.540 --> 00:28:21.850 Now, if you want to compute your value at time 2h based on value 00:28:21.850 --> 00:28:29.760 h, in this picture, I told you that this point 00:28:29.760 --> 00:28:33.030 came from these two points. 00:28:33.030 --> 00:28:38.104 But when it's stochastic, it could depend on everything. 00:28:38.104 --> 00:28:39.520 You don't know where it came from. 00:28:39.520 --> 00:28:41.061 This point could have come from here. 00:28:41.061 --> 00:28:42.390 It could have came from here. 00:28:42.390 --> 00:28:44.960 It could have came from here, came from here. 00:28:44.960 --> 00:28:47.170 You don't really know. 00:28:47.170 --> 00:28:50.900 But what you know is you have a probability distribution. 00:28:50.900 --> 00:28:57.350 So what I'm trying to say is now, 00:28:57.350 --> 00:28:59.140 if you want to adapt this method, what 00:28:59.140 --> 00:29:04.916 you're going to do is take a sample Brownian motion path. 00:29:11.900 --> 00:29:15.260 That means just, according to the distribution 00:29:15.260 --> 00:29:21.010 of the Brownian motion, take one path and use that path. 00:29:21.010 --> 00:29:29.050 Once we fix a path, once a path is fixed, 00:29:29.050 --> 00:29:31.840 we can exactly know where each value comes from. 00:29:31.840 --> 00:29:35.230 We know how to backtrack. 00:29:42.360 --> 00:29:44.680 That means, instead of all these possibilities, 00:29:44.680 --> 00:29:49.170 we have one fixed possibility, like that. 00:29:52.470 --> 00:29:54.570 So just use that finite difference method 00:29:54.570 --> 00:29:57.990 with that fixed path. 00:29:57.990 --> 00:30:01.155 That will be the idea. 00:30:01.155 --> 00:30:03.975 Let me do it a little bit more formally. 00:30:10.170 --> 00:30:11.470 And here is how it works. 00:30:16.230 --> 00:30:30.020 If we have a fixed sample path for Brownian motion of B(t), 00:30:30.020 --> 00:30:41.570 then X at time i plus 1 of h is approximately equal to X 00:30:41.570 --> 00:30:54.930 at time a of h plus h times dx at that time i of h, 00:30:54.930 --> 00:30:59.750 just by the exact same Taylor expansion. 00:30:59.750 --> 00:31:08.966 And then d of X we know to be-- that is equal to mu 00:31:08.966 --> 00:31:18.855 of dt plus-- oh, mu of dt is h. 00:31:18.855 --> 00:31:22.678 So let me write it like that, sigma times d of X-- dB. 00:31:27.350 --> 00:31:37.350 And these mu depend on [? their paths, ?] x at i of h 00:31:37.350 --> 00:31:39.845 dt, sigma... 00:31:56.850 --> 00:31:59.520 With that, here to here is Taylor expansion. 00:31:59.520 --> 00:32:01.780 Here to here I'm going to use the differential 00:32:01.780 --> 00:32:05.213 equation d of X is equal to mu dt plus sigma dB(t). 00:32:05.213 --> 00:32:05.712 Yes? 00:32:05.712 --> 00:32:08.500 AUDIENCE: Do we need that h for [INAUDIBLE]? 00:32:08.500 --> 00:32:10.700 PROFESSOR: No, we don't actually. 00:32:10.700 --> 00:32:13.840 Oh, yeah, I was-- thank you very much. 00:32:13.840 --> 00:32:17.350 That was what confused me. 00:32:17.350 --> 00:32:20.460 Yes, thank you very much. 00:32:20.460 --> 00:32:22.830 And now we can compute everything. 00:32:22.830 --> 00:32:27.260 This one, we're assuming that we know the value. 00:32:27.260 --> 00:32:30.670 That one can be computed from these two coordinates. 00:32:30.670 --> 00:32:36.350 Because we now have a fixed path X, we know what X of i*h is. 00:32:36.350 --> 00:32:40.350 dt, we took it to be h, approximated as h, 00:32:40.350 --> 00:32:42.180 or time difference. 00:32:42.180 --> 00:32:44.230 Again, sigma can be computed. 