# Expectation of an integral w.r.t. Brownian Motion

I know the following statement:

if $f$ is a deterministic function and continuous, i.e. $f\in C^0([0,T],\mathbb{R})$, then $\int f(s)dW_s$ is normally distributed with mean zero and variance $\int f^2(s) ds$.

Now my first question is, is this result still true if we drop the continuity assumption, i.e. is it true for $f\in L^2([0,T])$? Can I deduce this from the continuous case? If not how to prove this?

My second question is about the following: if $\gamma$ is some nice process, i.e. predictable and locally bounded so that $\int \gamma_s dW_s$ makes sense and is a local martingale. Are there assumptions on $\gamma$ such that $E[\int \gamma_s dW_s]=0$? Even more, is the distribution of this integral also known? I guess to know the distribution one has to know at least the interplay between $\gamma$ and $W$. It is clear that in general it seem rather hard (if not impossible) to make a statement about the distribution of $\int\gamma dW$, since both, the integrator and the integrand are stochastic.

Of course if $\gamma \in L^2(W)$, i.e. $E[\int \gamma^2_s ds]<\infty$, we know that $\int \gamma_s dW_s$ is martingale, hence $E[\int \gamma_s dW_s]=0$ is trivial. Therefore I'm interested in the non trivial case.

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 every mean 0 square integrable fctn can be represented as $\int \gamma_s dW_s$, so that is no constraint on the distribution. Also, check out the Burkholder Davis Gundy inequalities. – mike Jan 18 at 17:11

Let $f \in L^2([0,T])$ and $\{f_n\}$ be a sequence in $C([0,T])$ such that $f_n$ converges to $f$ in $L^2([0,T])$. It is easily seen that $$I_n = \int_0^T f_n(t)\,dW_t$$ defines a Cauchy sequence in $L^2(\Omega)$ (hence is convergent) since $$E\left[(I_n - I_m)^2\right] = E\left[\left(\int_0^T (f_n(t)-f_m(t))\,dW_t\right)^2\right] = \int_0^T (f_n(t)-f_m(t))^2\,dt.$$
Notice that, by definition, $$\int_0^Tf(t)\, dW_t = \lim_{n\to\infty} I_n.$$ Here the limit is in $L^2(\Omega)$, hence the answer to your first question follows.
 thanks for this help. Could you explain why $I_n$ should be Cauchy in $L^2(\Omega)$? – hulik Jan 14 at 18:53 @hulik: $f_n$ converges to $f$ in $L^2$ and $\|f_n-f_m\|_2\leq \|f-f_n\|_2+\|f-f_m\|_2$ by triangle inequality. – Alex Jan 14 at 20:16 @Alex But we must have $\|I_n -I_m\| \le \epsilon$ which seems not so easy, since we have stochastic integrals! I guess you have to use that $I_n$ is normally distributed with kown mean and variance – hulik Jan 15 at 7:53 @hulik: it is enough to use the "isometry" property of stochastic integration. See my answer above. – Ju'x Jan 15 at 8:56 @Ju'x Nice! thanks for the update. – hulik Jan 15 at 9:04