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PROFESSOR: Let's begin.
Today we're going to continue
the discussion on Ito calculus.
I briefly introduced you
to Ito's lemma last time,
but let's begin by
reviewing it and stating it
in a slightly more general form.
Last time what we did was we
did the quadratic variation
of Brownian motion,
Brownian process.
We defined the Brownian
process, Brownian motion,
and then showed that it has
quadratic variation, which
can be written in this form--
dB square is equal to dt.
And then we used that to show
the simple form of Ito's lemma,
which says that if f is a
function on the Brownian
motion, then d of f is
equal to f prime of dB_t
plus f double prime of dt.
This additional term was a
characteristic of Ito calculus.
In classical calculus
we only have this term,
but we have this
additional term.
And if you remember,
this happened exactly
because of this
quadratic variation.
Let's review it, and let's do
it in a slightly more general
form.
As you know, we
have a function f
depending on two
variables, t and x.
Now we're interested
in-- we want to evaluate
our information on the
function f(t, B_t).
The second coordinate,
we're planning
to put in the
Brownian motion there.
Then again, let's do
the same analysis.
Can we describe d of f in terms
of these differentiations?
To do that, deflect this, let
me start from Taylor expansion.
f at a point t plus delta
t, x plus delta x by Taylor
expansion for two variables is
f of t of x plus partial of f
over partial of t at t
comma x of delta t plus...
x.
That's the first-order terms.
Then we have the
second-order terms.
Then the third-order
terms, and so on.
That's just Taylor expansion.
If you look at it,
we have a function f.
We want to look at the
difference of f when we change
the first variable a little
bit and the second variable
a little bit.
We start from f of t of x.
In the first-order terms, you
take the partial derivative,
so take del f over del
t, and then multiply
by the t difference.
Second term, you take
the partial derivative
with respect to the second
variable-- partial f
over partial x-- and
then multiply by del x.
That much is enough
for classical calculus.
But then, as we
have seen before,
we ought to look at
the second-order term.
So let's first write
down what it is.
That's exactly what happened
in Taylor expansion,
if you remember.
If you don't remember,
just believe me.
This 1 over 2 times, take the
second derivatives, partial.
Let's write it in
terms of-- yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Oh,
yeah, you're right.
Thank you.
Is it good now?
Let's write it as
dt, all these deltas.
I'll just write like that.
I'll just not
write down t and x.
And what we have is f plus
del f over del t dt plus del
f over del x dx plus
the second-order terms.
The only important terms--
first of all, these terms
are important.
But then, if you want
to use x equals B of
t-- so if you're now
interested in f t comma B of t.
Or more generally, if you're
interested in f t plus dt,
f B_t plus d of B_t, then
these terms are important.
If you subtract f of
t of B_t, what you get
is these two terms.
Del f over del t dt
plus del f over del
x-- I'm just writing
this as a second variable
differentiation-- at dB_t.
And then the second-order terms.
Instead of writing it all down,
dt square is insignificant,
and dt comma-- dt times
dB_t also is insignificant.
But the only thing that
matters will be this one.
This is dB_t square, which
you saw is equal to dt.
From the second-order term,
we'll have this term surviving.
1 over 2 partial f over partial
x second derivative, of dt.
That's it.
If you rearrange
it, what we get is
partial f over partial
t plus 1/2 this plus--
and that's the additional term.
If you ask me why these
terms are not important
and this term is important, I
can't really say it rigorously.
But if you think about dB_t
square equals dt, then d times
B_t is kind of like
square root of dt.
It's not a good
notation, but if you
do that-- these two terms are
significantly smaller than dt
because you're
taking a power of it.
dt square becomes a lot
smaller than dt, dt to the 3/2
is a lot smaller than dt.
But this one survives because
it's equal to dt here.
That's just the
high-level description.
That's a slightly more
sophisticated form
of Ito's lemma.
Let me write it down here.
And let's just fix it now.
If f of t of B_t-- that's d of
f is equal to-- Any questions?
Just remember, from the
classical calculus term,
we're only adding
this one term there.
Yes?
AUDIENCE: Why do
we have x there?
PROFESSOR: Because the second
variable is supposed to be x.
I don't want to write down
partial derivative with respect
to a Brownian motion here
because it doesn't look good.
It just means, take the
partial derivative with respect
to the second term.
