I stuck in determining the expected value of the following product
$E[W\_{t\_i}W\_{t\_{i+1}}^2]$ where $W\_{t\_i}$ and $W\_{t\_{i+1}}$ are Brownian with normal distribution, i.e. $W\_{t\_i}\sim N(0,t\_i)$. I would appreciate if anybody give me a hint about this.



Expected value of $W_{t_i} W^2_{t_{i+1}}$
Assuming that $t\_{i+1} \ge t\_i$, let $X = W\_{t\_i}$, $Y = W\_{t\_{i+1}} - W\_{t\_i}$. Note that $X,Y$ each have a centered normal distribution and are independent. Then the desired quantity is
$$\begin{align\*}E[X(X+Y)^2] &= E[X^3 + 2 X^2 Y + XY^2] \\
&= E[X^3] + 2 E[X^2] E[Y] + E[X] E[Y^2] && \text{by linearity and independence} \\
&= 0\end{align\*}$$
because $E[X^3] = E[Y] = E[X] = 0$ due to the symmetry of the normal distribution.


This is the usual trick for computing expectation of polynomial functions of Brownian motion at various times, or more generally of any collection of jointly Gaussian random variables: rewrite them as sums of independent increments. (For a jointly Gaussian family, this comes down to diagonalizing the covariance matrix.) Then expand and use independence. Odd moments of a centered Gaussian random variable all vanish, and there is [a simple formula](https://en.wikipedia.org/wiki/Normal_distribution#Moments) for even moments if you need them.