00:32:44.230 --> 00:32:47.470 dB now can be computed from B_t. 00:32:47.470 --> 00:32:50.180 Because we have a fixed path, again, we 00:32:50.180 --> 00:32:58.162 know that it's equal to B of i plus 1 of h minus B of i of h, 00:32:58.162 --> 00:33:00.560 with this fixed path. 00:33:00.560 --> 00:33:02.950 They're basically exactly the same, 00:33:02.950 --> 00:33:06.810 if you have a fixed path B. The problem is 00:33:06.810 --> 00:33:10.270 we don't have a fixed path B. That's where Monte Carlo 00:33:10.270 --> 00:33:11.220 simulation comes in. 00:33:15.450 --> 00:33:18.990 So Monte Carlo simulation is just a way to draw, 00:33:18.990 --> 00:33:22.900 from some probability distribution, a lot of samples. 00:33:22.900 --> 00:33:27.030 So now, if you know how to draw samples from the Brownian 00:33:27.030 --> 00:33:30.860 motions, then what you're going to do is draw a lot of samples. 00:33:30.860 --> 00:33:34.550 For each sample, do this to compute the value of X(0), 00:33:34.550 --> 00:33:37.410 can compute X of 1. 00:33:40.130 --> 00:33:43.080 So, according to a different B, you will get a different value. 00:33:43.080 --> 00:33:46.910 And in the end, you'll obtain a probability distribution. 00:33:46.910 --> 00:33:57.770 So by repeating the experiment, that means just 00:33:57.770 --> 00:33:59.760 redraw the path again and again, you'll 00:33:59.760 --> 00:34:01.930 get different values of X of 1. 00:34:01.930 --> 00:34:04.160 That means you get a distribution of X of 1, 00:34:04.160 --> 00:34:10.564 obtain distribution of X of 1. 00:34:10.564 --> 00:34:13.350 And that's it. 00:34:13.350 --> 00:34:18.800 And that will approach the real distribution of X of 1. 00:34:18.800 --> 00:34:21.880 So that's how you numerically solve a stochastic differential 00:34:21.880 --> 00:34:23.949 equation. 00:34:23.949 --> 00:34:26.389 Again, there's this finite difference method 00:34:26.389 --> 00:34:29.889 that can be used to solve differential equations. 00:34:29.889 --> 00:34:33.000 But the reason it doesn't apply to stochastic differential 00:34:33.000 --> 00:34:36.810 equations is because there's underlying uncertainty coming 00:34:36.810 --> 00:34:38.870 from Brownian motion. 00:34:38.870 --> 00:34:42.320 However, once you fix a Brownian motion, 00:34:42.320 --> 00:34:44.389 then you can use that finite difference method 00:34:44.389 --> 00:34:46.502 to compute X of 1. 00:34:46.502 --> 00:34:48.659 So based on that idea, you just draw 00:34:48.659 --> 00:34:52.090 a lot of samples of the Brownian path, 00:34:52.090 --> 00:34:53.750 compute a lot of values of X of 1, 00:34:53.750 --> 00:34:56.199 and obtain a probability distribution of X of 1. 00:34:58.710 --> 00:35:00.165 That's the underlying principle. 00:35:03.610 --> 00:35:06.832 And, of course, you can't do it by hand. 00:35:06.832 --> 00:35:07.665 You need a computer. 00:35:11.320 --> 00:35:13.341 Then, what is tree method? 00:35:13.341 --> 00:35:16.060 That's cool. 00:35:16.060 --> 00:35:20.560 Tree method is based on this idea. 00:35:20.560 --> 00:35:34.284 Remember, Brownian motion is a limit of simple random walk. 00:35:44.610 --> 00:35:47.170 This gives you a kind of approximate way 00:35:47.170 --> 00:35:50.844 to draw a sample from Brownian motions. 00:35:50.844 --> 00:35:51.760 How would you do that? 00:35:58.480 --> 00:36:00.900 At time 0, you have 0. 00:36:00.900 --> 00:36:06.400 At time really tiny h, you'll have plus 1 or minus 1 00:36:06.400 --> 00:36:09.130 with same probability. 00:36:09.130 --> 00:36:16.880 And it goes up or down again, up or down again, and so on. 00:36:16.880 --> 00:36:19.170 And you know exactly the probability distribution. 00:36:22.050 --> 00:36:25.250 So the problem is that it ends up here as 1/2, 00:36:25.250 --> 00:36:30.160 ends up here as 1/2, 1/4, 1/2, 1/4, and so on. 00:36:30.160 --> 00:36:33.030 So instead of drawing from this sample path, what you're 00:36:33.030 --> 00:36:36.614 going to do is just compute the value of our function 00:36:36.614 --> 00:36:37.280 at these points. 