Can Elog(x) Be Infinite for Some Distributions?

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The discussion centers on the behavior of the expectation of the logarithm of a random variable, specifically whether E(log(x)) can equal negative infinity for certain distributions. It is established that for a random variable x > 0 with a finite mean, E(log(x)) is finite due to Jensen's inequality, which states E(log(x)) ≤ log(E(x)). However, the case where x approaches 0 is highlighted, as log(x) approaches negative infinity, raising the question of whether E(log(x)) can indeed be -∞. An example using the distribution x = exp(-1/u) where u is uniformly distributed on (0,1) is provided to illustrate this point.

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Let x>0 be a random variable with some distribution with finite mean and let E denote the expectation with respect to that distribution.
By Jensen's inequality we have Elog(x) =< logE(x) < +inf

But, does this imply that -inf < Elog(x) too? Or is it possible that Elog(x) = -inf

Sorry if my question is stupid. Thx in advance
 
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If x> 0 then log(x) is always finite and so E(log(x)) must be finite.
 
But when x -> 0 , log(x) -> -inf
 
Try x=exp(-1/u) where u is uniform on (0,1).
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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