I've been asked to determine the distribution of $\int_0^t h(s) \, dW_s$ where $h$ is a deterministic function of one variable, $W$ is brownian.
My thoughts were
$$ \int_0^t h(s) \, dW_s = \lim_{n \to \infty} \sum_{\pi (n)} h(t_i)(W_{t_{i+1}} - W_{t_i}) = Z $$
Where $\pi (n)$ is some partition. Now I know that $W_{t_{i+1}} - W_{t_i}$ is normal distributed, and a sum (finite?) of normal distributed is normal.
I calculated $E[Z] = 0$ and for the variance I need $E[Z^2]$,
$$ E[Z^2] = E\left[\left(\int_{0}^{t} h(s) \, dW_s \right)^2 \right] $$
Now if $h \in L^2$ then by Ito Isometry
$$ E[Z^2] = E\left[\left(\int_0^t h(s) \, dW_s \right)^2 \right] = \int_0^t E[h^2(s)] \, ds = \int_0^t h^2(s) \, ds = V $$
Thus I want to conclude that $Z \sim N(0,V)$