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PROFESSOR: So far
we had a function,
9
00:00:23,570 --> 00:00:27,840
and then we differentiated to
get an equation of this type.
10
00:00:27,840 --> 00:00:29,810
But now, we're
given this equation.
11
00:00:29,810 --> 00:00:31,320
And we have to go
backwards, want
12
00:00:31,320 --> 00:00:35,706
to find the stochastic process
that satisfies this equation.
13
00:00:35,706 --> 00:00:45,700
So goal is to find
a stochastic process
14
00:00:45,700 --> 00:00:48,295
X(t) satisfying this equation.
15
00:00:55,910 --> 00:01:06,130
In other words, we want X of
t to be the integral of mu dS
16
00:01:06,130 --> 00:01:07,325
plus sigma...
17
00:01:17,460 --> 00:01:18,740
The goal is clear.
18
00:01:18,740 --> 00:01:19,340
We want that.
19
00:01:22,520 --> 00:01:26,505
And so these type
of equations are
20
00:01:26,505 --> 00:01:30,630
called differential equations,
I hope you already know that.
21
00:01:30,630 --> 00:01:33,220
Also for PDE, partial
differential equations,
22
00:01:33,220 --> 00:01:35,590
so even when it's
not stochastic,
23
00:01:35,590 --> 00:01:38,310
these are not easy problems.
24
00:01:38,310 --> 00:01:40,150
At least not easy
in the sense that,
25
00:01:40,150 --> 00:01:42,310
if you're given an
equation, typically you
26
00:01:42,310 --> 00:01:45,290
don't expect to have a
closed form solution.
27
00:01:45,290 --> 00:01:48,430
So even if you find this
X, most of the time,
28
00:01:48,430 --> 00:01:50,910
it's not in a very good form.
29
00:01:50,910 --> 00:01:55,130
Still, a very important
result that you should first
30
00:01:55,130 --> 00:01:58,790
note before trying to solve any
of the differential equation
31
00:01:58,790 --> 00:02:03,690
is that, as long as mu and
sigma are reasonable functions,
32
00:02:03,690 --> 00:02:05,740
there does exist a solution.
33
00:02:05,740 --> 00:02:07,730
And it's unique.
34
00:02:07,730 --> 00:02:10,630
So we have the same
correspondence with this PDE.
35
00:02:10,630 --> 00:02:13,080
You're given a PDE, or given
a differential equation, not
36
00:02:13,080 --> 00:02:14,970
a stochastic
differential equation,
37
00:02:14,970 --> 00:02:18,060
you know that, if you're given
a reasonable differential
38
00:02:18,060 --> 00:02:21,550
equation, then a
solution exists.
39
00:02:21,550 --> 00:02:22,730
And it's unique.
40
00:02:22,730 --> 00:02:26,416
So the same principle
holds in stochastic world.
41
00:02:26,416 --> 00:02:28,360
Now, let me state it formally.
42
00:02:33,070 --> 00:02:47,040
This stochastic equation star
has a solution that is unique,
43
00:02:47,040 --> 00:02:51,920
of course with boundary
condition-- has a solution.
44
00:02:51,920 --> 00:03:18,344
And given the
initial points-- so
45
00:03:18,344 --> 00:03:20,760
if you're given the initial
point of a stochastic process,
46
00:03:20,760 --> 00:03:23,520
then a solution is unique.
47
00:03:30,840 --> 00:03:43,210
Just check, yes-- as long as
mu and sigma are reasonable.
48
00:03:51,310 --> 00:03:55,180
One way it can be reasonable,
if it satisfies this conditions.
49
00:04:27,800 --> 00:04:29,300
These are very
technical conditions.
50
00:04:29,300 --> 00:04:31,258
But at least let me parse
to you what they are.
51
00:04:31,258 --> 00:04:36,730
They say, if you fix a time
coordinate and you change x,
52
00:04:36,730 --> 00:04:39,630
you look at the difference
between the values of mu
53
00:04:39,630 --> 00:04:44,650
when you change the second
variable and the sigma.
54
00:04:44,650 --> 00:04:51,460
Then the change in your function
is bounded by the distance
55
00:04:51,460 --> 00:04:56,570
between the two points
by some constant K.
56
00:04:56,570 --> 00:05:00,730
So mu and sigma cannot change
too much when you change
57
00:05:00,730 --> 00:05:04,660
your space variable
a little bit.
58
00:05:04,660 --> 00:05:07,130
It can only change up to
the distance of how much you
59
00:05:07,130 --> 00:05:08,540
change the coordinate.
60
00:05:08,540 --> 00:05:09,970
So that's the first condition.
61
00:05:09,970 --> 00:05:16,730
Second condition says,
when you grow your x,
62
00:05:16,730 --> 00:05:18,382
very similar condition.
63
00:05:18,382 --> 00:05:20,340
Essentially it says it
cannot blow up too fast,
64
00:05:20,340 --> 00:05:21,690
the whole thing.
65
00:05:21,690 --> 00:05:25,690
This one is something about the
difference between two values.
66
00:05:25,690 --> 00:05:31,309
This one is about how it expands
as your space variable grows.
67
00:05:31,309 --> 00:05:32,600
These are technical conditions.
68
00:05:32,600 --> 00:05:37,620
And many cases, they will hold.
69
00:05:37,620 --> 00:05:41,450
So don't worry too much about
the technical conditions.
70
00:05:41,450 --> 00:05:45,160
Important thing here is that,
given a differential equation,
71
00:05:45,160 --> 00:05:48,940
you don't expect to
have a good closed form.
72
00:05:48,940 --> 00:05:52,880
But you do expect to have
a solution of some form.
73
00:05:57,260 --> 00:06:01,060
OK, let's work
out some examples.
74
00:06:05,260 --> 00:06:08,240
Here is one of the few
stochastic differential
75
00:06:08,240 --> 00:06:09,590
equations that can be solved.
76
00:06:21,090 --> 00:06:22,360
This one can be solved.
77
00:06:22,360 --> 00:06:26,376
And you already know what X is.
78
00:06:26,376 --> 00:06:29,700
But let's pretend you
don't know what X is.
79
00:06:29,700 --> 00:06:31,316
And let's try to solve it.
80
00:06:34,490 --> 00:06:36,530
I will show you
an approach, which
81
00:06:36,530 --> 00:06:40,280
can solve some differential
equations, some SDEs.
82
00:06:40,280 --> 00:06:43,390
But this really won't
happen that much.
83
00:06:43,390 --> 00:06:48,890
Still, it's like
a starting point.
84
00:06:48,890 --> 00:06:50,280
There's that.
85
00:06:50,280 --> 00:06:55,500
And assume X(0) [INAUDIBLE].
86
00:06:55,500 --> 00:06:56,855
And mu, sigma are constants.
87
00:07:05,130 --> 00:07:07,080
Just like when solving
differential equations,
88
00:07:07,080 --> 00:07:10,430
first thing you'll do is
just guess, suppose, guess.
89
00:07:23,870 --> 00:07:45,252
If you want this to happen,
then d of X of t is-- OK.
90
00:07:48,210 --> 00:07:51,290
x is just the second variable.
91
00:07:51,290 --> 00:07:52,872
And then these
two have to match.
92
00:07:52,872 --> 00:07:54,080
That has to be equal to that.
93
00:07:54,080 --> 00:07:55,830
That has to be equal to that.
94
00:07:55,830 --> 00:08:10,320
So we know that del f
over del t, mu of X of t
95
00:08:10,320 --> 00:08:11,516
is equal to mu of f.
96
00:08:20,240 --> 00:08:24,260
So we assumed that this is a
solution then differentiated
97
00:08:24,260 --> 00:08:24,990
that.
98
00:08:24,990 --> 00:08:28,470
And then, if that's a solution,
these all have to match.
99
00:08:28,470 --> 00:08:30,210
You get this equation.
100
00:08:30,210 --> 00:08:32,669
If you look at
that, that tells you
101
00:08:32,669 --> 00:08:36,429
that f is an exponential
function in the x variable.