00:36:41.770 --> 00:36:44.267 But then the probability distribution, 00:36:44.267 --> 00:36:46.100 because we know the probability distribution 00:36:46.100 --> 00:36:49.060 that the path will end up at these points, 00:36:49.060 --> 00:36:54.530 suppose that you computed all these values here. 00:36:57.810 --> 00:37:00.080 I draw too many, 1, 2, 3, 4, 5. 00:37:00.080 --> 00:37:03.015 This 1 or 32 probability here. 00:37:03.015 --> 00:37:10.362 5 choose 1, 5 over 32, 5 choose 2 is 10 over 32. 00:37:13.560 --> 00:37:17.200 Suppose that some stochastic process, after following this, 00:37:17.200 --> 00:37:23.162 has value 1 here, 2 here, 3 here, 4, 5, and 6 here. 00:37:23.162 --> 00:37:27.190 Then, approximately, if you take a Brownian motion, 00:37:27.190 --> 00:37:28.790 it will have 1 with probability 1 00:37:28.790 --> 00:37:32.346 over 32, 2 with probability 5 over 32, and so on. 00:37:34.640 --> 00:37:36.140 Maybe I didn't explain it that well. 00:37:36.140 --> 00:37:38.060 But basically, tree method just says, 00:37:38.060 --> 00:37:42.950 you can discretize the outcome of the Brownian motion, 00:37:42.950 --> 00:37:47.535 based on the fact that it's a limit of simple random walk. 00:37:47.535 --> 00:37:52.050 So just do the exact same method for simple random walk instead 00:37:52.050 --> 00:37:54.180 of Brownian motion. 00:37:54.180 --> 00:37:56.195 And then take it to the limit. 00:37:56.195 --> 00:37:57.070 That's the principle. 00:38:00.778 --> 00:38:01.278 Yeah. 00:38:04.219 --> 00:38:06.260 Yeah, I don't know what's being used in practice. 00:38:06.260 --> 00:38:10.150 But it seems like these two are the more important ones. 00:38:10.150 --> 00:38:12.540 This is more like if you want to do it by hand. 00:38:12.540 --> 00:38:15.400 Because you can't really do every single possibility. 00:38:15.400 --> 00:38:18.125 That makes you only a finite possibility. 00:38:22.150 --> 00:38:24.630 Any questions? 00:38:24.630 --> 00:38:25.130 Yeah. 00:38:25.130 --> 00:38:28.790 AUDIENCE: So here you said, by repeating 00:38:28.790 --> 00:38:30.498 the experiment we get [INAUDIBLE] 00:38:30.498 --> 00:38:32.938 distribution for X(1). 00:38:32.938 --> 00:38:36.490 I was wondering if we could also get the distribution for not 00:38:36.490 --> 00:38:38.210 just X(1) but also for X(i*h). 00:38:38.210 --> 00:38:40.060 PROFESSOR: All the intermediate values? 00:38:40.060 --> 00:38:40.735 AUDIENCE: Yeah. 00:38:40.735 --> 00:38:41.990 PROFESSOR: Yeah, but the problem is 00:38:41.990 --> 00:38:43.448 we're taking different values of h. 00:38:43.448 --> 00:38:45.720 So h will be smaller and smaller. 00:38:45.720 --> 00:38:47.650 But for those values that we took, 00:38:47.650 --> 00:38:49.770 yeah, we will get some distribution. 00:38:49.770 --> 00:38:52.330 AUDIENCE: Right, so we might have distributions 00:38:52.330 --> 00:38:56.002 for X of d for many different points, right? 00:38:56.002 --> 00:38:56.668 PROFESSOR: Yeah. 00:38:56.668 --> 00:38:58.090 AUDIENCE: Yeah. 00:38:58.090 --> 00:39:02.692 So maybe we could uh-- right, OK. 00:39:02.692 --> 00:39:04.650 PROFESSOR: But one thing you have to be careful 00:39:04.650 --> 00:39:08.650 is let's suppose you take h equal 1 over 100. 00:39:08.650 --> 00:39:11.030 Then, this will give you a pretty fairly good 00:39:11.030 --> 00:39:12.551 approximation for X of 1. 00:39:12.551 --> 00:39:14.300 But it won't give you a good approximation 00:39:14.300 --> 00:39:17.620 for X of 1 over 50. 