Is Elog(x) Always Finite?

[St41n]

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

 

[HallsofIvy]

If x> 0 then log(x) is always finite and so E(log(x)) must be finite.

 

[St41n]

But when x -> 0 , log(x) -> -inf

 

[bpet]

Try x=exp(-1/u) where u is uniform on (0,1).

 

[blue_raver22]

No, your question is not stupid. The answer to whether Elog(x) is always finite depends on the distribution of the random variable x.

Jensen's inequality tells us that for a concave function like log(x), the expectation of the function is always less than or equal to the logarithm of the expectation. This means that in most cases, Elog(x) will be finite. However, there are some distributions where Elog(x) can be -inf, such as when x follows a Cauchy distribution with a location parameter of 0.

In general, the expectation of a logarithmic function will be finite as long as the underlying distribution has a finite mean. But it is possible for Elog(x) to be -inf in certain cases. Therefore, it is important for scientists to understand the underlying distribution of their data when using logarithmic functions.