Compute expectation of the cube of stochastic integral I want to compute:
$$\mathbb{E} \left(  \left(\int_0^tudW_u \right)^3  \mid \mathcal{F_s} \right),$$
hence I write $$\int_0^t \text{as} \int_0^s + \int_s^t.$$
Then I need to compute:
$$\mathbb{E} \left(  \left(\int_s^tudW_u \right)^3  \mid \mathcal{F_s} \right),$$
and
$$\mathbb{E} \left(  \left(\int_0^sudW_u \right)^3  \mid \mathcal{F_s} \right).$$
Here I stuck. Moreover, if I have even bigger power, what shall I do?
Is it possible to write:
$$\mathbb{E} \left(  \left(\int_s^tudW_u \right)^3  \mid \mathcal{F_s} \right) =  \left(\sum_0^n u_i (W_{u_{i+1}}-W_{u_{i}}) \right)^3 = \sum_0^n u_i^3 \left(W_{u_{i+1}}-W_{u_{i}} \right)^3 = 0?$$
because $W_{u_{i+1}}-W_{u_{i}}$ is a normal with $\sim N(0,1)$ and 3rd moment by wiki is 0?
 A: First of all: If you split up the integral into two parts, then you don't only have to calculate $\mathbb{E}((\int_s^t u W_u)^3 \mid \mathcal{F}_s)$ and $\mathbb{E}((\int_0^s u \, dW_u \mid \mathcal{F}_s)$, but also the mixed terms:
$$(a+b)^3 = a^3+3a^2b + 3ab^2 + b^3$$
with $$a = \int_0^s u \, dW_u \qquad \text{and} \qquad b = \int_s^t u \, dW_u.$$
Recall that
$$X_t := \int_0^t u \, dW_u$$
is a Gaussian process with mean $0$ and variance $\frac{t^3}{3}$ (see e.g. this question). Moreover, it is well-known that $X_t$ is $\mathcal{F}_t$-measurable and $X_t-X_s$ is independent of $\mathcal{F}_s$. Therefore,
$$\begin{align*} \mathbb{E} \left( \left( \int_s^t u \, dW_u \right)^3 \mid \mathcal{F}_s \right) &= \mathbb{E}((X_t-X_s)^3 \mid \mathcal{F}_s) \\ &= \mathbb{E}((X_t-X_s)^3) \\ &= 0. \end{align*}$$
where we have used in the last step that for a Gaussian process with mean $0$ the odd moments equal $0$. Similarly, we find
$$\begin{align*} \mathbb{E} \left( \left( \int_0^s u \, dW_u \right)^2 \left( \int_s^t u \, dW_u \right) \mid \mathcal{F}_s \right) &= \left( \int_0^s u \, dW_u \right)^2 \mathbb{E} \left( \int_s^t u \, dW_u \mid \mathcal{F}_s \right) \\ &= \left( \int_0^s u \, dW_u \right)^2 \mathbb{E}(X_t-X_s) \\ &= 0. \end{align*}$$
The third one is a bit different, but we can use a similar easoning:
$$\begin{align*} \mathbb{E} \left( \left( \int_0^s u \, dW_u \right) \left( \int_s^t u \, dW_u \right)^2 \mid \mathcal{F}_s \right) &= \int_0^s u \, dW_u \mathbb{E} \left( \left( \int_s^t u \, dW_u \right)^2 \mid \mathcal{F}_s \right) \\ &= \int_0^s u \, dW_u \mathbb{E}((X_t-X_s)^2 \mid \mathcal{F}_s) \\ &= \int_0^s u \, dW_u \mathbb{E}((X_t-X_s)^2). \end{align*}$$
By Itô's isometry, we have
$$\begin{align*} \mathbb{E}((X_t-X_s)^2) &= \mathbb{E} \left( \left( \int_s^t u \, dW_u \right)^2 \right)  = \int_s^t u^2 \, du = \frac{t^3}{3}-\frac{s^3}{3}. \end{align*}$$
Finally, because $X_s = \int_0^s u \, dW_u$ is $\mathcal{F}_s$-measurable, we have
$$\mathbb{E} \left( \left( \int_0^s u \, dW_u \right)^3 \mid \mathcal{F}_s \right) = \left( \int_0^s u \, dW_u \right)^3.$$
Combining the above calculations yields
$$\mathbb{E}(X_t \mid \mathcal{F}_s) = 0 + 3 \cdot 0 + (t^3-s^3) \int_0^s u \, dW_u  + \left( \int_0^s u \, dW_u \right)^3.$$
A: I have found another option, if somebody is like me ( afraid of dealing with $\int_0^tudW_u$ you can rewrite it:
$$d(uW_u) = udW_u + du W_u$$
$$d(uW_u) - du W_u= udW_u $$
$$ \int d(uW_u) -  \int W_u du= \int udW_u $$
$$ \left( tW_t -  \int_0^t W_u du \right)^3= \left( \int_0^t udW_u \right)^3 $$
using 
$$\int_0^t W_u du  = \int_0^s W_u du  + \int_s^t W_u du $$
we again do the same( opening the brackets, compute expectations), you will receive even more terms, but without nasty $\int_0^tudW_u$.
$$ \left( tW_t -  \int_0^s W_u du  - \int_s^t W_u du \right)^3= \left( \int_0^t udW_u \right)^3 $$
Moreover: Keep in mind, $d(uW_u) = udW_u + du W_u$ should be done by Ito's. So it is: $$d(uW_u) = udW_u + du W_u + \frac{1}{2}0 du,$$
i.e. it is not simple product rule.