00:39:17.620 --> 00:39:21.260 So probably you can also get distribution for X of 1 over 3, 00:39:21.260 --> 00:39:22.550 1 over 4. 00:39:22.550 --> 00:39:28.720 But at some point, the approximation will be very bad. 00:39:28.720 --> 00:39:31.820 So the key is to choose a right h. 00:39:31.820 --> 00:39:33.660 Because if you pick h to be too small, 00:39:33.660 --> 00:39:35.940 you will have a very good approximation 00:39:35.940 --> 00:39:37.190 to your distribution. 00:39:37.190 --> 00:39:39.987 But at the same time, it will take too much time 00:39:39.987 --> 00:39:40.570 to compute it. 00:39:45.470 --> 00:39:48.330 Any remarks from a more practical side? 00:39:51.650 --> 00:39:54.200 OK, so that's actually all I wanted 00:39:54.200 --> 00:39:56.990 to say about stochastic differential equations. 00:39:56.990 --> 00:39:59.860 Really the basic principle is there 00:39:59.860 --> 00:40:02.490 is such thing called stochastic differential equation. 00:40:02.490 --> 00:40:04.660 It can be solved. 00:40:04.660 --> 00:40:08.210 But most of the time, it won't have a closed form formula. 00:40:08.210 --> 00:40:09.870 And if you want to do it numerically, 00:40:09.870 --> 00:40:12.920 here are some possibilities. 00:40:12.920 --> 00:40:18.250 But I won't go any deeper inside. 00:40:18.250 --> 00:40:22.670 So the last math lecture I will conclude with heat equation. 00:40:26.269 --> 00:40:28.654 Yeah. 00:40:28.654 --> 00:40:33.440 AUDIENCE: The mean computations of [INAUDIBLE], 00:40:33.440 --> 00:40:37.720 some of the derivatives are sort of path-independent, 00:40:37.720 --> 00:40:39.220 or have path-independent solutions, 00:40:39.220 --> 00:40:40.850 so that you basically are looking 00:40:40.850 --> 00:40:44.980 at say the distribution at the terminal value 00:40:44.980 --> 00:40:48.760 and that determines the price of the derivative. 00:40:48.760 --> 00:40:51.820 There are other derivatives where things really 00:40:51.820 --> 00:40:57.510 are path-dependent, like with options where you have 00:40:57.510 --> 00:41:00.240 early exercise possibilities. 00:41:00.240 --> 00:41:02.420 When do you exercise, early or not? 00:41:02.420 --> 00:41:06.590 Then the tree methods are really good because at each element 00:41:06.590 --> 00:41:10.650 of the tree you can condition on whatever the path was. 00:41:10.650 --> 00:41:13.690 So keep that in mind, that when there's 00:41:13.690 --> 00:41:15.600 path dependence in the problem, you'll 00:41:15.600 --> 00:41:17.858 probably want to use one of these methods. 00:41:17.858 --> 00:41:18.608 PROFESSOR: Thanks. 00:41:18.608 --> 00:41:21.637 AUDIENCE: I know that if you're trying to break it down 00:41:21.637 --> 00:41:26.854 into simple random walks you can only use [INAUDIBLE]. 00:41:26.854 --> 00:41:30.000 But I've heard of people trying to use, instead 00:41:30.000 --> 00:41:31.740 of a binomial, a trinomial tree. 00:41:31.740 --> 00:41:35.420 PROFESSOR: Yes, so this statement actually 00:41:35.420 --> 00:41:37.986 is quite a universal statement. 00:41:37.986 --> 00:41:42.670 Brownian motion is a limit of many things, not just 00:41:42.670 --> 00:41:43.620 simple random walk. 00:41:43.620 --> 00:41:47.230 For example, if you take plus 1, 0, or minus 1 00:41:47.230 --> 00:41:49.190 and take it to the limit, that will also 00:41:49.190 --> 00:41:51.670 converge to the Brownian motion. 00:41:51.670 --> 00:41:54.980 That will be the trinomial and so on. 00:41:54.980 --> 00:41:57.090 And as Peter said, if you're going 00:41:57.090 --> 00:41:59.320 to use tree method to compute something, 00:41:59.320 --> 00:42:03.475 that will increase accuracy, if you take more possibilities 00:42:03.475 --> 00:42:05.890 at each step. 00:42:05.890 --> 00:42:07.530 Now, there is two ways to increase 00:42:07.530 --> 00:42:14.020 the accuracy is take more possibilities at each step 00:42:14.020 --> 00:42:16.210 or take smaller time scales. 