Compute expectation of the cube of stochastic integral

Question

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?

Answer 1

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

$$\mathbb{E}(X_t \mid \mathcal{F}_s) = 0 + 3 \cdot 0 + (t^3-s^3) \int_0^s u \, dW_u + \left( \int_0^s u \, dW_u \right)^3.$$

Answer 2

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.