102
00:08:36,429 --> 00:08:40,640
So it's e to the sigma times x.
103
00:08:40,640 --> 00:08:44,750
And then we have a constant
term, plus some a times
104
00:08:44,750 --> 00:08:45,730
a function of t.
105
00:08:50,680 --> 00:08:54,770
The only way it can happen
is if it's in this form.
106
00:08:54,770 --> 00:08:57,000
So it's exponential
function in x.
107
00:08:57,000 --> 00:09:02,130
And in the time variable,
it's just some constant.
108
00:09:02,130 --> 00:09:04,540
When you fix a t,
it's a constant.
109
00:09:04,540 --> 00:09:06,310
And when you fix
a t and change x,
110
00:09:06,310 --> 00:09:08,590
it has to look like an
exponential function.
111
00:09:08,590 --> 00:09:11,064
It has to be in this form,
just by the second equation.
112
00:09:13,730 --> 00:09:16,590
Now, go back to this equation.
113
00:09:16,590 --> 00:09:25,520
What you get is partial f over
partial t is now a times g
114
00:09:25,520 --> 00:09:30,350
prime of t, f...
115
00:09:43,580 --> 00:09:44,580
AUDIENCE: Excuse me.
116
00:09:44,580 --> 00:09:46,200
PROFESSOR: Mm-hm.
117
00:09:46,200 --> 00:09:49,420
AUDIENCE: That one in
second to last line, yeah.
118
00:09:49,420 --> 00:09:52,790
So why is it minus
mu there at the end?
119
00:09:52,790 --> 00:09:55,364
PROFESSOR: It's equal.
120
00:09:55,364 --> 00:09:56,364
AUDIENCE: Oh, all right.
121
00:09:56,364 --> 00:09:57,030
PROFESSOR: Yeah.
122
00:10:03,860 --> 00:10:06,260
OK, and then let's plug it in.
123
00:10:06,260 --> 00:10:13,270
So we have a of g prime
t, f, plus 1/2 of sigma
124
00:10:13,270 --> 00:10:17,094
square f equals mu of f.
125
00:10:17,094 --> 00:10:25,525
In other words, a of g
prime of t is mu minus...
126
00:10:25,525 --> 00:10:26,025
square.
127
00:10:29,590 --> 00:10:36,502
g(t) is mu times some
constant c_1 plus c_2.
128
00:10:44,380 --> 00:10:48,980
OK, and then what we got is
original function f of t of x
129
00:10:48,980 --> 00:10:54,980
is e to the sigma*x plus mu
minus 1 over 2 sigma square t
130
00:10:54,980 --> 00:10:57,590
plus some constant.
131
00:10:57,590 --> 00:11:01,520
And that constant can
be chosen because we
132
00:11:01,520 --> 00:11:09,270
have the initial
condition, x of 0 equals 0.
133
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.
134
00:11:14,203 --> 00:11:16,510
That has to be x_0.
135
00:11:16,510 --> 00:11:21,701
In different words, this is just
x_0 times e to the sigma*x plus
136
00:11:21,701 --> 00:11:25,550
mu minus 1 over 2
sigma square of t.
137
00:11:25,550 --> 00:11:27,802
Just as we expected,
we got this.
138
00:11:33,210 --> 00:11:35,940
So some stochastic
differential equations
139
00:11:35,940 --> 00:11:39,920
can be solved by analyzing it.
140
00:11:39,920 --> 00:11:43,010
But I'm not necessarily saying
that this is a better way
141
00:11:43,010 --> 00:11:45,020
to do it than just guessing.
142
00:11:45,020 --> 00:11:46,830
Just looking at
it, and, you know,
143
00:11:46,830 --> 00:11:50,027
OK, it has to be
exponential function.
144
00:11:50,027 --> 00:11:50,610
You know what?
145
00:11:50,610 --> 00:11:54,430
I'll just figure
out what these are.
146
00:11:54,430 --> 00:11:56,030
And I'll come up
with this formula
147
00:11:56,030 --> 00:11:59,110
without going through
all those analysis.
148
00:11:59,110 --> 00:12:01,435
I'm not saying that's a
worse way than actually going
149
00:12:01,435 --> 00:12:02,310
through the analysis.
150
00:12:02,310 --> 00:12:05,250
Because, in fact, what we
did is we kind of already
151
00:12:05,250 --> 00:12:08,780
knew the answer and are
fitting into that answer.
152
00:12:12,940 --> 00:12:14,790
Still, it can be
one approach where
153
00:12:14,790 --> 00:12:17,590
you don't have a reasonable
guess of what the X_t has
154
00:12:17,590 --> 00:12:21,772
to be, maybe try to break it
down into pieces like this
155
00:12:21,772 --> 00:12:24,410
and backtrack to figure
out the function.
156
00:12:24,410 --> 00:12:26,260
Let me give you one
more example where we
157
00:12:26,260 --> 00:12:28,580
do have an explicit solution.
158
00:12:28,580 --> 00:12:30,700
And then I'll move
on and show you
159
00:12:30,700 --> 00:12:33,367
how to do when there
is no explicit solution
160
00:12:33,367 --> 00:12:35,700
or when you don't know how
to find an explicit solution.
161
00:12:38,304 --> 00:12:39,710
Maybe let's keep that there.
162
00:12:52,240 --> 00:12:55,781
Second equation is called this.
163
00:13:16,655 --> 00:13:18,080
What's the difference?
164
00:13:18,080 --> 00:13:22,830
The only difference is that
you don't have an X here.
165
00:13:22,830 --> 00:13:26,560
So previously, our
main drift term
166
00:13:26,560 --> 00:13:30,330
also was proportional
to the current value.
167
00:13:30,330 --> 00:13:33,980
And the error was also
dependent on the current value
168
00:13:33,980 --> 00:13:36,350
or is proportional
to the current value.
169
00:13:36,350 --> 00:13:38,580
But here now the drift
term is something
170
00:13:38,580 --> 00:13:42,610
like an exponential--
minus exponential.
171
00:13:42,610 --> 00:13:45,460
But still, it's proportional
to the current value.
172
00:13:45,460 --> 00:13:49,120
But the error term
is just some noise.
173
00:13:49,120 --> 00:13:51,690
Irrelevant of what
the value is, this has
174
00:13:51,690 --> 00:13:54,070
the same variance as the error.
175
00:13:54,070 --> 00:13:58,550
So it's a slightly
different-- oh, what is it?
176
00:13:58,550 --> 00:14:00,770
It's a slightly
different process.
177
00:14:00,770 --> 00:14:09,100
And it's known as
Ornstein-Uhlenbeck process.
178
00:14:18,530 --> 00:14:23,410
And this is used to model
a mean-reverting stochastic
179
00:14:23,410 --> 00:14:24,980
process.
180
00:14:24,980 --> 00:14:30,230
For alpha greater than 0, notice
that if X deviates from 0,
181
00:14:30,230 --> 00:14:33,940
this gives a force that drives
the stochastic process back
182
00:14:33,940 --> 00:14:36,940
to 0, that's
negatively proportional
183
00:14:36,940 --> 00:14:37,902
to your current value.
184
00:14:40,740 --> 00:14:44,330
So yeah, this is used to model
some mean-reverting stochastic
185
00:14:44,330 --> 00:14:45,130
processes.
186
00:14:45,130 --> 00:14:50,785
And they first used it to
study the behavior of gases.
187
00:14:57,230 --> 00:14:58,330
I don't exactly see why.
188
00:14:58,330 --> 00:15:01,810
But that's what they say.
189
00:15:01,810 --> 00:15:03,710
Anyway, so this is
another thing that
190
00:15:03,710 --> 00:15:05,510
can be solved by doing
similar analysis.
191
00:15:10,490 --> 00:15:14,590
But if you try the same
method, it will fail.
192
00:15:14,590 --> 00:15:34,550
So as a test function, your
guess, initial guess, will be--
193
00:15:34,550 --> 00:15:36,576
or a(0) is equal to 1.
194
00:15:40,878 --> 00:15:44,070
Now, honestly, I don't know
how to come up with this guess.
