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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.