So just view this as a
function f of t of x,
evaluate it, and then plug
in x equal B_t in the end,
because I don't want to
write down partial B_t here.
Other questions?
Consider a stochastic process
X of t such that d of X
is equal to mu times d of t
plus sigma times d of B_t.
This is almost like
a Brownian motion,
but you have this
additional term.
This is called a drift term.
Basically, this happens if X_t
is equal to mu*t plus sigma
of B_t.
Mu and sigma are constants.
From now on, what
we're going to study
is stochastic process of
this type, whose difference
can be written in terms of drift
term and the Brownian motion
term.
We want to do a slightly
more general form
of Ito's lemma, where we
want f of t of X_t here.
That will be the
main object of study.
I'll finally state the
strongest Ito's lemma
that we're going to use.
f is some smooth function and
X_t is a stochastic process
like that.
X_t satisfies...
where B_t is a Brownian motion.
Then df of t, X_t
can be expressed
as-- it's just getting
more and more complicated.
But it's based on this one
simple principle, really.
It all happened because
of quadratic variation.
Now I'll show you why this form
deviates from this form when
we replace B to an X.
Remember here all other
terms didn't matter,
that the only term that mattered
was partial square of f...
of dx square.
To prove this, note that df
is partial f over partial t
dt plus partial f over
partial x d of X_t
plus 1/2 of d of x squared.
Just exactly the same, but I've
replaced the dB-- previously,
what we had dB, I'm
replacing to dX.
Now what changes is dX_t
can be written like that.
If you just plug it
in, what you get here
is partial f over partial
x mu dt plus sigma of dB_t.
Then what you get here
is 1/2 of partials
and then mu plus
sigma dB_t square.
Out of those three terms here
we get mu square dt square
plus 2 times mu sigma d mu dB
plus sigma square dB square.
Only this was survives,
just as before.
These ones disappear.
And then you just
collect the terms.
So dt-- there's one dt here.
There's mu times that here,
and that one will become a dt.
It's 1/2 of sigma
square partial square...
of dt.
And there's only
one dB_t term here.
Sigma-- I made a mistake, sigma.
This will be a form that
you'll use the most,
because you want to evaluate
some stochastic process--
some function that
depends on time
and that stochastic process.
You want to understand
the difference, df.
The X would have
been written in terms
of a Brownian motion
and a drift term,
and then that's the
Ito lemma for you.
But if you want to
just-- if you just
see this for the first time,
it just looks too complicated.
You don't understand where
all the terms are coming from.
But in reality, what
it's really doing
is just take this
Taylor expansion.
Remember these two
classical terms,
and remember that there's
one more term here.
You can derive it
if you want to.
Really try to know
where it all comes from.
It all started from this one
fact, quadratic variation,
because that made some of the
second derivative survive,
and because of those,
you get these kind
of complicated terms.
Questions?
Let's do some examples.
That's too much.
Sorry, I'm going to use it
a lot, so let me record it.
Example number one.
Let f of x be equal to
x square, and then you
want to compute d of f at B_t.
I'll give you three minutes
just to try a practice.
Did you manage to do this?
It's a very simple example.
Assume it's just the
function of two variables,
but it doesn't depend on t.
You don't have to do that,
but let me just do that.
Partial f over partial t is 0.
Partial f over partial
x is equal to 2x,
and the second derivative
equal to 2 at t, x.
Now we just plug in
t comma B_t, and what
you have is mu equals
0, sigma equals 1,
if you want to write
it in this formula.
What you're going to have
is 2 times B_t of dB_t
plus 1 over 2 times 2dt.
If you write it down.
You can either use
these parameters
and just plug in each of
them to figure it out.
Or a different way
to do it is really
write down, remember the proof.
This is partial f
over partial t dt
plus partial f over partial x
dx plus 1/2-- remember this one.
And x is dB_t here.
That one is 0, that one
was 2x, so 2B_t dB_t.
Use it one more
time, so you get dt.
Make sense?
Let's do a few more examples.
And you want to compute
d of f at t comma B of t.
Let's do it this time.
Again, partial f over
partial t dt plus partial f
over partial x dB_t.
That's the first-order terms.
The second-order term
is 1/2 partial square f
over partial x square of dB_t
square, which is equal to dt.
Let's do it.
Partial f over partial
t, you get mu times f.
This one is just
equal to mu times f.