195
00:15:49,137 --> 00:15:50,720
Probably, if you're
really experienced
196
00:15:50,720 --> 00:15:52,870
with stochastic
differential equations,
197
00:15:52,870 --> 00:15:56,510
you'll see some
form, like you'll
198
00:15:56,510 --> 00:16:00,340
have some feeling on how
this process will look like.
199
00:16:00,340 --> 00:16:03,330
And then try this, try that,
and eventually something
200
00:16:03,330 --> 00:16:05,266
might succeed.
201
00:16:05,266 --> 00:16:06,890
That's the best
explanation I can give.
202
00:16:06,890 --> 00:16:08,264
I can't really
give you intuition
203
00:16:08,264 --> 00:16:10,590
why that's the right guess.
204
00:16:10,590 --> 00:16:13,570
Given some stochastic
differential equation,
205
00:16:13,570 --> 00:16:15,356
I don't know how to
say that you should
206
00:16:15,356 --> 00:16:16,730
start with this
kind of function,
207
00:16:16,730 --> 00:16:18,722
this kind of function.
208
00:16:18,722 --> 00:16:20,430
And it was the same
when, if you remember
209
00:16:20,430 --> 00:16:22,980
how we solved ordinary
differential equations
210
00:16:22,980 --> 00:16:25,680
or partial differential
equations, most of the time
211
00:16:25,680 --> 00:16:27,206
there is no good guess.
212
00:16:27,206 --> 00:16:31,690
It's only when your given
formula has some specific form
213
00:16:31,690 --> 00:16:33,334
such a thing happens.
214
00:16:33,334 --> 00:16:37,250
So let's see what happens here.
215
00:16:37,250 --> 00:16:39,310
That was given.
216
00:16:39,310 --> 00:16:42,990
Now, let's do exactly
the same as before.
217
00:16:42,990 --> 00:16:46,940
Differentiate it,
and let me go slow.
218
00:16:46,940 --> 00:16:49,240
So we have a prime of t.
219
00:16:49,240 --> 00:16:54,440
By chain rule, a prime
of t and that value,
220
00:16:54,440 --> 00:16:59,240
that part will be equal
to X(t) over a(t).
221
00:16:59,240 --> 00:17:00,440
This is chain rule.
222
00:17:00,440 --> 00:17:03,390
So I differentiate
that to get a prime t.
223
00:17:03,390 --> 00:17:05,580
That stays just as it was.
224
00:17:05,580 --> 00:17:09,609
But that can be rewritten
as X(t) divided by a(t).
225
00:17:09,609 --> 00:17:15,329
And then plus a(t) times the
differential of that one.
226
00:17:15,329 --> 00:17:17,109
And that is just b(t)*dB(t).
227
00:17:29,180 --> 00:17:31,870
You don't have to differentiate
that once more, even
228
00:17:31,870 --> 00:17:34,630
though it's stochastic
calculus, because that's
229
00:17:34,630 --> 00:17:35,500
a very subtle point.
230
00:17:35,500 --> 00:17:39,090
And there's also one exercise
about it in your homework.
231
00:17:39,090 --> 00:17:42,300
But when you have a given
stochastic process written
232
00:17:42,300 --> 00:17:46,190
already in this integral
form, if we remember
233
00:17:46,190 --> 00:17:49,410
the definition of an integral,
at least how I defined it,
234
00:17:49,410 --> 00:17:52,560
is that it was an inverse
operation of a differential.
235
00:17:52,560 --> 00:17:55,660
So when you differentiate
this, you just get that term.
236
00:17:55,660 --> 00:17:57,290
What I'm trying to
say is, there is
237
00:17:57,290 --> 00:18:02,830
no term, no term where you have
to differentiate this one more.
238
00:18:07,060 --> 00:18:11,780
Prime dt, something like
that, we don't have this term.
239
00:18:15,180 --> 00:18:16,170
This can be confusing.
240
00:18:16,170 --> 00:18:17,360
But think about it.
241
00:18:22,580 --> 00:18:26,690
Now, we laid it out
and just compare.
242
00:18:26,690 --> 00:18:32,110
So minus alpha of X of
t is equal to a prime t
243
00:18:32,110 --> 00:18:36,260
over a(t) X(t).
244
00:18:36,260 --> 00:18:42,590
And your second term,
sigma*dB(t) is equal to a(t)
245
00:18:42,590 --> 00:18:43,400
times b(t).
246
00:18:47,240 --> 00:18:48,660
But these two cancel.
247
00:18:51,240 --> 00:18:57,746
And we see that a(t) has to
be e to the minus alpha t.
248
00:19:06,510 --> 00:19:09,785
This says that is
an exponential.
249
00:19:09,785 --> 00:19:10,660
Now, plug it in here.
250
00:19:10,660 --> 00:19:11,201
You get b(t).
251
00:19:17,772 --> 00:19:18,940
And that's it.
252
00:19:18,940 --> 00:19:21,096
So plug it back in.
253
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
254
00:19:32,085 --> 00:19:33,067
to the alpha*s.
255
00:19:56,640 --> 00:20:02,670
So this is a variance,
expectation is 0,
256
00:20:02,670 --> 00:20:05,210
because that's a
Brownian motion.
257
00:20:05,210 --> 00:20:08,420
This term, as we
expected, as time passes,
258
00:20:08,420 --> 00:20:12,776
goes to 0, exponential decay.
259
00:20:12,776 --> 00:20:18,576
And that is kind of hinted by
this fact, the mean reversion.
260
00:20:18,576 --> 00:20:21,610
So if you start from some
value, at least the drift term
261
00:20:21,610 --> 00:20:25,380
will go to 0 quite quickly.
262
00:20:25,380 --> 00:20:29,120
And then the important
term will be the noise term
263
00:20:29,120 --> 00:20:30,200
or the variance term.
264
00:20:35,237 --> 00:20:35,820
Any questions?
265
00:20:45,750 --> 00:20:49,991
And I'm really emphasizing
a lot of times today,
266
00:20:49,991 --> 00:20:51,490
but really you can
forget about what
267
00:20:51,490 --> 00:20:53,370
I did in the past two
boards, this board
268
00:20:53,370 --> 00:20:54,789
and the previous board.
269
00:20:54,789 --> 00:20:56,705
Because most of the
times, it will be useless.
270
00:20:59,220 --> 00:21:01,970
And so now I will
describe what you'll
271
00:21:01,970 --> 00:21:05,090
do if you're given a stochastic
differential equation,
272
00:21:05,090 --> 00:21:06,991
and you have a computer
in front of you.
273
00:21:15,860 --> 00:21:18,038
What if such method fails?
274
00:21:30,030 --> 00:21:31,570
And it will fail
most of the time.
275
00:21:36,210 --> 00:21:38,100
That's when we use
these techniques called
276
00:21:38,100 --> 00:21:53,840
finite difference method,
Monte Carlo simulation, or tree
277
00:21:53,840 --> 00:21:54,340
method.
278
00:22:04,130 --> 00:22:06,250
The finite difference
method, you probably
279
00:22:06,250 --> 00:22:09,610
already saw it, if you took a
differential equation course.
280
00:22:09,610 --> 00:22:10,740
But let me review it.
281
00:22:19,520 --> 00:22:24,230
This is for PDEs or
ODEs, for ODE, PDE, not
282
00:22:24,230 --> 00:22:26,636
stochastic
differential equations.
283
00:22:26,636 --> 00:22:29,010
But it can be adapted to work
for stochastic differential
284
00:22:29,010 --> 00:22:30,640
equations.
285
00:22:30,640 --> 00:22:32,540
So I'll work with an example.
286
00:22:36,090 --> 00:22:46,445
Let u of t be u prime
of t plus-- u prime of t
287
00:22:46,445 --> 00:22:53,220
be u(t) plus 2,
u_0 is equal to 0.
288
00:22:53,220 --> 00:22:54,630
Now, this has an exact solution.
289
00:22:54,630 --> 00:22:57,095
But let's pretend that
there is no exact solution.
