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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?

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  • $\begingroup$ is the answer ok for you? $\endgroup$
    – gunes
    May 3, 2020 at 20:02

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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 $\boldsymbol\alpha$.

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    $\begingroup$ 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, 2020 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, 2020 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, 2020 at 8:02

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