What is the conditional expectation of the exponential functional? Consider the function $g(W)=-e^{-W}$, where $W$ is some random variable s.t.$W=X+YZ$. Furthermore, it holds that all the random variables $X,Y,Z$ follow the normal distribution with the following properties:


*

*$X\sim N(x,\sigma_x^{2})$, where $x$ denotes the constant mean of the r.v. $X$ and its variance is $\sigma_x^{2}>0$

*$Y\sim N(0,\sigma_y^{2})$ and $Z\sim N(0,\sigma_z^{2})$

*$X$ and $Z$ are correlated and we denote their covariance as $\sigma_{x,z}\neq 0$, while $\sigma_{y,z}=0$, $\sigma_{x,y}=0$.

*furthermore, $I=\{S,Y\}$ is an information set, where $S=X-x+U$, where $U$ is independently normally distributed with mean zero and variance $\sigma_u^{2}$, i.e. $U\sim N(0,\sigma_u^{2})$
If $(X,Y,Z,S)$ is jointly normally distributed, what is the conditional expectation$$E(g(W)|I)=E[-e^{-(X+YZ)}|\{S,Y\}]$$
 A: Reproducing this earlier question on the forum,

For ${\boldsymbol Y} \sim \mathcal{N}(\boldsymbol\mu, \Sigma)$,
  consider partitioning $\boldsymbol\mu$ and ${\boldsymbol Y}$ into 
  $$\boldsymbol\mu
 = \begin{bmatrix}  \boldsymbol\mu_1 \\  \boldsymbol\mu_2 \end{bmatrix} $$ $${\boldsymbol Y}=\begin{bmatrix}{\boldsymbol y}_1 \\  {\boldsymbol
 y}_2 \end{bmatrix}$$
with a similar partition of $\Sigma$ into $$  \begin{bmatrix}
 \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} $$
  Then, $f({\boldsymbol y}_1|{\boldsymbol y}_2={\boldsymbol a})$, the
  conditional distribution of the first partition given the second, is 
  $\mathcal{N}(\overline{\boldsymbol\mu},\overline{\Sigma})$, with mean 
  $$
\overline{\boldsymbol\mu}=\boldsymbol\mu_1+\Sigma_{12}{\Sigma_{22}}^{-1}({\boldsymbol a}-\boldsymbol\mu_2)
$$
  and covariance matrix
  $$
\overline{\Sigma}=\Sigma_{11}-\Sigma_{12}{\Sigma_{22}}^{-1}\Sigma_{21}$$

Therefore, since the vector $(X,Z,S)$ is Gaussian with zero mean, without loss of generality since $$\mathbb E[\exp\{X\}]=\mathbb E[\exp\{X-\mu+\mu\}]=\mathbb E[\exp\{X-\mu\}]\exp\{\mu\},\tag{1}$$ and arbitrary covariance matrix $\Sigma$, with the vector being independent from $Y$, the conditional distribution of $(X,Z)$ conditional on $(S,Y)$ is Gaussian with mean $\Sigma_{12}{\Sigma_{22}}^{-1}S$ and variance $\overline{\Sigma}$ (see above). This implies that the expectation of interest which is also the MGF of $X+YZ$ at $t=-1$ is
\begin{align*}\mathbb E[\exp\{W\}]&=\mathbb E[\mathbb E[\exp\{-\underbrace{(X+YZ)}_{(1\ Y)\,(X\ Z)^\text{T}}\}|S,Y]]\\
&=\exp\left\{-(1\ Y)\Sigma_{12}{\Sigma_{22}}^{-1}S + 
\frac{1}{2}(1\ Y) \overline{\Sigma} (1\ Y)^\text{T} \right\}
\end{align*}
since
$$(1\ Y)\,(X\ Z)^\text{T} | S,Y \sim \mathcal N\left((1\ Y)\Sigma_{12}{\Sigma_{22}}^{-1}S,(1\ Y) \overline{\Sigma} (1\ Y)^\text{T}\right)$$
A: A little more analytical I believe this is the case, following the solution of @Xi'an...Suppose that $W=X+ZY\sim N(\mu,\Sigma)$ is a partition where
$$\mu
 = \begin{bmatrix} \mu_X \\ \mu_Z \\ \mu_Y \end{bmatrix} = \begin{bmatrix} x \\ 0 \\ 0 \end{bmatrix}$$
and $$\Sigma=\begin{bmatrix} \sigma_x^2 & \sigma_{xz} & 0 \\ \sigma_{xz} & \sigma_z^2 & 0 \\ 0 & 0 & \sigma_{y}^2 \end{bmatrix}$$
Since, $(X,Z,S)$ is jointly normal vector, then we can claim that $\begin{bmatrix} X \\ Z  \end{bmatrix}|I=\{S,Y\}$ is also normal since $X\in I$ and $Z\sim N(0,\sigma_z^2)$, where the lemma gives you the $\mu_{\begin{bmatrix} X \\ Z\end{bmatrix}|I}=\mu_{xz}+(\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2},\dfrac{\sigma_{z}^2}{\sigma_{x}^2+\sigma_{u}^2})S=\begin{pmatrix}x + \dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2}S & \dfrac{\sigma_{z}^2}{\sigma_{x}^2+\sigma_{u}^2}S \end{pmatrix}^T$ and $\Sigma_{\begin{bmatrix} X \\ Z  \end{bmatrix}|I}=\begin{pmatrix}(1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{x}^2 & (1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{xz}\\
(1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{xz} & \sigma_{z}^2-\dfrac{\sigma_{xz}^2}{\sigma_{x}^2+\sigma_{u}^2}\end{pmatrix}$. Moreover $Y|I$ is also a normally distributed, since $Y\in I$ and in particular $\mu_{Y|I}=Y$ and $\Sigma_{Y|I}=0$. As a consequence, $W|I=X+ZY|I=1X+ZY|I=\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X \\Z\end{pmatrix}|I=\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I\sim N(\mathbb{E}[W|I], \mathbb{V}ar[W|I])$
Specifically, the conditional expectaion of the r.v. $W|I$ is also a mgf of a Gaussian normal distribution for $t=-1$, that it:
$$M_W(t)=\mathbb{E}[\exp\{-W\}]=\mathbb{E}\left(\mathbb{E}[\exp\{-(X+YZ)\}|I]\right)=\mathbb{E}[\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T\}|I]$$
Since, $Y \in I$ and $\begin{pmatrix}Z & X\end{pmatrix}^T|I\sim N\left(\mu_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}, \Sigma_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)$
we obtain that
$$\mathbb{E}\left(\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\mathbb{E}[\exp\{-\begin{pmatrix}X & Z\end{pmatrix}^T\}|I]\right)=\\ =\mathbb{E}\left(\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\exp\{\underbrace{-\mu_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}-\dfrac{1}{2}\Sigma_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}}\}_{mgf\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)=\\
=\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\exp\{-\mathbb{E}\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\mathbb{V}ar\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\}=\\
=\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\mathbb{E}\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\begin{pmatrix}1 & Y \end{pmatrix}\mathbb{V}ar\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\begin{pmatrix}1 & Y \end{pmatrix}^T\}=\\
=\exp\{-\mathbb{E}\left({\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\mathbb{V}ar\left({\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\}$$
Thus, $W|I$ is the Gasussian normal with $\mu_{W|I}=\mathbb{E}[W|I]$ and $\Sigma_{W|I}=\mathbb{V}ar\left[W|I\right]$
I believe this is the full solution...it took me some hours to understand, but there is nothing complex here...