Maybe I'm going too quick.
Mu times e to the
mu t plus dx, dt.
Partial f over partial
x is sigma times e
to the mu t plus
dx, and then dB_t
plus-- if you take
the second derivative,
you do that again,
what you get is
1/2, and then sigma square
times e to the mu t plus dx, dt.
Yes?
AUDIENCE: In the original
equation that you just wrote,
isn't it 1/2 times
sigma squared,
and then the second derivative?
Up there.
PROFESSOR: Here?
AUDIENCE: Yes.
PROFESSOR: 1/2?
AUDIENCE: Times sigma squared.
PROFESSOR: Oh, sigma-- OK,
that's a good question.
But that sigma is different.
That's if you plug in X_t here.
If you plug in X_t
where X_t is equal to mu
prime dt plus sigma
prime d of B_t,
then that sigma prime will
become a sigma prime square
here.
But here the function
is mu and sigma,
so maybe it's not
a good notation.
Let me use a and b here instead.
The sigma here is
different from here.
AUDIENCE: Yeah, that
makes a lot more sense.
PROFESSOR: If you
replace a and b,
but I already wrote down
all mu's and sigma's.
That's a good point, actually.
But that's when you
want to consider
a general stochastic
process here
other than Brownian motion.
But here it's just
a Brownian motion,
so it's the most simple form.
And that's what you get.
Mu plus 1/2 sigma square-- and
these are just all f itself.
That's the good thing
about exponential.
f times dt plus
sigma times d of B_t.
Make sense?
And there's a reason I
was covering this example.
It's because-- let's come
back to this question.
You want to model stock price
using Brownian motion, Brownian
process, S of t.
But you don't want S_t
to be a Brownian motion.
What you want is a
percentile difference
to be a Brownian motion, so you
want this percentile difference
to behave like a Brownian
motion with some variance.
The question was, is S_t equal
to e to the sigma times B of t
in this case?
And I already told you last
time that no, it's not true.
We can now see
why it's not true.
Take this function, S_t
equals e to the sigma
B_t, that's exactly where
mu is equal to 0 here.
What we got here was d of S_t,
in this case, is equal to-- mu
is 0, so we get 1/2 of sigma
square times dt plus sigma
times d of B_t.
We originally were
targeting sigma times dB_t,
but we got this
additional term which we
didn't want in the first hand.
In other words, we
have this drift.
I wasn't really clear
in the beginning,
but our goal was to
model stock price
where the expected
value is 0 at all times.
Our guess was to take
e to the sigma B_t,
but it turns out
that in this case
we have a drift,
if you just take
natural e to the sigma of B_t.
To remove that drift,
what you can do
is subtract that term somehow.
If you can get rid of that
term then you can see,
if you add this mu to be
minus 1 over 2 sigma square,
you can remove that term.
That's why it doesn't work.
So instead use S of t equals
e to the minus 1 over 2 sigma
square t plus sigma of B_t.
That's the geometric Brownian
motion without drift.
And the reason it has no
drift is because of that.
If you actually do
the computation,
the dt term disappears.
Question?
So far we have been
discussing differentiation.
Now let's talk
about integration.
Yes?
AUDIENCE: Could you we do get
this solution as [INAUDIBLE].
Could you also
describe what it means?
What does it mean,
this solution of B_t?
Does that mean if we
have a sample path B_t,
then we could get a
sample path for S_t?
PROFESSOR: Oh, what
this means, yes.
Whenever you have the B_t
value, just at each time
take the exponential value.
Because-- why we want to express
this in terms of a Brownian
motion is, for
Brownian motion we
have a pretty good
understanding.
It's a really good process
you understand fairly well,
and you have good control on it.
But the problem is you want to
have a process whose percentile
difference behaves
like a Brownian motion.
And this gives you a
way of describing it
in terms of Brownian motion, as
an exponential function of it.
Does that answer your question?
AUDIENCE: Right,
distribution means
that if we have a
sample path B_t,
that would be the corresponding
sample path for S of t?
Is it a pointwise evaluation?
PROFESSOR: That's a
good question, actually.
Think of it as a
pointwise evaluation.
That is not always
correct, but for most
of the things that
we will cover,
it's safe to think
about it that way.
But if you think about it
path-wise all the time,
eventually it fails.
But that's a very
advanced topic.