290
00:22:57,095 --> 00:23:00,890
And if you want to
do it numerically,
291
00:23:00,890 --> 00:23:04,180
you want to find the value of
u equals-- u(1) numerically.
292
00:23:07,155 --> 00:23:08,660
And here's what
you're going to do.
293
00:23:08,660 --> 00:23:10,660
You're going to chop up
the interval from 0 to 1
294
00:23:10,660 --> 00:23:12,590
into very fine pieces.
295
00:23:12,590 --> 00:23:17,050
So from 0 to 1, chop it
down into tiny pieces.
296
00:23:20,460 --> 00:23:22,100
And since I'm in
front of a blackboard,
297
00:23:22,100 --> 00:23:25,190
my tiniest piece will
be 1 over 2 and 1.
298
00:23:25,190 --> 00:23:27,550
I'll just take two steps.
299
00:23:27,550 --> 00:23:29,400
But you should think
of it as really
300
00:23:29,400 --> 00:23:33,090
repeating this a lot of times.
301
00:23:33,090 --> 00:23:35,480
I'll call my step to be h.
302
00:23:35,480 --> 00:23:40,800
So in my case, I'm increasing my
steps by 1 over 2 at each time.
303
00:23:40,800 --> 00:23:44,710
So what is u of 1 over 2?
304
00:23:44,710 --> 00:23:47,135
Approximately, by
Taylor's formula,
305
00:23:47,135 --> 00:23:53,844
it's u(0) plus 1/2
times u prime of 0.
306
00:23:53,844 --> 00:23:55,010
That's Taylor approximation.
307
00:23:57,590 --> 00:23:59,910
OK, u(0) we already know.
308
00:23:59,910 --> 00:24:02,650
It's given to be equal to 0.
309
00:24:02,650 --> 00:24:05,700
u prime of 0, on the other hand,
is given by this differential
310
00:24:05,700 --> 00:24:07,520
equation.
311
00:24:07,520 --> 00:24:12,830
So it's 1 over 2 times
5 times u(0) plus 2.
312
00:24:12,830 --> 00:24:13,930
u(0) is 0.
313
00:24:13,930 --> 00:24:15,040
So we get equal to 1.
314
00:24:17,940 --> 00:24:20,932
Like we have this value
equal to 1, approximately.
315
00:24:20,932 --> 00:24:22,015
I don't know what happens.
316
00:24:22,015 --> 00:24:24,410
But it will be close to 1.
317
00:24:24,410 --> 00:24:28,170
And then for the next thing, u1.
318
00:24:28,170 --> 00:24:32,300
This one is, again by Taylor
approximation, is u of 1 over 2
319
00:24:32,300 --> 00:24:37,720
plus 1 over 2 u
prime of 1 over 2.
320
00:24:37,720 --> 00:24:40,520
And now you know the value of u
of 1 over 2, approximate value,
321
00:24:40,520 --> 00:24:41,277
by this.
322
00:24:41,277 --> 00:24:41,860
So plug it in.
323
00:24:41,860 --> 00:24:48,552
You have 1 plus 1 over 2 and,
again, 5 times u(1/2) plus 2.
324
00:24:48,552 --> 00:24:50,000
If you want to do
the computation,
325
00:24:50,000 --> 00:24:51,510
it should give 9 over 2.
326
00:24:55,740 --> 00:24:58,240
It's really simple.
327
00:24:58,240 --> 00:25:01,790
The key idea here
is just u prime
328
00:25:01,790 --> 00:25:05,720
is given by an
equation, this equation.
329
00:25:05,720 --> 00:25:09,267
So you can compute it once
you know the value of u
330
00:25:09,267 --> 00:25:09,850
at that point.
331
00:25:12,610 --> 00:25:16,030
And basically, the method is
saying take h to be very small,
332
00:25:16,030 --> 00:25:17,590
like 1 over 100.
333
00:25:17,590 --> 00:25:19,540
Then you just repeat
it 1 over 100 times.
334
00:25:19,540 --> 00:25:23,760
So the equation is the
i plus 1 step value
335
00:25:23,760 --> 00:25:30,232
can be approximated from
the i-th value plus h times
336
00:25:30,232 --> 00:25:33,064
u prime of h.
337
00:25:33,064 --> 00:25:34,560
Now repeat it and repeat it.
338
00:25:34,560 --> 00:25:36,570
And you reach u of 1.
339
00:25:36,570 --> 00:25:38,250
And there is a
theorem saying, again,
340
00:25:38,250 --> 00:25:42,060
if the differential
equation is reasonable, then
341
00:25:42,060 --> 00:25:44,355
that will approach the
true value as you take h
342
00:25:44,355 --> 00:25:45,520
to be smaller and smaller.
343
00:25:48,350 --> 00:25:50,360
That's called the
finite difference method
344
00:25:50,360 --> 00:25:52,921
for differential equations.
345
00:25:52,921 --> 00:25:55,170
And you can do the exact
same thing for two variables,
346
00:25:55,170 --> 00:25:55,669
let's say.
347
00:26:07,240 --> 00:26:10,050
And what we showed was for one
variable, finite difference
348
00:26:10,050 --> 00:26:18,412
method, we want to find the
value of u, function u of t.
349
00:26:18,412 --> 00:26:23,040
We took values at 0, h, 2h, 3h.
350
00:26:23,040 --> 00:26:26,680
Using that, we did some
approximation, like that,
351
00:26:26,680 --> 00:26:29,670
and found the value.
352
00:26:29,670 --> 00:26:33,340
Now, suppose we want
to find, similarly,
353
00:26:33,340 --> 00:26:38,920
a two-variable function,
let's say v of t and x.
354
00:26:38,920 --> 00:26:42,340
And we want to find
the value of v of 1, 1.
355
00:26:42,340 --> 00:26:44,292
Now the boundary
conditions are these.
356
00:26:44,292 --> 00:26:45,666
We already know
these boundaries.
357
00:26:49,650 --> 00:26:51,520
I won't really show
you by example.
358
00:26:51,520 --> 00:26:55,560
But what we're going to do
now is compute this value
359
00:26:55,560 --> 00:26:59,150
based on these two variables.
360
00:26:59,150 --> 00:27:00,300
So it's just the same.
361
00:27:00,300 --> 00:27:01,970
Taylor expansion
for two variables
362
00:27:01,970 --> 00:27:05,610
will allow you to compute this
value from these two values.
363
00:27:05,610 --> 00:27:09,072
Then compute this from these
two, this from these two,
364
00:27:09,072 --> 00:27:11,230
and just fill out the
whole grid like that,
365
00:27:11,230 --> 00:27:14,380
just fill out layer by layer.
366
00:27:14,380 --> 00:27:16,510
At some point, you're
going to reach this.
367
00:27:16,510 --> 00:27:20,320
And then you'll have an
approximate value of that.
368
00:27:20,320 --> 00:27:24,080
So you chop up your
domain into fine pieces
369
00:27:24,080 --> 00:27:25,970
and then take the limit.
370
00:27:25,970 --> 00:27:27,500
And most cases, it will work.
371
00:27:31,160 --> 00:27:33,495
Why does it not work for
stochastic differential
372
00:27:33,495 --> 00:27:33,995
equations?
373
00:27:38,690 --> 00:27:40,490
Kind of works, but
the only problem
374
00:27:40,490 --> 00:27:44,490
is we don't know which
value we're looking at,
375
00:27:44,490 --> 00:27:47,670
we're interested in.
376
00:27:47,670 --> 00:27:51,050
So let me phrase it a
little bit differently.
377
00:27:54,770 --> 00:27:58,140
You're given a differential
equation of the form dX
378
00:27:58,140 --> 00:28:12,930
equals mu dt plus t dB of t
and time variable and space
379
00:28:12,930 --> 00:28:15,540
variable.
380
00:28:15,540 --> 00:28:21,850
Now, if you want to compute your
value at time 2h based on value
381
00:28:21,850 --> 00:28:29,760
h, in this picture, I
told you that this point
382
00:28:29,760 --> 00:28:33,030
came from these two points.
