Find conditional Expectation of $E\left[e^{X} | \; (Y, Z) \right]$.

Question

I have the following problem: \begin{equation*} \text{Let } (X, Y, Z) \sim N \left( \left(\begin{array}{cccc} 0\\ 0\\ 0 \end{array} \right), \left( \begin{array}{cccc} 4 & 1 & 0\\ 1 & 3 & 1\\ 0 & 1 & 2 \end{array} \right) \right). \end{equation*} ( A gaussian random vector with zero mean and given covariance matrix). Find the conditional expectation of $E\left[e^{X} | \; (Y, Z) \right]$.

My attempt:

We know, that \begin{equation*} E\left[e^{X} | \; (Y, Z) \right] = \int e^{x} \cdot p_{X | (Y, Z)} (x | (y, z)) \; dx, \end{equation*} where the conditional density \begin{equation*} p_{X | (Y, Z)} \; (x | (y, z)) = \frac{p_{X, Y, Z}(x, y, z)}{p_{Y, Z}(y, z)} \cdot I \left[p_{Y, Z}(y, z) > 0\right] \end{equation*} Here, $p_{X, Y, Z}(x, y, z)$ and $p_{Y, Z}(y, z)$ are the joint density functions. Also, we can calculate, for example, $p_{X, Y, Z}(x, y, z)$ by using the formula, that if $\xi \sim N(a, \Sigma)$, and $\Sigma$ is positive definite, we have \begin{equation*} p_{\xi}({\bf{t}}) = \frac{1}{2 \pi ^ {n / 2} \cdot \sqrt{det{\; \Sigma}}} \cdot exp \left(-1/2 \cdot \langle \Sigma^{-1} (\bf{t} - a), (\bf{t} - a) \rangle \right) \end{equation*} (Here $n$ is the length of $a$).

However, by this method the computations are quite ugly, and I wasn't able to finish them. Also, I don`t think that we can easily find the desired integral. So my question is, is this approach correct, and is there a better method?

Answer 1

Here is a method which can work in a more general setting. Let $V:= X-a Y-bZ$, where the constants $a$ and $b$ are chosen in such a way that $\operatorname{cov}(V,Y)=\operatorname{cov}(V,Z)=0$. Then $V$ is independent of $(Y,Z)$ and $$ \mathbb E\left[\exp(X)\mid Y,Z\right]=\mathbb E\left[\exp(V+aY+bZ)\mid Y,Z\right]=\exp(aY+bZ)\mathbb E\left[\exp(V)\mid Y,Z\right]. $$

Answer 2

More generally, let $X$ and $Y$ be Gaussian vector taking values in $\mathbb{R}^k,$ and $\mathbb{R}^p,$ suppose that $(X,Y)$ is also a Gaussian vector, with mean $\begin{pmatrix}E[X]\\ E[Y]\end{pmatrix}$ and covariance matrix $\begin{pmatrix} R_X&R_{XY} \\ R_{YX}&R_Y\end{pmatrix}$ where $R_X,R_Y$ are the covariance matrices of $X$ and $Y,$ and respectively, and where $R_{XY}=E[(X-E[X])^t(Y-E[Y])],$ we assume moreover that $R_Y$ is positive defined (and thus invertible), than the conditional law of $X$ given $Y,$ is Gaussian with mean $E[X|Y]=E[X]+R_{XY}R^{-1}_Y(Y-E[Y])$ and covariance matrix $R_X-R_{XY}R^{-1}_YR_{YX}.$

Take $W=(Y,Z).$ $R_W=\begin{pmatrix}3& 1\\ 1 & 2\end{pmatrix},R_W^{-1}=\begin{pmatrix}2/5& -1/5\\ -1/5 & 3/5\end{pmatrix},$ $R_{XW}R^{-1}_{W}=\begin{pmatrix}1/5 & -1/5 \end{pmatrix},R_X-R_{XY}R^{-1}_YR_{YX}=\sigma^2=19/5$ then for $P_W$-almost all $(y,w) \in \mathbb{R}^2$ $$E[e^X|W=(y,w)]=\int_{\mathbb{R}}e^xdP_{X|W=(y,w)}(x)=\frac{1}{\sigma\sqrt{2\pi}}\int_{\mathbb{R}}e^{x-(x-(y-w)/5)^2/2\sigma^2}dx=e^{y/5-w/5+19/10},$$ so $E[e^X|Y,Z]=e^{Y/5-Z/5+19/10},$ notice that $a=1/5,b=-1/5,E[e^V]=e^{19/10}$ in the above answer.