So what this question
is, basically
B_t is a probability space,
it's a probability distribution
over paths.
For this equation, if you just
look at it, it looks right,
but it doesn't
really make sense,
because B_t-- if it's a
probability distribution, what
is e to the B_t?
Basically, what
it's saying is B_t
is a probability
distribution over paths.
If you take omega according
to-- a path according
to the Brownian motion sample
probability distribution,
and for this path it's well
defined, this function.
So the probability density
function of this path
is equal to the probability
density function of e
to the whatever that is
in this distribution.
Maybe it confused you more.
Just consider this as some path,
some well-defined function,
and you have a
well-defined function.
Integral definition.
I will first give you a
very, very stupid definition
of integration.
We say that we define
F as the integration...
if d of F is equal
to f dB_t plus-- we
define it as an inverse
of differentiation.
Because differentiation
is now well-defined--
we just defined integration
as the inverse of it,
just as in classical calculus.
So far, it doesn't
have that good meaning,
other than being
an inverse of it,
but at least it's well-defined.
The question is, does it exist?
Given f and g, does it exist,
does integration always exist,
and so on.
There's lots of
questions to ask,
but at least this
is some definition.
And the natural question is,
does there exist a Riemannian
sum type description?
That means-- if you remember
how we defined integral
in calculus, you have a
function f, integration
of f from a to b according to
the Riemannian sum description
was, you just chop the
interval into very fine pieces,
a_0, a_1, a_2, a_3,
dot, dot, dot--
and then sum the area of these
boxes, and take the limit.
And this is the limit
of Riemannian sums.
Slightly more, if you want, it's
the limit as n goes to infinity
of the function 1 over n times
the sum of i equal zero to t--
I'll just call it 0 to b-- f of
t*b over n minus f of t minus 1
over n.
Does this ring a bell?
Question?
AUDIENCE: [INAUDIBLE]
PROFESSOR: No, you're right.
Good point, no we don't.
Thanks.
Does integral
defined in this way
have this Riemannian sum type
description, is the question.
So keep that in mind.
I will come back to
this point later.
In fact, it turns out to
be a very deep question
and very important
question, this question,
because if you remember
like I hope you remember,
in the Riemannian sum, it
didn't matter which point you
took in this interval.
That was the whole point.
You have the function.
In the interval a_i to
a_(i+1), you take any point
in the middle and make a
rectangle according to that
point.
And then, no matter
which point you take,
when you go to the limit,
you had exactly the same sum
all the time.
That's how you define the limit.
But what's really
interesting here
is that it's no longer true.
If you take the left
point all the time,
and you take the right
point all the time,
the two limits are different.
And again, that's due to
the quadratic variation,
because that much of variance
can accumulate over time.
That's the reason we didn't
start with Riemannian sum type
definition of integral.
But I'll just make one remark.
Ito integral is the
limit of Riemannian sums
when always take the leftmost
point of each interval.
So you chop down this curve
at-- the time interval
into pieces, and
for each rectangle,
pick the leftmost point,
and use it as a rectangle.
And you take the limit.
That will be your
Ito integral defined.
It will be exactly equal to this
thing, the inverse of our Ito
differentiation.
I won't be able
to go into detail.
What's more
interesting is instead,
what happens if you take the
rightmost point all the time,
you get an equivalent
theory of calculus.
It's just like Ito's calculus.
It looks really, really similar
and it's coherent itself,
so there is no
logical flaw in it.
It all makes sense,
but the only difference
is instead of a plus in
the second-order term,
you get minuses.
Let me just make this
remark, because it's just
a theoretical part, this thing,
but I think it's really cool.
Remark-- there's this
and equivalent version.
Maybe equivalent is
not the right word,
but a very similar
version of Ito
calculus such that
basically, what
it says is d B_t square
is equal to minus dt.
Then that changed
a lot of things.
But this part, it's
not that important.
Just cool stuff.
Let's think about this a
little bit more, this fact.
Taking the leftmost
point all the time
means if you want to make
a decision for your time
interval-- so at time t of
i and time t of i plus 1,
let's say it's the stock price.
You want to say that you had
so many stocks in this time
interval.
Let's say you had so many
stocks in this time interval
according to the values
between this and this.
In real world, your only
choice you have is you
have to make the
decision at time t of i.
Your choice cannot depend
on the future time.
