How Does Ito's Lemma Simplify the Stochastic Integral of W^n?

[ra_forever8]

Prove the following result for the ito stochastic integral (n>1)
∫_(t_0)^t▒〖W^n dW = 1/(n+1) (W^(n+1) (t)- 〗 W^(n+1) (t_0 ))- n/2 ∫_(t_0)^t▒〖W^(n-1 ) dt〗 Hint: apply ito differentiation rule f(W) = W^(n+1) to express W^n dW via dW^(n+1) and W^(n-1) dt (analogue of integration by parts for stochastic calculus)

=> stochastic differential equation: dW_t= A(t,W_t)dt +B(t,W_t)dW

∂f(t,W_t)= (∂f/∂t)* dt + ∂f/∂W *(dW_t) + 1/2 ∂^2/∂W^2 * (dW_t ^2)

We have f(W) = W^(n+1)
A=0 and B=1

Can some one help me after this to prove the equation

 

[Ray Vickson]

Prove the following result for the ito stochastic integral (n>1)
∫_(t_0)^t▒〖W^n dW = 1/(n+1) (W^(n+1) (t)- 〗 W^(n+1) (t_0 ))- n/2 ∫_(t_0)^t▒〖W^(n-1 ) dt〗 Hint: apply ito differentiation rule f(W) = W^(n+1) to express W^n dW via dW^(n+1) and W^(n-1) dt (analogue of integration by parts for stochastic calculus)

=> stochastic differential equation: dW_t= A(t,W_t)dt +B(t,W_t)dW

∂f(t,W_t)= (∂f/∂t)* dt + ∂f/∂W *(dW_t) + 1/2 ∂^2/∂W^2 * (dW_t ^2)

We have f(W) = W^(n+1)
A=0 and B=1

Can some one help me after this to prove the equation

Use Ito's Lemma:
$$dX_t = a(X_t,t) dt + b(X_t,t) dW_t \text{ and } Y_t = f(X_t,t)\\<br /> \text{implies}\\<br /> dY_t = f_x(X_t,t) dX_t + f_t(X_t,t) dt + \frac{1}{2} f_{xx}(X_t,t) (dX_t)^2 \\<br /> = f_x(a \, dt + b\, dW) + \frac{1}{2} f_{xx} b^2 dt = \left(a\, f_x + f_t + \frac{1}{2}b^2 \,f_{xx}\right)\, dt + f_x b \,dW$$
Apply this to $a = 0, b = 1, f(x) = x^{n+1}$.

 

[ra_forever8]

dY_t= af_x *dt + f_t * dt +1/2 b^2 f_xx *dt + f_x *b *dW
Applying a =0 and b=1
=0+ f_t * dt+ 1/2 b^2 f_xx *dt + f_x *b *dW
with f(x) = x^(n+1)
Now differentiating in terms of x
= 0+ n(n-1)/2 * x^(n-1) * dt + (n+1)* x^n *dW (f_t * dt =0)
= n(n-1)/2 * x^(n-1) * dt + (n+1)* x^n *dW
what happens to left side term dY_t ? what to do to prove the qs?

 

[Ray Vickson]

dY_t= af_x *dt + f_t * dt +1/2 b^2 f_xx *dt + f_x *b *dW
Applying a =0 and b=1
=0+ f_t * dt+ 1/2 b^2 f_xx *dt + f_x *b *dW
with f(x) = x^(n+1)
Now differentiating in terms of x
= 0+ n(n-1)/2 * x^(n-1) * dt + (n+1)* x^n *dW (f_t * dt =0)
= n(n-1)/2 * x^(n-1) * dt + (n+1)* x^n *dW
what happens to left side term dY_t ? what to do to prove the qs?

You should have written
$$dY_t \equiv d W^{n+1} = \frac{n(n+1)}{2} W^{n-1} \, dt + (n+1) W^n \, dW$$
When you realize that a stochastic DE is basically a shorthand notation for an integral result, you will see that you have everything you need.

 

[ra_forever8]

sorry I did my differentiation wrong
dW^(n+1) = n(n+1)/2 W^(n+1) dt + (n+1) W^n dW
dW^(n+1) = (n+1) ( n/2 *W^(n+1) dt + W^n dW)
dW^(n+1) / (n+1) = ( n/2 *W^(n+1) dt + W^n dW)
did I miss any terms? After that please?