00:42:20.090 --> 00:42:23.470 OK, so let's move on to the final topic, heat equation. 00:42:27.399 --> 00:42:29.570 Heat equation is not a stochastic differential 00:42:29.570 --> 00:42:32.630 equation, first of all. 00:42:32.630 --> 00:42:33.230 It's a PDE. 00:42:47.780 --> 00:42:50.650 That equation is known as a heat equation 00:42:50.650 --> 00:42:52.180 where t is like the time variable, 00:42:52.180 --> 00:42:54.570 x is like the space variable. 00:42:54.570 --> 00:42:56.940 And the reason we're interested in this heat 00:42:56.940 --> 00:42:59.340 equation in this course is, if you 00:42:59.340 --> 00:43:09.480 came to the previous lecture, maybe from Vasily last week, 00:43:09.480 --> 00:43:12.640 Black-Scholes equation, after change of variables, 00:43:12.640 --> 00:43:17.050 can be reduced to heat equation. 00:43:17.050 --> 00:43:19.900 That's one reason we're interested in it. 00:43:19.900 --> 00:43:22.310 And this is a really, really famous equation 00:43:22.310 --> 00:43:23.620 also in physics. 00:43:23.620 --> 00:43:27.150 So it was known before Black-Scholes equation. 00:43:27.150 --> 00:43:32.430 Particularly, this can be a model for-- equations 00:43:32.430 --> 00:43:33.590 that model this situation. 00:43:36.260 --> 00:43:42.544 So you have an infinite bar, very long and thin. 00:43:42.544 --> 00:43:43.585 It's perfectly insulated. 00:43:48.990 --> 00:43:53.850 So heat can only travel along the x-axis. 00:43:53.850 --> 00:43:58.080 And then at time 0, you have some heat distribution. 00:44:02.120 --> 00:44:07.076 At time 0, you know the heat distribution. 00:44:13.040 --> 00:44:17.230 Then this equation tells you the behavior of how the heat will 00:44:17.230 --> 00:44:20.020 be distributed at time t. 00:44:20.020 --> 00:44:25.570 So u of t of x, for fixed t, will be the distribution 00:44:25.570 --> 00:44:28.852 of the heat over the x-axis. 00:44:28.852 --> 00:44:30.560 That's why it's called the heat equation. 00:44:30.560 --> 00:44:31.935 That's where the name comes from. 00:44:35.050 --> 00:44:37.720 And this equation is very well understood. 00:44:37.720 --> 00:44:42.720 It does have a closed-form solution. 00:44:42.720 --> 00:44:44.960 And that's what I want to talk about. 00:44:48.250 --> 00:44:51.560 OK, so few observations before actually solving it. 00:44:58.450 --> 00:45:13.520 Remark one, if u_1 and u_2 satisfies heat equation, 00:45:13.520 --> 00:45:18.951 then u_1 plus u_2 also satisfies, also does. 00:45:21.850 --> 00:45:24.750 That's called linearity. 00:45:24.750 --> 00:45:25.840 Just plug it in. 00:45:25.840 --> 00:45:28.080 And you can figure it out. 00:45:28.080 --> 00:45:31.760 More generally that means, if you 00:45:31.760 --> 00:45:46.890 integrate a family of functions ds, where u_s all satisfy star, 00:45:46.890 --> 00:46:00.435 then this also satisfies star, as long 00:46:00.435 --> 00:46:02.300 as you use reasonable function. 00:46:02.300 --> 00:46:05.240 I'll just assume that we can switch the order of integration 00:46:05.240 --> 00:46:06.620 and differentiation. 00:46:06.620 --> 00:46:08.550 So it's the same thing. 00:46:08.550 --> 00:46:10.432 Instead of summation, I'm taking integration 00:46:10.432 --> 00:46:11.265 of lot of solutions. 00:46:14.410 --> 00:46:17.490 And why is that helpful? 00:46:17.490 --> 00:46:28.120 This is helpful because now it suffices 00:46:28.120 --> 00:46:36.160 to solve for-- what is it? 00:46:36.160 --> 00:46:44.345 Initial condition, u of t of x equals delta, delta function, 00:46:44.345 --> 00:46:44.950 of 0. 