You can't suddenly say, OK,
in this interval the stock
price increased a
lot, so I'll assume
that I had a lot of
stocks in this interval.
In this interval, I knew
it was going to drop,
so I'll just take the
rightmost interval.
I'll assume that I only
had this many stock.
You can't do that.
Your decision has to be
based on the leftmost point,
because the time.
You can't see the future.
And the reason Ito's calculus
works well in our setting is
because of this fact, because it
has inside it the fact that you
cannot see the future.
Every decision is made
based on the leftmost time.
If you want to make a decision
for your time interval,
you have to do it
in the beginning.
That intuition is hidden
inside of the theory,
and that's why it works so well.
Let me reiterate this
part a little bit more.
It's the definition
of these things
where you're only
allowed to-- at time t,
you're only allowed to use
the information up to time t.
Definition: delta t is an
adapted process-- sorry,
adapted to another
stochastic process X_t--
if for all values
of time variables
delta t depends only
on X_0 up to X_t.
There's a lot of vague
statements inside here,
but what I'm trying
to say is just
assume X is the Brownian
motion underlying stock price.
Your stock is changing.
You want to come
up with a strategy,
and you want to say
that mathematically
this strategy makes sense.
And what it's saying
is if your strategy
makes your decision
at time t is only
based on the past values
of your stock price,
then that's an adapted process.
This defines the processes
that are reasonable,
that cannot see future.
And these are all--
in terms of strategy,
if delta_t is a
portfolio strategy,
these are the only meaningful
strategies that you can use.
And because of what I said
before, because we're always
taking the leftmost
point, adaptive
processes just also fit very
well with Ito's calculus.
They'll come into
play altogether.
Just a few examples.
First, a very stupid example.
X_t is adapted to X_t.
Of course, because
at time, X_t really
depends on only
X_t, nothing else.
Two, X_(t+1) is
not adapted to X_t.
This is maybe a
little bit vague,
so we'll call it
Y_t equals X_(t+1).
Y_t is the value at t
plus 1, and it's not based
on the values up to time t.
Just a very artificial example.
Another example, delta
t equals minimum...
is adapted.
And I'll let you think about it.
The fourth is quite interesting.
Suppose T is fixed,
some large integer,
or some large real number.
Then you let delta t to be the
maximum where X of s, where...
It's not adapted.
What is this?
This means at time T,
I'm going to take at it
this value, the
maximum of all value
inside this part, the future.
This refers to the future.
It's not an adapted process.
Any questions?
Now we're ready to talk
about the properties
of Ito's integral.
Let's quickly
review what we have.
First, I defined Ito's lemma--
that means differentiation
in Ito calculus.
Then I defined integration using
differentiation-- integration
was an inverse operation
of the differentiation.
But this integration also had
an alternative description
in terms of
Riemannian sums, where
you're taking just the
leftmost point as the reference
point for each interval.
And then, as you
see, this naturally
had this concept of
using the leftmost point.
And to abstract
that concept, we've
come up with this adapted
process, very natural process,
which is like the
real-life procedures,
real-life strategies
we can think of.
Now let's see what
happens when you
take the integral of
adapted processes.
Ito integral has
really cool properties.
The first thing is about
normal distribution.
B_t has normal
distribution of 0 up to t.
So your Brownian
motion at time t
has normal
distribution with 0, t.
That means if your stochastic
process is some constant times
B of t, of course, then
you have 0 and c square t.
It's still a normal variable.
That means if you
integrate, that's
the integration of some sigma.
That's the integration
of sigma of dB_t.
If sigma is a fixed
constant, when
you take the Ito
integral of sigma times
dB_t, this constant,
at each time
you get a normal distribution.
And this is like saying the
sum of normal distribution
is also normal distribution.
It has this hidden
fact, because integral
is like sum in the limit.
And this can be generalized.
If delta t is a process
depending only on the time
variable-- so it does not depend
on the Brownian motion-- then
the process X of t equals the
integration of delta t dB_t
has normal distribution at
all time, just like this.
We don't know the
exact variance yet;
the variance will
depend on the sigmas.
But still, it's like a
sum of normal variables,
so we'll have
normal distribution.
In fact, it just gets
better and better.
The second fact is
called Ito isometry.
That was cool.
Can we compute the variance?
Yes?
AUDIENCE: Can you
put that board up?
PROFESSOR: Sure.
