Computing expectation of: $\small E\left[f(Z,U)e^{\frac{V^2-(V+W)^2}{2}} \right] $

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Suppose we have three mutually independent random variables $U,V,W$ where $W \sim \mathcal{N}(0,1)$, $V \sim \mathcal{N}(0,c)$ and $E[U]=0$, $E[U^2]=1$.

Lets define $Z=U+V+W$.

Can we compute (or simplify) the following quantity \begin{align} E\left[f(Z,U)e^{\frac{V^2-(V+W)^2}{2}} \right] \end{align} where $f$ is some deterministic function. For now I don't want to make any assumptions about $f$ but if we need to here are some assumptions that we can make:

  1. $f(Z,U)>0$

  2. $0 \le E\left[f(Z,U)\right] \le 1$

I was thinking some thing like this

\begin{align} E\left[f(Z,U)e^{\frac{V^2-(V+W)^2}{2}} \right]&= E\left[ E\left[f(Z,U)e^{\frac{V^2-(V+W)^2}{2}} \Big| Z,U\right] \right]\\ &=E\left[ f(Z,U)E\left[e^{\frac{V^2-(V+W)^2}{2}} \Big| Z,U\right] \right] \end{align}

Now the question is how to compute the quantity $E\left[e^{\frac{V^2-(V+W)^2}{2}} \Big| Z,U\right]$. The good think is that $V$ and $W$ is are Gaussian.

Note that we can rewrite $E\left[e^{\frac{V^2-(V+W)^2}{2}} \Big| Z,U\right]=E\left[e^{\frac{-W^2-2VW}{2}} \Big| Z,U\right]$. The above looks like a moment generating function (but conditioned) and is related to this question.

Thank you in advance for any help.