Can Elog(x) Be Infinite for Some Distributions?

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Discussion Overview

The discussion revolves around the expectation of the logarithm of a random variable, specifically whether Elog(x) can be infinite for certain distributions. The scope includes theoretical considerations and implications of Jensen's inequality in the context of probability and statistics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that by Jensen's inequality, Elog(x) is less than or equal to logE(x), which is finite if the mean is finite.
  • Another participant asserts that since x > 0, log(x) must always be finite, implying E(log(x)) must also be finite.
  • However, a different viewpoint is presented that as x approaches 0, log(x) approaches -infinity, raising the possibility of Elog(x) being -infinity.
  • A further suggestion is made to consider the distribution x = exp(-1/u) where u is uniform on (0,1) as a potential counterexample.

Areas of Agreement / Disagreement

Participants express differing views on whether Elog(x) can be -infinity, indicating that the discussion remains unresolved with competing perspectives on the implications of the logarithmic behavior of x.

Contextual Notes

There are limitations regarding the assumptions about the distribution of x and the behavior of log(x) as x approaches 0, which are not fully explored in the discussion.

St41n
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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).
 

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