383
00:28:33,030 --> 00:28:38,104
But when it's stochastic, it
could depend on everything.
384
00:28:38,104 --> 00:28:39,520
You don't know
where it came from.
385
00:28:39,520 --> 00:28:41,061
This point could
have come from here.
386
00:28:41,061 --> 00:28:42,390
It could have came from here.
387
00:28:42,390 --> 00:28:44,960
It could have came from
here, came from here.
388
00:28:44,960 --> 00:28:47,170
You don't really know.
389
00:28:47,170 --> 00:28:50,900
But what you know is you have
a probability distribution.
390
00:28:50,900 --> 00:28:57,350
So what I'm trying
to say is now,
391
00:28:57,350 --> 00:28:59,140
if you want to adapt
this method, what
392
00:28:59,140 --> 00:29:04,916
you're going to do is take a
sample Brownian motion path.
393
00:29:11,900 --> 00:29:15,260
That means just, according
to the distribution
394
00:29:15,260 --> 00:29:21,010
of the Brownian motion, take
one path and use that path.
395
00:29:21,010 --> 00:29:29,050
Once we fix a path,
once a path is fixed,
396
00:29:29,050 --> 00:29:31,840
we can exactly know where
each value comes from.
397
00:29:31,840 --> 00:29:35,230
We know how to backtrack.
398
00:29:42,360 --> 00:29:44,680
That means, instead of
all these possibilities,
399
00:29:44,680 --> 00:29:49,170
we have one fixed
possibility, like that.
400
00:29:52,470 --> 00:29:54,570
So just use that finite
difference method
401
00:29:54,570 --> 00:29:57,990
with that fixed path.
402
00:29:57,990 --> 00:30:01,155
That will be the idea.
403
00:30:01,155 --> 00:30:03,975
Let me do it a little
bit more formally.
404
00:30:10,170 --> 00:30:11,470
And here is how it works.
405
00:30:16,230 --> 00:30:30,020
If we have a fixed sample path
for Brownian motion of B(t),
406
00:30:30,020 --> 00:30:41,570
then X at time i plus 1 of h
is approximately equal to X
407
00:30:41,570 --> 00:30:54,930
at time a of h plus h times
dx at that time i of h,
408
00:30:54,930 --> 00:30:59,750
just by the exact
same Taylor expansion.
409
00:30:59,750 --> 00:31:08,966
And then d of X we know to
be-- that is equal to mu
410
00:31:08,966 --> 00:31:18,855
of dt plus-- oh, mu of dt is h.
411
00:31:18,855 --> 00:31:22,678
So let me write it like that,
sigma times d of X-- dB.
412
00:31:27,350 --> 00:31:37,350
And these mu depend on
[? their paths, ?] x at i of h
413
00:31:37,350 --> 00:31:39,845
dt, sigma...
414
00:31:56,850 --> 00:31:59,520
With that, here to here
is Taylor expansion.
415
00:31:59,520 --> 00:32:01,780
Here to here I'm going
to use the differential
416
00:32:01,780 --> 00:32:05,213
equation d of X is equal
to mu dt plus sigma dB(t).
417
00:32:05,213 --> 00:32:05,712
Yes?
418
00:32:05,712 --> 00:32:08,500
AUDIENCE: Do we need
that h for [INAUDIBLE]?
419
00:32:08,500 --> 00:32:10,700
PROFESSOR: No, we
don't actually.
420
00:32:10,700 --> 00:32:13,840
Oh, yeah, I was--
thank you very much.
421
00:32:13,840 --> 00:32:17,350
That was what confused me.
422
00:32:17,350 --> 00:32:20,460
Yes, thank you very much.
423
00:32:20,460 --> 00:32:22,830
And now we can
compute everything.
424
00:32:22,830 --> 00:32:27,260
This one, we're assuming
that we know the value.
425
00:32:27,260 --> 00:32:30,670
That one can be computed
from these two coordinates.
426
00:32:30,670 --> 00:32:36,350
Because we now have a fixed path
X, we know what X of i*h is.
427
00:32:36,350 --> 00:32:40,350
dt, we took it to be
h, approximated as h,
428
00:32:40,350 --> 00:32:42,180
or time difference.
429
00:32:42,180 --> 00:32:44,230
Again, sigma can be computed.
430
00:32:44,230 --> 00:32:47,470
dB now can be computed from B_t.
431
00:32:47,470 --> 00:32:50,180
Because we have a
fixed path, again, we
432
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,
433
00:32:58,162 --> 00:33:00,560
with this fixed path.
434
00:33:00,560 --> 00:33:02,950
They're basically
exactly the same,
435
00:33:02,950 --> 00:33:06,810
if you have a fixed
path B. The problem is
436
00:33:06,810 --> 00:33:10,270
we don't have a fixed path
B. That's where Monte Carlo
437
00:33:10,270 --> 00:33:11,220
simulation comes in.
438
00:33:15,450 --> 00:33:18,990
So Monte Carlo simulation
is just a way to draw,
439
00:33:18,990 --> 00:33:22,900
from some probability
distribution, a lot of samples.
440
00:33:22,900 --> 00:33:27,030
So now, if you know how to
draw samples from the Brownian
441
00:33:27,030 --> 00:33:30,860
motions, then what you're going
to do is draw a lot of samples.
442
00:33:30,860 --> 00:33:34,550
For each sample, do this to
compute the value of X(0),
443
00:33:34,550 --> 00:33:37,410
can compute X of 1.
444
00:33:40,130 --> 00:33:43,080
So, according to a different B,
you will get a different value.
445
00:33:43,080 --> 00:33:46,910
And in the end, you'll obtain
a probability distribution.
446
00:33:46,910 --> 00:33:57,770
So by repeating the
experiment, that means just
447
00:33:57,770 --> 00:33:59,760
redraw the path again
and again, you'll
448
00:33:59,760 --> 00:34:01,930
get different values of X of 1.
449
00:34:01,930 --> 00:34:04,160
That means you get a
distribution of X of 1,
450
00:34:04,160 --> 00:34:10,564
obtain distribution of X of 1.
451
00:34:10,564 --> 00:34:13,350
And that's it.
452
00:34:13,350 --> 00:34:18,800
And that will approach the
real distribution of X of 1.
453
00:34:18,800 --> 00:34:21,880
So that's how you numerically
solve a stochastic differential
454
00:34:21,880 --> 00:34:23,949
equation.
455
00:34:23,949 --> 00:34:26,389
Again, there's this
finite difference method
456
00:34:26,389 --> 00:34:29,889
that can be used to solve
differential equations.
457
00:34:29,889 --> 00:34:33,000
But the reason it doesn't apply
to stochastic differential
458
00:34:33,000 --> 00:34:36,810
equations is because there's
underlying uncertainty coming
459
00:34:36,810 --> 00:34:38,870
from Brownian motion.
460
00:34:38,870 --> 00:34:42,320
However, once you fix
a Brownian motion,
461
00:34:42,320 --> 00:34:44,389
then you can use that
finite difference method
462
00:34:44,389 --> 00:34:46,502
to compute X of 1.
463
00:34:46,502 --> 00:34:48,659
So based on that
idea, you just draw
464
00:34:48,659 --> 00:34:52,090
a lot of samples of
the Brownian path,
465
00:34:52,090 --> 00:34:53,750
compute a lot of
values of X of 1,
466
00:34:53,750 --> 00:34:56,199
and obtain a probability
distribution of X of 1.
467
00:34:58,710 --> 00:35:00,165
That's the underlying principle.
468
00:35:03,610 --> 00:35:06,832
And, of course, you
can't do it by hand.
469
00:35:06,832 --> 00:35:07,665
You need a computer.
470
00:35:11,320 --> 00:35:13,341
Then, what is tree method?
471
00:35:13,341 --> 00:35:16,060
That's cool.
472
00:35:16,060 --> 00:35:20,560
Tree method is
based on this idea.
473
00:35:20,560 --> 00:35:34,284
Remember, Brownian motion is
a limit of simple random walk.
