Let $0<\sigma_s<M$ be a bounded continuous stochastic process independent from a Brownian motion $W$. Let $\mathcal{F}_{\sigma}$ be the sigma-algebra generated by $\left\{\sigma_s|0\leq s\leq 1\right\}$.
I need to prove that
$$
\mathbb{E}\left[\left(\int_0^1\sigma_s\,dW_s\right)^k\left|\right.\mathcal{F}_{\sigma}\right] = \mathbb{E}\left[U^k\right]\left(\int_0^1\sigma_s^2\,ds\right)^{\frac{k}{2}}
$$
where $U\stackrel{d}{=}\text{N}\left(0,1\right)$.
Best Answer
Since $\sigma$ is bounded and continuous, it is not difficult to see that
$$\sum_{j=0}^{n-1} \sigma_{j/n} (W_{(j+1)/n}-W_{j/n}) \xrightarrow[]{n \to \infty} \int_0^1 \sigma_s \, dW_s \quad \text{in probability.} \tag{$\star$}$$
This implies
$$\begin{align*} \mathbb{E} \left( \exp \left[ i \xi \int_0^1 \sigma_s \, dW_s \right] \mid \mathcal{F}_{\sigma} \right) &= \lim_{n \to \infty} \mathbb{E} \left( \exp \left[ i \xi \sum_{j=0}^{n-1} \sigma_{j/n} (W_{(j+1)/n}-W_{j/n}) \right] \mid \mathcal{F}_{\sigma} \right) \\ &= \lim_{n \to \infty} \mathbb{E} \left( \prod_{j=0}^{n-1} \exp \left[ i \xi \sigma_{j/n} (W_{(j+1)/n}-W_{j/n}) \right] \mid \mathcal{F}_{\sigma} \right). \end{align*}$$
Since $(W_t)_{t \in [0,1]}$ and $\mathcal{F}_{\sigma}$ are independent, we obtain that
$$\mathbb{E} \left( \exp \left[ i \xi \int_0^1 \sigma_s \, dW_s \right] \mid \mathcal{F}_{\sigma} \right) = \lim_{n \to \infty} \mathbb{E} \left( \prod_{j=0}^{n-1} \exp \left( i \xi x_j (W_{(j+1)/n}-W_{j/n}) \right) \right] \bigg|_{x_j = \sigma_{j/n}};$$
using that $W_{(j+1)/n}-W_{j/n} \sim N(0,1/n$, $j=0,\ldots,n-1$ are independent, we find that
$$\begin{align*} \mathbb{E} \left( \exp \left[ i \xi \int_0^1 \sigma_s \, dW_s \right] \mid \mathcal{F}_{\sigma} \right) &= \lim_{n \to \infty} \prod_{j=0}^{n-1} \exp \left(- \frac{1}{2}\xi^2 x_j^2 \frac{1}{n} \right)\bigg|_{x_j = \sigma_{j/n}} \\ &= \lim_{n \to \infty} \exp \left(- \frac{1}{2} \xi^2 \sum_{j=1}^n \sigma_{j/n}^2 \frac{1}{n} \right). \end{align*}$$
The expression on the right-hand side is a Riemann sum and therefore we conclude that
$$ \mathbb{E} \left( \exp \left[ i \xi \int_0^1 \sigma_s \, dW_s \right] \mid \mathcal{F}_{\sigma} \right) = \exp \left(- \frac{1}{2}\xi^2 \int_0^1 \sigma_s^2 \, ds \right). \tag{1}$$ Finally, we note that
$$i^k \mathbb{E}(X^k \mid \mathcal{F}) =\frac{d^k}{d\xi^k} \mathbb{E}(e^{i \xi X} \mid \mathcal{F}_{\sigma}) \bigg|_{\xi=0}$$
for any real-valued random variable $X$ such that $\mathbb{E}(|X|^k)<\infty$; differentiating $(1)$ $k$ times with respect to $\xi$ and evaluating at $\xi=0$ we conclude that
$$ \mathbb{E} \left( \left[ \int_0^1 \sigma_s \, dW_s \right]^k \mid \mathcal{F}_{\sigma} \right)=\mathbb{E}(U^k) \left( \int_0^1 \sigma_s^2 \, ds \right)^{k/2}$$
for $U \sim N(0,1)$.
Remark: Instead of calculating the conditional characteristic function, you can also show that the convergence in $(\star)$ holds in $L^k$ and then use this convergence to conclude that $$\mathbb{E} \left( \left( \int_0^1 \sigma_s \, dW_s \right)^k \mid \mathcal{F}_{\sigma} \right) = \lim_{n \to \infty} \left( \left( \sum_{j=0}^{n-1} \sigma_{j/n} (W_{(j+1)/n}-W_{j/n}) \right)^k \mid \mathcal{F}_{\sigma} \right);$$ then you can proceed with similar arguments as above.