Assume that $\text{E}[\log(X)] $ is given, can I derive $\text{E}[X]$ in a closed form format?
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$\begingroup$ Do you know the distribution of log x or x? Eg. is is $x$ normally distributed? In the most general case with $x$ following some arbitrary distribution, you won't be able to say much. $\endgroup$– Matthew GunnCommented Sep 19, 2016 at 15:41
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$\begingroup$ No. Only the mean value of the log(x) is given. $\endgroup$– TheodenCommented Sep 19, 2016 at 15:43
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11$\begingroup$ Pretty much all you can say is that $\mathrm{E}[\log(x)] \leq \log \left( \mathrm{E}[x] \right)$ due to Jensen's inequality. $\endgroup$– Matthew GunnCommented Sep 19, 2016 at 15:55
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No.
For example, if $X$ follows a log normal distribution, where $\log(X) \sim N(\mu,\sigma)$, then $E[\log(x)] = \mu$ and is independent of $\sigma$. However, its mean is $E[X] = \exp \left(\mu + \frac{\sigma^2}{2} \right)$. Clearly, you cannot derive a $\sigma$ dependent number from a $\sigma$ independent number.
answered Sep 19, 2016 at 15:42
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4$\begingroup$ Or for a super-simple example, consider discrete random variables X (taking value 1 with probability 1/3 and 10 with probability 2/3), and Y (taking value 1 with probability 2/3 and 100 with probability 1/3). Then E[log X] = E[log Y] = 1/3 (or some other number if you prefer your logarithms taken to a more sensible base than 10, but still equal since in any base log 100 = 2 log 10). However E[X] != E[Y]. $\endgroup$ Commented Sep 19, 2016 at 19:32