Proof: How do we use Ito's formula

[squenshl]

Homework Statement

For all integers n >= 1 define z_n = E[W(1)]ⁿ. Use Ito's formula to prove that z_n+2 = (n+1)z_n. Compute z_n for all integers n >= 1. z = mu.

Homework Equations

The Attempt at a Solution

How do we use Ito's formula, do we use it directly?

 

[Kreizhn]

It looks like you're doing some work in stochastic here, but it's impossible to tell what you're trying to do. Could you define a few more things?

I assume E[] is the expectation, what is W(1)?

 

[squenshl]

Sorry.
I have proved the Ito formula for the special case W(1)³ = 3*integral from 0 to 1 of W(1)² dW(t) + 3*integral from 0 to 1 of W(1) dt

 

[squenshl]

I was told to try doing it by induction but I have never done induction before. Please help.

 

[Ray Vickson]

Homework Statement

For all integers n >= 1 define z_n = E[W(1)]ⁿ. Use Ito's formula to prove that z_n+2 = (n+1)z_n. Compute z_n for all integers n >= 1. z = mu.

Homework Equations

The Attempt at a Solution

How do we use Ito's formula, do we use it directly?

If by {W(t)} you mean a standard Wiener process with W(0)=0, then calculation of E[W(1)^n] does not need Ito's formula (although you could use Ito, but that would be doing it the *hard* way). Your problem, as written, does not involve stochastic integration, but is just a simple problem in elementary probability theory! Think about the nature of W(t): what is its probability distribution for any fixed value of t?

RGV

 

[Stephen Tashi]

Both the problem and the title of the original post say to use Ito's formula, so we should consider how to do this.

Does $E(W(1))^n$ mean $E(W(1)^n)$

I'm not an expert on the stochastic calculus. So let me ask if $E(W(t)^n)$ , as a function of T, is equal to $\int_0^T W(t)^n dW_t$? If so, perhaps section 1.8 of these class notes is relevant: http://www.google.com/url?sa=t&sour...sg=AFQjCNG1sNKK8JK-WNvfqVq8MHAz6dXmug&cad=rja

 

[Ray Vickson]

Both the problem and the title of the original post say to use Ito's formula, so we should consider how to do this.

Does $E(W(1))^n$ mean $E(W(1)^n)$

I'm not an expert on the stochastic calculus. So let me ask if $E(W(t)^n)$ , as a function of T, is equal to $\int_0^T W(t)^n dW_t$? If so, perhaps section 1.8 of these class notes is relevant: http://www.google.com/url?sa=t&sour...sg=AFQjCNG1sNKK8JK-WNvfqVq8MHAz6dXmug&cad=rja

The answer to your question is NO. If f is a C^2 function, and W(t) is a standard Wiener process, the Ito formula says: for $$Y(t) = f(W(t))$$ we have
$$\displaystyle dY = f'(W) dW + \frac{1}{2} f''(W) dt$$
Apply this to f(W) = W^n and use the fact that {W(t)} is non-anticipatory (i.e., events in {W(s), s > t} are independent of those in {W(s), s <= t}) to simplify the expectation E[d(W(t)^n)] = dE[W(t)^n] and get an ODE connecting Fn(t) to F_{n-2}(t), where Fk(t) = E[W(t)^k]. Then use the known values of F1(t) and F2(t) to get all the Fn(t), at least in principle. As I said, this is doing it the hard way.

RGV