Deriving a Brownian Motion Integral with Itô's Lemma

# How to perform basic integrations with the Ito integral?


# How to perform basic integrations with the Ito integral?












From the text book Quantitative Finance for Physicists: An Introduction (Academic Press Advanced Finance) I have this excercise:

Prove that

$$ \int_{t_1}^{t_2}W(s)^ndW(s)=\frac{1}{n+1}[W(t_2)^{n+1}-W(t_1)^{n+1}]-\frac{n}{2}\int_{t_1}^{t_2}W(s)^{n-1}ds $$

Hint: Calculate $d(W^{n+1})$ using Ito's lemma.

This is my calculation:

I use Ito's Lemma and use, as the text book does, the simplified case $\mu=0$, $\sigma=1$. So Ito's lemma reduces to:

$$ dF=dt+2WdW $$

Now I use the Ito Lemma here like this:

$$ \int_{t_1}^{t_2}W(s)^ndW(s)=\int_{t_1}^{t_2}W(s)^{n-1}W(s)dW(s) $$

Because $\mu=0$, $\sigma=1$, we have $F=W^2$ and therefore the integral equals: $$ \int_{t_1}^{t_2}F^\frac{n-1}{2}\frac{dF}{2}-\int_{t_1}^{t_2}F^\frac{n-1}{2}\frac{dt}{2} $$ Further simplification: $$ \frac{1}{2}\frac{2}{n+1}F^\frac{n+1}{2}]_{t_1}^{t_2}-\frac{1}{2}W^{n-1}(t)dt=\frac{1}{n+1}W^{n+1}]_{t_1}^{t_2}-\frac{1}{2}W(t)^{n-1}(t)dt $$ Putting everything together obviously yields: $$ \int_{t_1}^{t_2}W(s)^ndW(s)=\frac{1}{n+1}[W(t_2)^{n+1}-W(t_1)^{n+1}]-\frac{1}{2}\int_{t_1}^{t_2}W(s)^{n-1}ds $$

The second term lacks a factor $n$.

What am I doing wrong? Or is the book wrong? (By the way for $n=1$ this is consistent with the book. For $n=1$ I was also able to calculate the integral using summation $\lim_n\sum$.... But this seems too complicated for general case.)

Besides I have problems understanding why I have to use the differential equation. Or do I have to see $W$ as $W=W(F,t)$?

Edit: Removed some lines...

## Answer by SRKX (score 6, accepted)

https://quant.stackexchange.com/a/2153

I think you should see the hint as follows:

$$d(W_t^{n+1})=d(f(W_t))$$ with $$f(x)=x^{n+1}$$

Apply Ito:

$$d(W_t^{n+1}) = f'(W_t)dW_t + \frac{1}{2} f''(W_t) d<W>_t$$

$$d(W_t^{n+1}) = (n+1) W_t^n dW_t + \frac{1}{2} n (n+1) W_t^{n-1} dt$$

If you integrate, you get:

$$W_{t_2}^{n+1}-W_{t_1}^{n+1}=(n+1) \int_{t_1}^{t_2} W_t^n dW_t+ \frac{1}{2} n (n+1) \int_{t_1}^{t_2} W_t^{n-1} dt$$

You divide both sides by $(n+1)$ and you're done.

I'm not sure how you use your Ito's lemma, maybe you've been fooled by some simplified version of some books. I'd recommend Elementary Stochastic Calculus with Finance in View if you want a good introduction to the field.

As for you assumptions, $\mu=0$ and $\sigma=1$ are the properties of a Brownian motion after just a single step $W_1$, by definition.

Here the purpose of the exercise is in my opinion to show you that by applying Ito's lemma, you might end up finding interesting properties (one of the most famous one was shown in this post).