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.
$\begingroup$
$\endgroup$
7
-
$\begingroup$ this average is zero, isn't it? $\endgroup$– Carlo BeenakkerCommented Oct 11, 2018 at 14:29
-
$\begingroup$ 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). $\endgroup$– user129994Commented Oct 11, 2018 at 14:36
-
$\begingroup$ 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". $\endgroup$– Thomas KojarCommented Oct 11, 2018 at 15:53
-
$\begingroup$ 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. $\endgroup$– Carlo BeenakkerCommented Oct 11, 2018 at 15:58
-
1$\begingroup$ For a one line answer, notice that Brownian motion is symmetric under reflection and your function is odd. $\endgroup$– Anthony QuasCommented Oct 11, 2018 at 17:49
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
0
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 for even moments if you need them.
answered Oct 11, 2018 at 17:16