00:46:52.470 --> 00:46:55.970 That one is a little bit subtle. 00:46:55.970 --> 00:46:58.980 The Dirac delta function is just like an infinite ass 00:46:58.980 --> 00:47:00.100 at x equals 0. 00:47:00.100 --> 00:47:03.660 It's 0 everywhere else. 00:47:03.660 --> 00:47:05.950 And basically, in this example, what you're saying is, 00:47:05.950 --> 00:47:09.680 at time 0, you're putting like a massive amount of heat 00:47:09.680 --> 00:47:11.430 at a single point. 00:47:11.430 --> 00:47:14.800 And you're observing what's going to happen afterwards, 00:47:14.800 --> 00:47:17.380 how this heat will spread out. 00:47:17.380 --> 00:47:19.240 If you understand that, you can understand 00:47:19.240 --> 00:47:21.220 all initial conditions. 00:47:21.220 --> 00:47:23.160 Why is that? 00:47:23.160 --> 00:47:38.840 Because if u sub s t, x-- u_0-- is such solution, 00:47:38.840 --> 00:47:45.030 then integration of-- let me get it right-- 00:47:45.030 --> 00:48:10.546 u of t of s minus x ds is a solution with initial condition 00:48:10.546 --> 00:48:11.046 x(0, x). 00:48:16.006 --> 00:48:16.998 What is it? 00:48:37.334 --> 00:48:38.930 Sorry about that. 00:48:38.930 --> 00:48:41.720 So this is really the key. 00:48:41.720 --> 00:48:46.120 If you have a solution to the Dirac delta initial condition, 00:48:46.120 --> 00:48:49.340 then you can superimpose a lot of those solutions 00:48:49.340 --> 00:48:51.935 to obtain a solution for arbitrary initial condition. 00:48:55.030 --> 00:48:58.860 So this is based on that principle, because each of them 00:48:58.860 --> 00:48:59.950 is now a solution. 00:48:59.950 --> 00:49:04.890 If you superpose this, then that is a solution. 00:49:04.890 --> 00:49:06.640 And then if you plug it in, you figure out 00:49:06.640 --> 00:49:09.331 that actually it has satisfied this initial condition. 00:49:14.050 --> 00:49:18.260 That was my first observation. 00:49:18.260 --> 00:49:27.160 Second observation, second remark, 00:49:27.160 --> 00:49:44.930 is for the initial value u(0, x) equals a Dirac delta function, 00:49:44.930 --> 00:49:59.246 u of-- is a solution. 00:50:05.710 --> 00:50:08.320 So we know the solution for the Dirac delta part. 00:50:08.320 --> 00:50:10.950 First part, we figured out that if we know the solution 00:50:10.950 --> 00:50:12.800 for the Dirac delta function, then 00:50:12.800 --> 00:50:15.260 we can solve it for every single initial value. 00:50:15.260 --> 00:50:17.160 And for the initial value Dirac delta, 00:50:17.160 --> 00:50:23.300 that is the solution that solves the differential equation. 00:50:23.300 --> 00:50:26.490 So let me say a few words about this equation, actually 00:50:26.490 --> 00:50:29.002 one word. 00:50:29.002 --> 00:50:30.986 Have you seen this equation before? 00:50:35.090 --> 00:50:36.865 It's the p.d.f. of normal distribution. 00:50:36.865 --> 00:50:37.740 So what does it mean? 00:50:37.740 --> 00:50:41.820 It means, in this example, if you have a heat traveling 00:50:41.820 --> 00:50:45.280 along the x-axis, perfectly insulated, 00:50:45.280 --> 00:50:49.180 if you put a massive heat at this 0, at one point, 00:50:49.180 --> 00:50:53.090 at time 0, then at time t your heat 00:50:53.090 --> 00:50:55.640 will be distributed according to the normal distribution. 00:50:55.640 --> 00:50:58.014 In other words, assume that you have a bunch of particle. 00:50:58.014 --> 00:51:00.140 Heat is just like a bunch of particles, 00:51:00.140 --> 00:51:04.590 say millions of particles at a single point. 00:51:04.590 --> 00:51:07.410 And then you grab it. 00:51:07.410 --> 00:51:09.908 And then time t equals 0 you release it. 00:51:14.547 --> 00:51:16.880 Now the particle at time t will be distributed according 00:51:16.880 --> 00:51:18.190 to a normal distribution. 