# Expected value of $W_{t_i} W^2_{t_{i+1}}$

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.

• this average is zero, isn't it? – Carlo Beenakker Oct 11 at 14:29
• I am not sure. Any reason? If the subs were the same, i.e. $t_i$ which makes the whole thing $W_{t_i}^3$ then expected value of the new guy would be zero (skewness). – user129994 Oct 11 at 14:36
• Brownian motion states are jointly gaussian, so we can compute this expected value explicitly to get zero. For higher moments it becomes harder and there are formulas "From moments of sum to moments of product". – Thomas Kojar Oct 11 at 15:53
• the average is zero because $W_{t_{i+1}}=W_{t_i}+\delta W_{t_i}$ and the increment $\delta W$ is independent of $W$, hence all terms in $W_{t_i}(W_{t_i}+\delta W_{t_i})^2=W_{t_i}^3+2W_{t_i}^2\delta W_{t_i}+W_{t_i}\delta W_{t_i}^2$ average to zero. – Carlo Beenakker Oct 11 at 15:58
• For a one line answer, notice that Brownian motion is symmetric under reflection and your function is odd. – Anthony Quas Oct 11 at 17:49

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.