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Is the logarithm of an expectation the same as the expectation of the logarithm?

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  • $\begingroup$ A special case of this statement, if true, would be "the arithmetic mean of positive numbers always equals their geometric mean." That contradicts a very well known inequality: see en.wikipedia.org/wiki/…. $\endgroup$ – whuber May 29 '19 at 17:58
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No, it's not true. For a counter-example, we have Jensen's inequality, which implies $f(E[X])\geq E[f(X)]$, for concave functions, in which $\log x $ is also concave as in your case.

Or, for a concrete example, take $X\in\{1,2\}$ with equal probability. $\log(E[X])=\log(3/2)$ while $E[\log(X)]=\frac{1}{2}\log 2$, which are not equal.

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  • $\begingroup$ +1, note that the inequality is strict when $f(x)$ is strictly concave (unless $X$ is degenerate). $\endgroup$ – knrumsey May 29 '19 at 16:57
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    $\begingroup$ @knrumsey The idea's correct, but it's not quite that simple. Consider the case where $X$ is a Uniform variable, for instance and $f$ is linear on $[0,1]$ but strictly concave outside that range. $\endgroup$ – whuber May 29 '19 at 17:56
  • $\begingroup$ @whuber, I don't see that as a counterexample to my statement. The function that you posit is not strictly concave. I agree it's not the "whole picture" however. Consider the function $f(x) = x^2 I(x > 0)$. This function is not strictly concave, but the inequality will still be strict for a Uniform variable. $\endgroup$ – knrumsey May 29 '19 at 19:35
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    $\begingroup$ @knrumsey That's part of my point: sometimes the inequality will be strict, sometimes not, even when $f$ is strictly concave in the sense of being strictly concave somewhere. It sounds like you intended it to be strictly concave everywhere, which is fine now that your meaning is clear. $\endgroup$ – whuber May 29 '19 at 19:42
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No by Jensen's inequality we have $\log(\mathbb{E}[X])\geq \mathbb{E}[\log(X)] $ as $\log(\cdot)$ is a concave function.

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