474
00:35:44,610 --> 00:35:47,170
This gives you a kind
of approximate way
475
00:35:47,170 --> 00:35:50,844
to draw a sample from
Brownian motions.
476
00:35:50,844 --> 00:35:51,760
How would you do that?
477
00:35:58,480 --> 00:36:00,900
At time 0, you have 0.
478
00:36:00,900 --> 00:36:06,400
At time really tiny h,
you'll have plus 1 or minus 1
479
00:36:06,400 --> 00:36:09,130
with same probability.
480
00:36:09,130 --> 00:36:16,880
And it goes up or down again,
up or down again, and so on.
481
00:36:16,880 --> 00:36:19,170
And you know exactly the
probability distribution.
482
00:36:22,050 --> 00:36:25,250
So the problem is that
it ends up here as 1/2,
483
00:36:25,250 --> 00:36:30,160
ends up here as 1/2,
1/4, 1/2, 1/4, and so on.
484
00:36:30,160 --> 00:36:33,030
So instead of drawing from
this sample path, what you're
485
00:36:33,030 --> 00:36:36,614
going to do is just compute
the value of our function
486
00:36:36,614 --> 00:36:37,280
at these points.
487
00:36:41,770 --> 00:36:44,267
But then the probability
distribution,
488
00:36:44,267 --> 00:36:46,100
because we know the
probability distribution
489
00:36:46,100 --> 00:36:49,060
that the path will end
up at these points,
490
00:36:49,060 --> 00:36:54,530
suppose that you computed
all these values here.
491
00:36:57,810 --> 00:37:00,080
I draw too many, 1, 2, 3, 4, 5.
492
00:37:00,080 --> 00:37:03,015
This 1 or 32 probability here.
493
00:37:03,015 --> 00:37:10,362
5 choose 1, 5 over 32, 5
choose 2 is 10 over 32.
494
00:37:13,560 --> 00:37:17,200
Suppose that some stochastic
process, after following this,
495
00:37:17,200 --> 00:37:23,162
has value 1 here, 2 here,
3 here, 4, 5, and 6 here.
496
00:37:23,162 --> 00:37:27,190
Then, approximately, if
you take a Brownian motion,
497
00:37:27,190 --> 00:37:28,790
it will have 1
with probability 1
498
00:37:28,790 --> 00:37:32,346
over 32, 2 with probability
5 over 32, and so on.
499
00:37:34,640 --> 00:37:36,140
Maybe I didn't
explain it that well.
500
00:37:36,140 --> 00:37:38,060
But basically, tree
method just says,
501
00:37:38,060 --> 00:37:42,950
you can discretize the outcome
of the Brownian motion,
502
00:37:42,950 --> 00:37:47,535
based on the fact that it's a
limit of simple random walk.
503
00:37:47,535 --> 00:37:52,050
So just do the exact same method
for simple random walk instead
504
00:37:52,050 --> 00:37:54,180
of Brownian motion.
505
00:37:54,180 --> 00:37:56,195
And then take it to the limit.
506
00:37:56,195 --> 00:37:57,070
That's the principle.
507
00:38:00,778 --> 00:38:01,278
Yeah.
508
00:38:04,219 --> 00:38:06,260
Yeah, I don't know what's
being used in practice.
509
00:38:06,260 --> 00:38:10,150
But it seems like these two
are the more important ones.
510
00:38:10,150 --> 00:38:12,540
This is more like if you
want to do it by hand.
511
00:38:12,540 --> 00:38:15,400
Because you can't really do
every single possibility.
512
00:38:15,400 --> 00:38:18,125
That makes you only
a finite possibility.
513
00:38:22,150 --> 00:38:24,630
Any questions?
514
00:38:24,630 --> 00:38:25,130
Yeah.
515
00:38:25,130 --> 00:38:28,790
AUDIENCE: So here you
said, by repeating
516
00:38:28,790 --> 00:38:30,498
the experiment we
get [INAUDIBLE]
517
00:38:30,498 --> 00:38:32,938
distribution for X(1).
518
00:38:32,938 --> 00:38:36,490
I was wondering if we could also
get the distribution for not
519
00:38:36,490 --> 00:38:38,210
just X(1) but also for X(i*h).
520
00:38:38,210 --> 00:38:40,060
PROFESSOR: All the
intermediate values?
521
00:38:40,060 --> 00:38:40,735
AUDIENCE: Yeah.
522
00:38:40,735 --> 00:38:41,990
PROFESSOR: Yeah,
but the problem is
523
00:38:41,990 --> 00:38:43,448
we're taking
different values of h.
524
00:38:43,448 --> 00:38:45,720
So h will be
smaller and smaller.
525
00:38:45,720 --> 00:38:47,650
But for those
values that we took,
526
00:38:47,650 --> 00:38:49,770
yeah, we will get
some distribution.
527
00:38:49,770 --> 00:38:52,330
AUDIENCE: Right, so we
might have distributions
528
00:38:52,330 --> 00:38:56,002
for X of d for many
different points, right?
529
00:38:56,002 --> 00:38:56,668
PROFESSOR: Yeah.
530
00:38:56,668 --> 00:38:58,090
AUDIENCE: Yeah.
531
00:38:58,090 --> 00:39:02,692
So maybe we could
uh-- right, OK.
532
00:39:02,692 --> 00:39:04,650
PROFESSOR: But one thing
you have to be careful
533
00:39:04,650 --> 00:39:08,650
is let's suppose you
take h equal 1 over 100.
534
00:39:08,650 --> 00:39:11,030
Then, this will give
you a pretty fairly good
535
00:39:11,030 --> 00:39:12,551
approximation for X of 1.
536
00:39:12,551 --> 00:39:14,300
But it won't give you
a good approximation
537
00:39:14,300 --> 00:39:17,620
for X of 1 over 50.
538
00:39:17,620 --> 00:39:21,260
So probably you can also get
distribution for X of 1 over 3,
539
00:39:21,260 --> 00:39:22,550
1 over 4.
540
00:39:22,550 --> 00:39:28,720
But at some point, the
approximation will be very bad.
541
00:39:28,720 --> 00:39:31,820
So the key is to
choose a right h.
542
00:39:31,820 --> 00:39:33,660
Because if you pick
h to be too small,
543
00:39:33,660 --> 00:39:35,940
you will have a very
good approximation
544
00:39:35,940 --> 00:39:37,190
to your distribution.
545
00:39:37,190 --> 00:39:39,987
But at the same time, it
will take too much time
546
00:39:39,987 --> 00:39:40,570
to compute it.
547
00:39:45,470 --> 00:39:48,330
Any remarks from a
more practical side?
548
00:39:51,650 --> 00:39:54,200
OK, so that's
actually all I wanted
549
00:39:54,200 --> 00:39:56,990
to say about stochastic
differential equations.
550
00:39:56,990 --> 00:39:59,860
Really the basic
principle is there
551
00:39:59,860 --> 00:40:02,490
is such thing called stochastic
differential equation.
552
00:40:02,490 --> 00:40:04,660
It can be solved.
553
00:40:04,660 --> 00:40:08,210
But most of the time, it won't
have a closed form formula.
554
00:40:08,210 --> 00:40:09,870
And if you want to
do it numerically,
555
00:40:09,870 --> 00:40:12,920
here are some possibilities.
556
00:40:12,920 --> 00:40:18,250
But I won't go
any deeper inside.
557
00:40:18,250 --> 00:40:22,670
So the last math lecture I will
conclude with heat equation.
558
00:40:26,269 --> 00:40:28,654
Yeah.
559
00:40:28,654 --> 00:40:33,440
AUDIENCE: The mean
computations of [INAUDIBLE],
560
00:40:33,440 --> 00:40:37,720
some of the derivatives are
sort of path-independent,
561
00:40:37,720 --> 00:40:39,220
or have path-independent
solutions,
562
00:40:39,220 --> 00:40:40,850
so that you
basically are looking
563
00:40:40,850 --> 00:40:44,980
at say the distribution
at the terminal value
564
00:40:44,980 --> 00:40:48,760
and that determines the
price of the derivative.