AUDIENCE: Does it go up?
PROFESSOR: This
one doesn't go up.
That's bad.
I wish it did go up.
This has a name
called Ito isometry.
Can be used to
compute the variance.
B_t is a Brownian
motion, delta t
is adapted to a Brownian motion.
Then the expectation
of your Ito integral--
that's the Ito integral
of your adapted process.
That's the variance-- we
take the square of it--
is equal to something cool.
The square just comes in.
Quite nice, isn't it?
I won't prove it, but
let me tell you why.
We already saw this
phenomenon before.
This is basically
quadratic variation.
And the proof also uses it.
If you take delta s
equals to 1-- sorry,
I was using Korean-- 1 at all
time, then what we have is
here you get a
Brownian motion, B_t.
So on the left you get like
expectation of B_t square,
and on the right,
what you get is t.
Because when delta
s is equal to 1
at all time, when you have
to get from 0 to t you get t,
and you have t on
the right hand side.
That's what it's saying.
And that was the content
of quadratic variation,
if you remember.
We're summing the squares--
maybe not exactly this,
but you're summing the
squares over small intervals.
So that's a really
good fact that you can
use to compute the variance.
You have an Ito integral,
you know the square,
can be computed this simple way.
That's really cool.
And one more property.
This one will be
really important.
You'll see it a lot
in future lectures.
It's that when is Ito
integral a martingale?
What's a martingale?
Martingale meant if you
have a stochastic process,
at any time t, whatever happens
after that, the expected value
at time t is equal to 0.
It doesn't have any natural
tendency to go up or go down.
No matter which point
you stop your process
and you see your future, it
doesn't have a natural tendency
to go up or go down.
In formal language, it can
be defined as where F_t
is the events X_0 up to X_t.
So if you take the
conditional expectation
based on whatever
happened up to time t,
that expectation will
just be whatever value
you have at that time.
Intuitively, that just means you
don't have any natural tendency
to go up or go down.
Question is, when is an
Ito integral a martingale?
Adapted to B of t, then...
is a martingale.
As long as g is not
some crazy function,
as long as g is reasonable--
one way can be reasonable if its
L^2-norm is bounded.
If you don't know what it
means, you can safely ignore it.
Basically, if g doesn't-- it's
not a crazy function if it
doesn't grow too fast, then
in most cases this integral is
always a martingale.
If you flip it--
remember, integral
was defined as the inverse
of differentiation.
So if dX_t is equal to
some function mu, that
depends on both t and
B_t, times dt, plus sigma
of dB_t, what this means
is X_t is a martingale
if that is 0 at
all time, always.
And if it's not 0,
you have a drift,
so it's not a martingale.
That gives you some
classification.
Now, if you look at a
differential equation
of this stochastic--
this is called
a stochastic differential
equation-- if you know
stochastic process, if you look
at a stochastic differential
equation, if it doesn't have a
drift term, it's a martingale.
If it has a drift term,
it's not a martingale.
That'll be really useful
later, so try to remember it.
The whole point is
when you write down
a stochastic process in
terms of something times dt,
something times dB_t,
really this term
contributes towards the
tendency, the slope of whatever
is going to happen
in the future.
And this is like
the variance term.
It adds some variance to
your stochastic process.
But still, it doesn't add
or subtract value over time,
it fairly adds variation.
Remember that.
That's very important fact.
You're going to use it a lot.
For example, you're going to
use it for pricing theory.
In pricing theory, you come up
with this stochastic process
or some strategy.
You look at its value.
Let's say X_t is your value
of your portfolio over time.
If that portfolio has-- then you
match it with your financial--
let me go over it slowly again.
First you have a financial
derivative, like option
of a stock.
Then you have your
portfolio strategy.
Assume that you have
some strategy that,
at the expiration
time, gives you
the exact value of the option.
Now you look at the
difference between these two
stochastic processes.
Basically what the thing is,
when your variance goes to 0,
your drift also has to go to 0.
So when you look
at the difference,
if you can somehow get rid
of this variance term, that
means no matter
what you do, that
will govern the value
of your portfolio.
If it's positive, that means
you can always make money,
because there's no variance.
Without variance,
you make money.
That's called arbitrage,
and you cannot have that.
But I won't go
into further detail
because Vasily will
cover it next time.
But just remember that flavor.
