Ito's Lemma

[Cheng]

Hi there,

I would appreciate if someone could explain further on Ito's integral and Ito's lemma. I'm very confused with the materials given in the core reading.

1) in page 15, how do I get the Taylor's theorem to second order? I could not connect the theorem given in pg 15, to the Taylor's series (two variables) as given in section 1.2 of the Tables.

2) Also, how do I get formula 7.3 in page 19? Is f(t,Xt) = Xt?
given in formula 7.2, dXt= Yt dBt + At dt, do I partial differentiate eqn 7.2 with respect to x, to get eqn 7.3?

3) also in page 19, how do I get the Taylor's formula to second-order? and what's the difference between this formula (in pg19) as compare to the one given earlier in pg 15?

Thanks in advance!

 

[Oxymoron]

Thanks for bringing this up Cheng. I'd appreciate help in this too. I'm really going cul-de-sacs in part II of CT-8.

 

[Mike Lewry]

From Taylor to Ito

This topic can be tricky the first time you meet it, but I hope the following helps.

1) in page 15, how do I get the Taylor's theorem to second order? I could not connect the theorem given in pg 15, to the Taylor's series (two variables) as given in section 1.2 of the Tables.

It's the one-variable version shown on p15, so we need to start with the top formula on p3 of the Tables, which can be written:

f(x+h)-f(x)=hf'(x)+0.5(h^2)f''(x)+...

This is the same as what we have in the notes if we map x to b(t) and h to delta.b(t)

2) Also, how do I get formula 7.3 in page 19? Is f(t,Xt) = Xt?
given in formula 7.2, dXt= Yt dBt + At dt, do I partial differentiate eqn 7.2 with respect to x, to get eqn 7.3?

No, the steps to get Equation (7.3), which is Ito's lemma, are described in the bullet points on p19.

Since we want an SDE for f(t,X(t)), which has 2 variables, we need to start with Taylor in 2 variables. This is stated on p19 and comes from the second formula on p3 of the Tables, making the additional mappings y to t and k to dt.

Next, we substitute in Equation (7.2) for dX(t).

Note [dX(t)]^2 = Y(t)^2.dt, using Ito's multiplication table.

Finally, grouping the dB(t) and dt terms together gives Equation (7.3).

3) also in page 19, how do I get the Taylor's formula to second-order? and what's the difference between this formula (in pg19) as compare to the one given earlier in pg 15?

I've answered this above, by giving you the additional mappings required. Here we need Taylor for 2 variables (t and X(t)), but on page 15 there was only one variable, b(t).

So, in the exam, always check whether you need to produce an SDE in 1 or 2 variables, then write down the appropriate version of Taylor, work out the (partial) derivatives and use Ito's multiplication table to simplify the second order terms.

 

[Cheng]

Thanks for the reply,

but how do I construct/use the Ito's multiplication table?

also, (dXt)^2 does not always equal to dt?

 

[Mike Lewry]

Ito's "multiplicatiob table"

Thanks for the reply,

but how do I construct/use the Ito's multiplication table?

also, (dXt)^2 does not always equal to dt?

Ito's multiplication table (or grid) is given on page 19 of Chapter 8 and it shows how to multiply deterministic/stochastic infinitesimal increments together. It shows increments of the basic building blocks of SDEs - time and standard Brownian motion.

You don't need to derive/construct this table; it's fine just to remember it, so "(dBt)^2=dt, everything else is zero".

If you're looking at increments of other processes, such as the general Xt, you need to expand out the terms.

So, if dXt = Adt + YdBt, then:

dXt.dt = A(dt)^2 + YdBt.dt = 0

(dXt)^2 = A^2.dt^2 + 2A.Y.dt.dBt + Y^2.(dBt)^2
= 0 + 0 + Y^2.dt

If you work through a few of the Q&A Bank questions on Ito, you should get the hang of it. It just takes a little practice.

 

[Cheng]

thanks alot for the reply.

In writing Taylor's theorem to second order (p15), I get
f(bt+del.bt)=f(bt) + del.bt*f'(bt) + 0.5(del.bt)^2.f''(bt)+..
f(bt)+f(del.bt)=f(bt)+...
why, in the 2nd step as above, I can split f(bt+del.bt)=f(bt)+f(del.bt) ?

No, the steps to get Equation (7.3), which is Ito's lemma, are described in the bullet points on p19.

Since we want an SDE for f(t,X(t)), which has 2 variables, we need to start with Taylor in 2 variables. This is stated on p19 and comes from the second formula on p3 of the Tables, making the additional mappings y to t and k to dt.

Next, we substitute in Equation (7.2) for dX(t).

Note [dX(t)]^2 = Y(t)^2.dt, using Ito's multiplication table.

Finally, grouping the dB(t) and dt terms together gives Equation (7.3).

I tried mapping y to t, k to dt, h to dXt and x to x in the Taylor series (2 variables) formula, and since (dt)^2=0, dt.dXt=0, I get

f(x+dt,t+dt)-f(x,y)= (del.f/del.x).dXt + (del.f/del.t).dt +0.5(del^2 f/del.x^2).(dXt)^2

From there, how do I get the LHS equal to df(t,Xt)?

 

[Mike Lewry]

In writing Taylor's theorem to second order (p15), I get
f(bt+del.bt)=f(bt) + del.bt*f'(bt) + 0.5(del.bt)^2.f''(bt)+..
f(bt)+f(del.bt)=f(bt)+...
why, in the 2nd step as above, I can split f(bt+del.bt)=f(bt)+f(del.bt) ?

You can't - it's not true

I tried mapping y to t, k to dt, h to dXt and x to x in the Taylor series (2 variables) formula, and since (dt)^2=0, dt.dXt=0, I get

f(x+dt,t+dt)-f(x,y)= (del.f/del.x).dXt + (del.f/del.t).dt +0.5(del^2 f/del.x^2).(dXt)^2

From there, how do I get the LHS equal to df(t,Xt)?

There are a couple of slips in your formula. Applying the mappings you give above to the 2-variable Taylor series on p3 of the Tables, I get:
f(X+dX,t+dt)=f(X,t) + dX.(del.f/del.x) + dt.(del.f/del.t) + 0.5(dXt)^2.(del^2 f/del.x^2)

Taking the first term on the RHS over to the left gives:
f(X+dX,t+dt)-f(X,t) = dX.(del.f/del.x) + dt.(del.f/del.t) + 0.5(dXt)^2.(del^2 f/del.x^2)

The LHS represents the change in f when X changes by dX and t changes by dt and so can be written df(X,t), as required.

 

[kylie jane]

A few more technicalities!

It's the one-variable version shown on p15, so we need to start with the top formula on p3 of the Tables, which can be written:

f(x+h)-f(x)=hf'(x)+0.5(h^2)f''(x)+...

This is the same as what we have in the notes if we map x to b(t) and h to delta.b(t)

If you do this mapping and letting b(t) = b

f(b+ delta.b)-f(b) = delta.b f'(b) + 0.5 (delta.b)^2 f''(b) +...

dividing through by delta.t,

LHS = [f(b+delta.b)-f(b)]/delta.t

how does this equal df(b)/dt in the limit of delta.t ->0 ??

I thought: [f(b+delta.t)-f(b)] / delta.t = df(b)/dt as delta.t ->0

Also in the formal definition of Ito's lemma (page 19) I find the use of x very confusing, does x=Xt??