2
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ is the answer ok for you? $\endgroup$ – gunes May 3 '20 at 20:02
1
$\begingroup$

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 $\boldsymbol\alpha$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ 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 $\sigma^2 = Var[\boldsymbol{\alpha}]'\mathbf{b}^2$). $\endgroup$ – ben Apr 28 '20 at 22:35
  • $\begingroup$ Yes, you can use a log transform, but this assumes that the error terms are multiplicative (like percent errors). This usually makes sense for economic data, but is it true for your data? Likely true, but not definitely so. $\endgroup$ – AJKOER Apr 29 '20 at 0:02
  • $\begingroup$ @ben that's correct, you can do it for specific cases, but in general you can't. $\endgroup$ – gunes Apr 29 '20 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.