# 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...

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your assumptions about $\mu$ and $\sigma$ are unclear to me. but see my answer – SRKX♦ Oct 12 '11 at 20:00
The text book does in that chapter most calculations with this simplified case. – Philip Oct 12 '11 at 20:07
also, for latter readers, you might want to tell us what textbook you're using. – SRKX♦ Oct 12 '11 at 20:11
the formula you use for ito is the particular case $f(x)=x^2$ – SRKX♦ Oct 12 '11 at 20:13
I added the reference. I see, in fact the book isn't really clear about Ito. It starts by deriving it using the Taylor expansion of $df$ and then continues simplifying it step by step. – Philip Oct 12 '11 at 20:25

## 1 Answer

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).

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Thanks a lot! Though I still have to find my error... – Philip Oct 12 '11 at 20:05
@Philip: just added more details to my answer because honestly I'm not sure what you're doing exactly in your way to tackle the problem. – SRKX♦ Oct 12 '11 at 20:10
@Philip: found your mistake, see the last comment on the question. – SRKX♦ Oct 12 '11 at 20:20
Great, thanks again, also for the references. – Philip Oct 12 '11 at 20:26
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