Limit of stochastic integrals?

Question

Let $(W_t)t$ be a Wiener process. I want to find the limit for $\epsilon\to 0$ of $$\frac{W_t^2}{2\epsilon}\chi_{(-\epsilon,\epsilon)}(W_t)-\int_0^t \frac{W_s\chi_{(-\epsilon,\epsilon)}(W_s)}{\epsilon}\,dW_s$$ More precisely I want to prove that this limit is a non decreasing process non identically zero.I would appreciate it if anyone could give me any idea. Thank you in advance.

Answer 1

I am not so sure about your result I would have said that it converges to 0 in probability $\forall t>0$.

First observe that $\frac{W_t^2}{2\epsilon}\chi_{(-\epsilon,\epsilon)}(W_t)\le \frac{\epsilon^2}{2\epsilon}=\frac{\epsilon}{2}$ so the first term goes to 0 almost surely.

As for the second term using Itô's isometry we see that : $$E[(\int_0^t W_s\frac{\chi_{(-\epsilon,\epsilon)}(W_s)}{\epsilon}.dW_s )^2]= E[\int_0^t W_s^2\frac{\chi_{(-\epsilon,\epsilon)}(W_s)}{\epsilon^2}ds]=\int_0^t E[\frac{W_s^2}{\epsilon^2}\chi_{(-\epsilon,\epsilon)}(W_s)]ds$$

$$ \le \int_0^t E[\chi_{(-\epsilon,\epsilon)}(W_s)] ds ~~as~~(\frac{W_s^2}{\epsilon^2}\le 1)$$

Finally it is easy to see analytically (using properties of the cumulative density function of Gaussian random variables) that the dominating term goes to 0 as $\epsilon\to 0$.

So the second term converges to 0 in $L^2$, finally the sum of the two processes tends to 0 (at least) in probability.

I think that what you wanted to prove is convergence to the local time at 0 of the Brownian motion so there must be a mistake in the line of reasoning that lead you to this question (or maybe in mine).

Best regards.