Let $X=(X_1,\ldots,X_N)$ be multivariate normally distributed. What is $\mathbb E(e^{X_i})?$

Question

The question is in the title. I am wondering that if we know that $X=(X_1,\ldots,X_N)$ is multivariate normally distributed with mean zero.

What can we say about $\mathbb E(e^{X_i})?$ The one-dimensional case is solved here on stackexchange

Answer 1

If $Y$ has normal distribution then $Ee^{Y}=e^{\mu} e^{\sigma^{2}/2}$ where $\sigma^{2}$ is the variance and $\mu$ is the mean of $Y$. From the joint normal distribution you have to find out the mean and variance of $X_i$ and apply this formula.

The mean of $X_i$ is simply the $i$-th coordinate of the mean vector and the variance of $Y_i$ is just $\Sigma_{ii}$ where $\Sigma$ is the variance-covariance matrix.

Answer 2

A random vector $(X_1,\ldots,X_N)$ has a multivariate normal distribution precisely if for every sequence $a_1,\ldots,a_N$ of constants (i.e. non-random scalars) the scalar-valued random variable $$ a_1 X_1 + \cdots + a_N X_N $$ has a univariate normal distribution. Notice the word "every" above. That means that in particular, it is true of the sequence $a_1=0,$ $a_2=0,$ $\ldots,$ $a_{i-1}=0,$ $a_i=1,$ $a_{i+1}=0,$ $\ldots,$ $a_N=0.$ So $X_i$ has a univariate normal distribution. That case, as you mentioned, was already posted.