So when you write something down
in a stochastic differential
equation form, that
term is a drift term,
that term is a variance term.
And if you don't have
drift, it's a martingale.
That is very important.
Any questions?
That's kind of the
basics of Ito calculus.
I will give you some
exercises on it,
mostly just basic computation
exercises, so that you'll
get familiar with it.
Try to practice it.
And let me cover one more
thing called Girsanov theorem.
It's related, but
these are really
basics of the Ito
calculus, so if you
have any questions on
this, please ask me
right now before I move
on to the next topic.
The last thing I want
to talk about today.
Here is an underlying question.
Suppose you have two
Brownian motions.
This is without drift.
And you have another B tilde,
Brownian motion with drift.
These are two probability
distributions over paths.
According to B_t, you're
more likely to have
some Brownian motion
that has no drift.
That's a sample path.
According to B tilde,
you have some drift.
Your Brownian motion will--
A typical path will follow this
line and will follow that line.
The question is
this-- can we switch
from this distribution
to this distribution
by a change of measure?
Can we switch between
the two measures
to probability distributions
by a change of measure?
Let me go a little bit
more what it really means.
Assume that you're just
looking at a Brownian motion
from time 0 up to time t,
some fixed time interval.
Then according to B_t, let's
say this is a sample path omega.
You have some probability
of omega-- this is a p.d.f.
given by this Brownian
motion B. And then you
have another p.d.f., P tilde
of omega, which is a p.d.f.
given by B of t.
The question is,
does there exist a Z
depending on omega
such that P of omega
is equal to Z times P tilde?
Do you understand the question?
Clearly, if you just look at
it, they're quite different.
The path that you get
according to the distributions
are quite different.
It's not clear why we
should expect it at all.
You'll see the answer soon.
But let me discuss all this
in a different context.
Just forget about all the
Brownian motion and everything
just for a moment.
In this concept, changing from
one probability distribution
to another distribution,
it's a very important concept
in analysis and probability
just in general, theoretically.
And there's a name for this
Z, for this changing measure.
If Z exists, it's called the
Radon-Nikodym derivative.
Before doing that, let me
talk a little bit more.
Suppose P is a probability
distribution over omega.
It's a probability distribution.
So this is some set, and P
describes the probability
that you have each
element in the set.
And you have another probability
distribution, P tilde.
We define P and P tilde to be
equivalent if the probability
that A is greater than
zero if and only if...
For all...
These probability distributions
describe the probability
of the subsets.
Think about a very simple case.
Sigma is equal to 1, 2, and 3.
P gives 1/3 probability
to 1, 1/3 probability
to 2, 1/3 probability to 3.
P tilde gives 2/3 probability
to 3, 1 over 6 probability
to 2, 1 over 6 probability to 3.
We have two probability
distribution over some space.
They are equivalent
if, whenever you
take a subset of your
background set-- let's say 1, 2.
When A is equal
to 1, 2, according
to probability distribution
P, the probability
you fall into this
set A is equal to 2/3.
According to P
tilde, you have 5/6.
They're not the same.
The probability itself
is not the same,
but this condition is
satisfied when it's 0.
And when it's not 0, it's not 0.
And you can just check that it's
always true, because they're
all positive probabilities.
On the other hand, if
you take instead, say,
1/3 and 0, now you
take your A to be 3.
Then you have 1/3 equal to 0.
This means, according to
probability distribution P,
there is some probability
that you'll get 3.
But according to probability
distribution P tilde,
you don't have any
probability of getting 3.
So they're not
equivalent in this case.
If you think about it,
then it's really clear.
The theorem says-- this is
a very important theorem
in analysis, actually.
The theorem-- there exists a Z
such that P of omega is equal
to...
If and only if P and P
tilde are equivalent.
You can change from
one probability measure
to another probability
measure just
in terms of multiplication, if
and only if they're equivalent.
And you can see that it's
not the case for this
when they're not equivalent.
You can't make a zero
probability to 1/3 probability
by multiplication.
So in the finite world this is
very just intuitive theorem,
but what this is saying is
it's true for all probability
spaces.
And these are called the
Radon-Nikodym derivative.
Our question is, are these two
Brownian motions equivalent?
The paths that this Brownian
motion without drift
takes and the Brownian
motion with drift
takes, are they kind of
the same but just skewed
in distribution, or are they
really fundamentally different?
