# Expected value of a sum of random variables raised by $e$?

There is a function $$y$$ defined

$$y=\exp(-\boldsymbol{\alpha}'\mathbf{b})\:\:;\:\:\:\:y\in(0,\infty)$$

where $$\boldsymbol{\alpha}$$ is a vector of random variables and $$\mathbf{b}$$ is a vector of non-random parameters.

So, the expected value of $$\ln y$$ is just

$$E[\ln y] = -E[\boldsymbol{\alpha}]'\mathbf{b}$$

My question is, how to go about finding a computable expression for $$E[y]$$ in terms of the means (and maybe standard deviations?) of $$\boldsymbol{\alpha}$$?

It seems like the law of the unconscious statistician should be of some use here. But I have been unable to make use of it.

Also, if it is known that $$y$$ is lognormally distributed ($$\ln y \sim N(\mu,\sigma^2)$$), does this facilitate things?

• is the answer ok for you? May 3 '20 at 20:02

Let $$\mathbf b=-\mathbf t$$, then $$\mathbb E[Y]=\mathbb E[e^{\mathbf t^T\boldsymbol \alpha}]$$ which is the moment generating function (MGF) for multivariate case, which is in general cannot be written by only first and second moments of $$\boldsymbol\alpha$$.
• Thanks for this answer. True, but actually, it just dawned on me that, if $\ln y \sim N(\mu, \sigma^2)$, then $E[y] = \exp(\mu + \sigma^2 / 2)$. Hence, $E[y]$ can be expressed in terms of means and standard deviations of $\boldsymbol{\alpha}$ (since $\mu = E[\ln y] = E[\boldsymbol{\alpha}]'\mathbf{b}$ and $\sigma^2 = Var[\boldsymbol{\alpha}]'\mathbf{b}^2$).