expected value problem for a normalized integral of sign Say we have a Gausian process $X_s=\int_0^sh(x)\,dW(x)$ where $W(x)$ is a Wiener process. Now define $$Z=\frac{\int_0^1\operatorname{sign}(X_s) \, dW(s)}{\int_0^1|X_s| \, ds}$$
Intuitively we have $EZ=0$. But I would like to justify it somehown, but have no clue where to start! I appreciate any thought or comments. Many thanks.
 A: (Can't comment, so write as an answer.)
Intuitively, I am not even sure if $Z$ has a finite expected value in the case when $h(t)=1$ (which I would definitely start with). A priori, whenever you divide something by something else that can possibly be small, bad things can happen.
Anyways, I would look at the problem from the following angle. Let's write a system of SDEs ($h(t)=1$)
$$
\begin{aligned}
&dX_1(t) = dW(t),\\
&dX_2(t) = \operatorname{sign}{(X_1(t))}\ dW(t),\\
&dX_3(t) = |X_1(t)|\ dt,
\end{aligned}
$$
and initial conditions $X_1(0) = X_2(0) = X_3(0) = 0.$ Then, $Z$ can be represented as $Z = \frac{X_2(1)}{X_3(1)}$, and all you need to know is CDF 
$$F(x_2, x_3; t) = P \{ X_2(t) \le x_2, X_3(t) \le x_3 \}.$$ You can try to find $F(x_2, x_3)$ using the Kolmogorov$-$Fokker$-$Planck equation, or more appropriately, the Pugachev equation (see [1], and on how to handle sign, [2]). Then, you be able to answer the question whether the expectation exists and to actually find it. I think some related information may also be found in [3].
