# Baldi - Stochastic Calculus - Exercise about construsction of Stratonovich integral

I need to prove that $$\lim_n \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) \rightarrow \frac{1}{2}W_1^2 - \frac{1}{2}$$ in $$L^2$$.

where $$W$$ is a standard Wiener process.

So I started computing

$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) )^2]$$

and I split have that, after expanding the square:

$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ))^2]=\frac{1}{2}$$

by independence the Gaussian distribution of independent increments.

Then, the double-product term $$\frac{1}{2} E[(1-W_1^2) \cdot \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )) ] = 0$$ since it's a standard Wiener process and by independence of increments

The last term to compute is

$$\frac{1}{4} E[(1-W_1^2)^2)] = \frac{1}{4} E[1+W_1^4 - 2 W_1^2 ] = \frac{1}{4} (1+3-2)=\frac{1}{2}$$

using that $$E[W_1^2] = 1$$ and $$E[W_1^4] = 3$$.

But in this way the sum is $$1$$ and not $$0$$... so what am I missing?

• I think that the error is in the computation of the expectation of the double product Commented Nov 27, 2019 at 22:22

You expectation numbers are incorrect, you could see that in your proof, your solution is independent of $$n$$...

First notice that

$$W_1^2= \sum_{i=0}^{2^n -1} W(\frac{i+1}{2^n})^2- W(\frac{i}{2^n})^2$$ $$W_1^2= \sum_{i=0}^{2^n -1}\left( W(\frac{i+1}{2^n})+ W(\frac{i}{2^n})\right)\left( W(\frac{i+1}{2^n})- W(\frac{i}{2^n})\right)$$

Therefore, by factorizing, we have $$\sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) =\frac{1}{2}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]$$

You want to calculate

$$E\left[\frac{1}{4}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]$$

The variance of a Brownian motion increment is known, we have
$$E\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right] =\sum_{i=0}^{2^n -1}\frac{1}{2^n }=1$$

Finally,

$$E\left[\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]=E\left[\sum_{i,j=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]$$

or $$\sum_{i,j=0, i \ne j}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]+\sum_{i=0}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^4\right]$$

Increments are independent , and you know the 4th moment of a Brownian motion, you can conclude when you make $$n$$ goes to infinity.

• Many thanks for your detailed answer! In the last line, I have a question about the two sums: the first sum I should get $\frac{1}{2}$ as sending $n$ to infinity, by using independence of increments. WWhile in the second sum I should get $3$, right? Commented Nov 27, 2019 at 23:01
• The first term of the last equation should converge to $1$ and the second one to $0$ Commented Nov 27, 2019 at 23:02
• That is what you should get $\frac{3}{(2^N)^2}*2^N$ Commented Nov 27, 2019 at 23:04
• Try it, the same approach works, there will be a different sign when you factorize. Commented Nov 27, 2019 at 23:11
• yes, it is the same Commented Nov 28, 2019 at 3:17