565
00:40:48,760 --> 00:40:51,820
There are other derivatives
where things really
566
00:40:51,820 --> 00:40:57,510
are path-dependent, like
with options where you have
567
00:40:57,510 --> 00:41:00,240
early exercise possibilities.
568
00:41:00,240 --> 00:41:02,420
When do you exercise,
early or not?
569
00:41:02,420 --> 00:41:06,590
Then the tree methods are really
good because at each element
570
00:41:06,590 --> 00:41:10,650
of the tree you can condition
on whatever the path was.
571
00:41:10,650 --> 00:41:13,690
So keep that in mind,
that when there's
572
00:41:13,690 --> 00:41:15,600
path dependence in
the problem, you'll
573
00:41:15,600 --> 00:41:17,858
probably want to use
one of these methods.
574
00:41:17,858 --> 00:41:18,608
PROFESSOR: Thanks.
575
00:41:18,608 --> 00:41:21,637
AUDIENCE: I know that if
you're trying to break it down
576
00:41:21,637 --> 00:41:26,854
into simple random walks you
can only use [INAUDIBLE].
577
00:41:26,854 --> 00:41:30,000
But I've heard of people
trying to use, instead
578
00:41:30,000 --> 00:41:31,740
of a binomial, a trinomial tree.
579
00:41:31,740 --> 00:41:35,420
PROFESSOR: Yes, so
this statement actually
580
00:41:35,420 --> 00:41:37,986
is quite a universal statement.
581
00:41:37,986 --> 00:41:42,670
Brownian motion is a limit
of many things, not just
582
00:41:42,670 --> 00:41:43,620
simple random walk.
583
00:41:43,620 --> 00:41:47,230
For example, if you take
plus 1, 0, or minus 1
584
00:41:47,230 --> 00:41:49,190
and take it to the
limit, that will also
585
00:41:49,190 --> 00:41:51,670
converge to the Brownian motion.
586
00:41:51,670 --> 00:41:54,980
That will be the
trinomial and so on.
587
00:41:54,980 --> 00:41:57,090
And as Peter said,
if you're going
588
00:41:57,090 --> 00:41:59,320
to use tree method
to compute something,
589
00:41:59,320 --> 00:42:03,475
that will increase accuracy,
if you take more possibilities
590
00:42:03,475 --> 00:42:05,890
at each step.
591
00:42:05,890 --> 00:42:07,530
Now, there is two
ways to increase
592
00:42:07,530 --> 00:42:14,020
the accuracy is take more
possibilities at each step
593
00:42:14,020 --> 00:42:16,210
or take smaller time scales.
594
00:42:20,090 --> 00:42:23,470
OK, so let's move on to the
final topic, heat equation.
595
00:42:27,399 --> 00:42:29,570
Heat equation is not a
stochastic differential
596
00:42:29,570 --> 00:42:32,630
equation, first of all.
597
00:42:32,630 --> 00:42:33,230
It's a PDE.
598
00:42:47,780 --> 00:42:50,650
That equation is known
as a heat equation
599
00:42:50,650 --> 00:42:52,180
where t is like
the time variable,
600
00:42:52,180 --> 00:42:54,570
x is like the space variable.
601
00:42:54,570 --> 00:42:56,940
And the reason we're
interested in this heat
602
00:42:56,940 --> 00:42:59,340
equation in this
course is, if you
603
00:42:59,340 --> 00:43:09,480
came to the previous lecture,
maybe from Vasily last week,
604
00:43:09,480 --> 00:43:12,640
Black-Scholes equation,
after change of variables,
605
00:43:12,640 --> 00:43:17,050
can be reduced to heat equation.
606
00:43:17,050 --> 00:43:19,900
That's one reason
we're interested in it.
607
00:43:19,900 --> 00:43:22,310
And this is a really,
really famous equation
608
00:43:22,310 --> 00:43:23,620
also in physics.
609
00:43:23,620 --> 00:43:27,150
So it was known before
Black-Scholes equation.
610
00:43:27,150 --> 00:43:32,430
Particularly, this can be
a model for-- equations
611
00:43:32,430 --> 00:43:33,590
that model this situation.
612
00:43:36,260 --> 00:43:42,544
So you have an infinite
bar, very long and thin.
613
00:43:42,544 --> 00:43:43,585
It's perfectly insulated.
614
00:43:48,990 --> 00:43:53,850
So heat can only travel
along the x-axis.
615
00:43:53,850 --> 00:43:58,080
And then at time 0, you
have some heat distribution.
616
00:44:02,120 --> 00:44:07,076
At time 0, you know
the heat distribution.
617
00:44:13,040 --> 00:44:17,230
Then this equation tells you the
behavior of how the heat will
618
00:44:17,230 --> 00:44:20,020
be distributed at time t.
619
00:44:20,020 --> 00:44:25,570
So u of t of x, for fixed
t, will be the distribution
620
00:44:25,570 --> 00:44:28,852
of the heat over the x-axis.
621
00:44:28,852 --> 00:44:30,560
That's why it's called
the heat equation.
622
00:44:30,560 --> 00:44:31,935
That's where the
name comes from.
623
00:44:35,050 --> 00:44:37,720
And this equation is
very well understood.
624
00:44:37,720 --> 00:44:42,720
It does have a
closed-form solution.
625
00:44:42,720 --> 00:44:44,960
And that's what I
want to talk about.
626
00:44:48,250 --> 00:44:51,560
OK, so few observations
before actually solving it.
627
00:44:58,450 --> 00:45:13,520
Remark one, if u_1 and u_2
satisfies heat equation,
628
00:45:13,520 --> 00:45:18,951
then u_1 plus u_2 also
satisfies, also does.
629
00:45:21,850 --> 00:45:24,750
That's called linearity.
630
00:45:24,750 --> 00:45:25,840
Just plug it in.
631
00:45:25,840 --> 00:45:28,080
And you can figure it out.
632
00:45:28,080 --> 00:45:31,760
More generally
that means, if you
633
00:45:31,760 --> 00:45:46,890
integrate a family of functions
ds, where u_s all satisfy star,
634
00:45:46,890 --> 00:46:00,435
then this also
satisfies star, as long
635
00:46:00,435 --> 00:46:02,300
as you use reasonable function.
636
00:46:02,300 --> 00:46:05,240
I'll just assume that we can
switch the order of integration
637
00:46:05,240 --> 00:46:06,620
and differentiation.
638
00:46:06,620 --> 00:46:08,550
So it's the same thing.
639
00:46:08,550 --> 00:46:10,432
Instead of summation,
I'm taking integration
640
00:46:10,432 --> 00:46:11,265
of lot of solutions.
641
00:46:14,410 --> 00:46:17,490
And why is that helpful?
642
00:46:17,490 --> 00:46:28,120
This is helpful
because now it suffices
643
00:46:28,120 --> 00:46:36,160
to solve for-- what is it?
644
00:46:36,160 --> 00:46:44,345
Initial condition, u of t of x
equals delta, delta function,
645
00:46:44,345 --> 00:46:44,950
of 0.
646
00:46:52,470 --> 00:46:55,970
That one is a little bit subtle.
647
00:46:55,970 --> 00:46:58,980
The Dirac delta function is
just like an infinite ass
648
00:46:58,980 --> 00:47:00,100
at x equals 0.
649
00:47:00,100 --> 00:47:03,660
It's 0 everywhere else.
650
00:47:03,660 --> 00:47:05,950
And basically, in this
example, what you're saying is,
651
00:47:05,950 --> 00:47:09,680
at time 0, you're putting
like a massive amount of heat
652
00:47:09,680 --> 00:47:11,430
at a single point.
653
00:47:11,430 --> 00:47:14,800
And you're observing what's
going to happen afterwards,
654
00:47:14,800 --> 00:47:17,380
how this heat will spread out.
655
00:47:17,380 --> 00:47:19,240
If you understand that,
you can understand
656
00:47:19,240 --> 00:47:21,220
all initial conditions.