That's the question.
And what Girsanov's theorem says
is that they are equivalent.
To me, it came as a
little bit non-intuitive.
I would imagine that it's
not equivalent, these two.
These paths have a
very natural tendency.
As it goes to infinity,
these paths and these paths
will really look
a lot different,
because when you go
really, really far,
the paths which have
drift will be just really
close to your line mu of
t, while the paths which
don't have drift will be
really close to the x-axis.
But still, they are equivalent.
You can change from
one to another.
I'll just state that
theorem without proof.
And this will also be
used in pricing theory.
I'm not an expert
enough to tell why,
but basically what
it's saying is,
you switch some
stochastic process
into a stochastic
process without drift,
thus making it
into a martingale.
And martingale has a lot of
meaning in pricing theory,
as you'll see.
This also has application.
That's why I'm trying to
cover it, although it's
quite a technical theorem.
Try to remember at least
a statement and the spirit
of what it means.
It just means these
two are equivalent,
you can change
from one to another
by a multiplicative function.
Let me just state
it in a simple form.
GUEST SPEAKER: If I could
just interject a comment.
PROFESSOR: Sure.
GUEST SPEAKER: With
these changes of measure,
it turns out that all of these
theories with continuous time
processes should have an
interpretation if you've
discretized time,
and should consider
sort of a finer and finer
discretization of the process.
And with this change of measure,
if you consider problems
in discrete stochastic
processes like random walks,
basically how-- say if you're
gambling against a casino
or against another
player, and you
look at how your winnings
evolve as a random walk,
depending on your
odds, your odds
could be that you
will tend to lose.
So there's basically
a drift in your wealth
as this random process evolves.
You can transform that process,
basically by taking out
your expected losses,
to a process which
has zero change in expectation.
And so you can convert
these gambling problems
where there's drift to a version
where the process, essentially,
has no drift and
is a martingale.
And the martingale theory in
stochastic process courses
is very, very powerful.
There's martingale
convergence theorems.
So you know that the
limit of the martingale
is-- there's a convergence
of the process,
and that applies here as well.
PROFESSOR: You will see some
surprising applications.
GUEST SPEAKER: Yeah.
PROFESSOR: And try to at
least digest the statement.
When the guest speaker comes
and says by Girsanov theorem,
they actually know what it is.
There's a spirit.
This is a very simple version.
There's a lot of
complicated versions,
but let me just do it.
So P is a probability
distribution over paths
from [0, T] to the infinity.
What this means is just paths
from that-- stochastic process
defined from time
0 to time T. These
are paths defined by a
Brownian motion with drift mu.
And then P tilde is a
probability distribution
defined by Brownian
motion without drift.
Then P and P tilde
are equivalent.
Not only are they
equivalent, we can actually
compute their
Radon-Nikodym derivative.
And the Radon-Nikodym
derivative Z
which is defined as T of--
which we denote like this
has this nice form.
That's a nice closed form.
Let me just tell you a
few implications of this.
Now, assume you have
some, let's say, value
of your portfolio over time.
That's the stochastic process.
And you measure it according to
this probability distribution.
Let's say it depends
on some stock price
as the stock price is
modeled using a Brownian
motion with drift.
What this is saying
is, now, instead
of computing this expectation
in your probability space--
so this is defined over
the probability space P,
our sigma-- (omega, P)
defined by this probability
distribution.
You can instead
compute it in-- you
can compute as expectation in
a different probability space.
You transform the problems
about Brownian motion with drift
into a problem about Brownian
motion without a drift.
And the reason I have
Z tilde instead of Z
here is because I flipped.
What you really should have is Z
tilde here as expectation of Z.
If you want to use this Z.
I don't expect you to really
be able to do computations
and do that just by looking
at this theorem once.
Just really trying to
digest what it means
and understand the flavor of
it, that you can transform
problems in one
probability space
to another probability space.
And you can actually do that
when the two distributions are
defined by Brownian motions
when one has drift and one
doesn't have a drift.
How we're going
to use it is we're
going to transform a
non-martingale process
into a martingale process.
When you change
into martingale it
has very good physical
meanings to it.
That's it for today.
And you only have one more
math lecture remaining
and maybe one or two
homeworks but if you have two,
the second one
won't be that long.
And you'll have a lot of
guest lectures, exciting guest
lectures, so try
not to miss them.