00:51:18.190 --> 00:51:24.300 In other words, each particle is like a Brownian motion. 00:51:24.300 --> 00:51:28.740 So for particle by particle, the location 00:51:28.740 --> 00:51:31.760 of its particle at time t will be kind of distributed 00:51:31.760 --> 00:51:33.400 like a Brownian motion. 00:51:33.400 --> 00:51:36.450 So if you have a massive amount of particles, 00:51:36.450 --> 00:51:38.590 altogether their distribution will look 00:51:38.590 --> 00:51:40.490 like a normal distribution. 00:51:40.490 --> 00:51:42.930 That's like its content. 00:51:42.930 --> 00:51:45.910 So that's also one way you see the appearance of a Brownian 00:51:45.910 --> 00:51:49.510 motion inside of this equation. 00:51:49.510 --> 00:51:52.570 It's like a bunch of Brownian motions happening together 00:51:52.570 --> 00:51:53.820 at the exact same time. 00:52:01.320 --> 00:52:06.841 And now we can just write down the solution. 00:52:06.841 --> 00:52:08.450 Let me be a little bit more precise. 00:52:18.600 --> 00:52:32.200 OK, for the heat equation delta u over delta t, 00:52:32.200 --> 00:52:44.590 with initial value u of 0, x equals some initial value, 00:52:44.590 --> 00:52:50.700 let's say v of x, and t greater than equal to 0. 00:53:00.510 --> 00:53:09.728 The solution is given by integration. 00:53:14.160 --> 00:53:46.524 u at t of x is equal to e to the minus-- let me get it right. 00:54:08.890 --> 00:54:12.680 Basically, I'm just combining this solution to there. 00:54:12.680 --> 00:54:18.090 Plugging in that here, you get that. 00:54:18.090 --> 00:54:20.280 So you have a explicit solution, no matter 00:54:20.280 --> 00:54:24.050 what the initial conditions are, initial conditions 00:54:24.050 --> 00:54:27.280 are given as, you can find an explicit solution 00:54:27.280 --> 00:54:32.090 at time t for all x. 00:54:32.090 --> 00:54:36.590 That means, once you change the Black-Scholes equation 00:54:36.590 --> 00:54:41.300 into the heat equation, you now have a closed-form solution 00:54:41.300 --> 00:54:41.976 for it. 00:54:45.550 --> 00:54:51.325 In that case, it's like a backward heat equation. 00:54:51.325 --> 00:54:53.200 And what will happen is the initial condition 00:54:53.200 --> 00:54:56.890 you should think of as a final payout function. 00:54:56.890 --> 00:54:58.700 The final payout function you integrate 00:54:58.700 --> 00:55:00.860 according to this distribution. 00:55:00.860 --> 00:55:03.486 And then you get the value at time t equals 0. 00:55:07.220 --> 00:55:09.910 The detail, one of the final project 00:55:09.910 --> 00:55:11.760 is to actually carry out all the details. 00:55:11.760 --> 00:55:13.990 So I will stop here. 00:55:13.990 --> 00:55:17.180 Anyway, we didn't see how the Black-Scholes equation actually 00:55:17.180 --> 00:55:18.410 changed into heat equation. 00:55:22.470 --> 00:55:24.890 If you want to do that project, it 00:55:24.890 --> 00:55:26.930 will be good to have this in mind. 00:55:26.930 --> 00:55:29.030 It will help. 00:55:29.030 --> 00:55:30.030 Any questions? 00:55:34.530 --> 00:55:38.938 OK, so I think that's all I have. 00:55:38.938 --> 00:55:41.800 I think I'll end a little bit early today. 00:55:41.800 --> 00:55:45.155 So that will be the last math lecture for the semester. 00:55:45.155 --> 00:55:48.390 From now on you'll only have application lectures. 00:55:48.390 --> 00:55:55.910 There are great lectures coming up, I hope, and I know. 00:55:55.910 --> 00:55:59.390 So you should come and really enjoy now. 00:55:59.390 --> 00:56:02.300 You went through all this hard work. 00:56:02.300 --> 00:56:04.465 Now it's time to enjoy.