657
00:47:21,220 --> 00:47:23,160
Why is that?
658
00:47:23,160 --> 00:47:38,840
Because if u sub s t, x--
u_0-- is such solution,
659
00:47:38,840 --> 00:47:45,030
then integration of--
let me get it right--
660
00:47:45,030 --> 00:48:10,546
u of t of s minus x ds is a
solution with initial condition
661
00:48:10,546 --> 00:48:11,046
x(0, x).
662
00:48:16,006 --> 00:48:16,998
What is it?
663
00:48:37,334 --> 00:48:38,930
Sorry about that.
664
00:48:38,930 --> 00:48:41,720
So this is really the key.
665
00:48:41,720 --> 00:48:46,120
If you have a solution to the
Dirac delta initial condition,
666
00:48:46,120 --> 00:48:49,340
then you can superimpose
a lot of those solutions
667
00:48:49,340 --> 00:48:51,935
to obtain a solution for
arbitrary initial condition.
668
00:48:55,030 --> 00:48:58,860
So this is based on that
principle, because each of them
669
00:48:58,860 --> 00:48:59,950
is now a solution.
670
00:48:59,950 --> 00:49:04,890
If you superpose this,
then that is a solution.
671
00:49:04,890 --> 00:49:06,640
And then if you plug
it in, you figure out
672
00:49:06,640 --> 00:49:09,331
that actually it has satisfied
this initial condition.
673
00:49:14,050 --> 00:49:18,260
That was my first observation.
674
00:49:18,260 --> 00:49:27,160
Second observation,
second remark,
675
00:49:27,160 --> 00:49:44,930
is for the initial value u(0, x)
equals a Dirac delta function,
676
00:49:44,930 --> 00:49:59,246
u of-- is a solution.
677
00:50:05,710 --> 00:50:08,320
So we know the solution
for the Dirac delta part.
678
00:50:08,320 --> 00:50:10,950
First part, we figured out
that if we know the solution
679
00:50:10,950 --> 00:50:12,800
for the Dirac delta
function, then
680
00:50:12,800 --> 00:50:15,260
we can solve it for every
single initial value.
681
00:50:15,260 --> 00:50:17,160
And for the initial
value Dirac delta,
682
00:50:17,160 --> 00:50:23,300
that is the solution that solves
the differential equation.
683
00:50:23,300 --> 00:50:26,490
So let me say a few words
about this equation, actually
684
00:50:26,490 --> 00:50:29,002
one word.
685
00:50:29,002 --> 00:50:30,986
Have you seen this
equation before?
686
00:50:35,090 --> 00:50:36,865
It's the p.d.f. of
normal distribution.
687
00:50:36,865 --> 00:50:37,740
So what does it mean?
688
00:50:37,740 --> 00:50:41,820
It means, in this example,
if you have a heat traveling
689
00:50:41,820 --> 00:50:45,280
along the x-axis,
perfectly insulated,
690
00:50:45,280 --> 00:50:49,180
if you put a massive heat
at this 0, at one point,
691
00:50:49,180 --> 00:50:53,090
at time 0, then at
time t your heat
692
00:50:53,090 --> 00:50:55,640
will be distributed according
to the normal distribution.
693
00:50:55,640 --> 00:50:58,014
In other words, assume that
you have a bunch of particle.
694
00:50:58,014 --> 00:51:00,140
Heat is just like a
bunch of particles,
695
00:51:00,140 --> 00:51:04,590
say millions of particles
at a single point.
696
00:51:04,590 --> 00:51:07,410
And then you grab it.
697
00:51:07,410 --> 00:51:09,908
And then time t equals
0 you release it.
698
00:51:14,547 --> 00:51:16,880
Now the particle at time t
will be distributed according
699
00:51:16,880 --> 00:51:18,190
to a normal distribution.
700
00:51:18,190 --> 00:51:24,300
In other words, each particle
is like a Brownian motion.
701
00:51:24,300 --> 00:51:28,740
So for particle by
particle, the location
702
00:51:28,740 --> 00:51:31,760
of its particle at time t
will be kind of distributed
703
00:51:31,760 --> 00:51:33,400
like a Brownian motion.
704
00:51:33,400 --> 00:51:36,450
So if you have a massive
amount of particles,
705
00:51:36,450 --> 00:51:38,590
altogether their
distribution will look
706
00:51:38,590 --> 00:51:40,490
like a normal distribution.
707
00:51:40,490 --> 00:51:42,930
That's like its content.
708
00:51:42,930 --> 00:51:45,910
So that's also one way you see
the appearance of a Brownian
709
00:51:45,910 --> 00:51:49,510
motion inside of this equation.
710
00:51:49,510 --> 00:51:52,570
It's like a bunch of Brownian
motions happening together
711
00:51:52,570 --> 00:51:53,820
at the exact same time.
712
00:52:01,320 --> 00:52:06,841
And now we can just
write down the solution.
713
00:52:06,841 --> 00:52:08,450
Let me be a little
bit more precise.
714
00:52:18,600 --> 00:52:32,200
OK, for the heat equation
delta u over delta t,
715
00:52:32,200 --> 00:52:44,590
with initial value u of 0,
x equals some initial value,
716
00:52:44,590 --> 00:52:50,700
let's say v of x, and t
greater than equal to 0.
717
00:53:00,510 --> 00:53:09,728
The solution is
given by integration.
718
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.
719
00:54:08,890 --> 00:54:12,680
Basically, I'm just combining
this solution to there.
720
00:54:12,680 --> 00:54:18,090
Plugging in that
here, you get that.
721
00:54:18,090 --> 00:54:20,280
So you have a explicit
solution, no matter
722
00:54:20,280 --> 00:54:24,050
what the initial conditions
are, initial conditions
723
00:54:24,050 --> 00:54:27,280
are given as, you can
find an explicit solution
724
00:54:27,280 --> 00:54:32,090
at time t for all x.
725
00:54:32,090 --> 00:54:36,590
That means, once you change
the Black-Scholes equation
726
00:54:36,590 --> 00:54:41,300
into the heat equation, you
now have a closed-form solution
727
00:54:41,300 --> 00:54:41,976
for it.
728
00:54:45,550 --> 00:54:51,325
In that case, it's like
a backward heat equation.
729
00:54:51,325 --> 00:54:53,200
And what will happen is
the initial condition
730
00:54:53,200 --> 00:54:56,890
you should think of as
a final payout function.
731
00:54:56,890 --> 00:54:58,700
The final payout
function you integrate
732
00:54:58,700 --> 00:55:00,860
according to this distribution.
733
00:55:00,860 --> 00:55:03,486
And then you get the
value at time t equals 0.
734
00:55:07,220 --> 00:55:09,910
The detail, one of
the final project
735
00:55:09,910 --> 00:55:11,760
is to actually carry
out all the details.
736
00:55:11,760 --> 00:55:13,990
So I will stop here.
737
00:55:13,990 --> 00:55:17,180
Anyway, we didn't see how the
Black-Scholes equation actually
738
00:55:17,180 --> 00:55:18,410
changed into heat equation.
739
00:55:22,470 --> 00:55:24,890
If you want to do
that project, it
740
00:55:24,890 --> 00:55:26,930
will be good to
have this in mind.
741
00:55:26,930 --> 00:55:29,030
It will help.
742
00:55:29,030 --> 00:55:30,030
Any questions?
743
00:55:34,530 --> 00:55:38,938
OK, so I think
that's all I have.
744
00:55:38,938 --> 00:55:41,800
I think I'll end a
little bit early today.
745
00:55:41,800 --> 00:55:45,155
So that will be the last math
lecture for the semester.
746
00:55:45,155 --> 00:55:48,390
From now on you'll only
have application lectures.
747
00:55:48,390 --> 00:55:55,910
There are great lectures
coming up, I hope, and I know.
748
00:55:55,910 --> 00:55:59,390
So you should come
and really enjoy now.
749
00:55:59,390 --> 00:56:02,300
You went through
all this hard work.
750
00:56:02,300 --> 00:56:04,465
Now it's time to enjoy.