Logarithmic mean of two positive real numbers is well defined in the literature, it has also been extended to more than two arguments in various papers. Is there any notion of logarithmic mean of random variables or functions? Thank you for your help and time.
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$\begingroup$ By the way, I know that logarithmic mean of convex functionals is there in the literature, but I want to know can we define it for random variables or just positive functions ? $\endgroup$– andyCommented Oct 31, 2013 at 23:07
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2 Answers
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You're looking for $e^{E \log X}$. It has all the nice properties you'd like it to.
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1$\begingroup$ Could you expand a bit on your answer? $\endgroup$– SuvritCommented Nov 1, 2013 at 1:27
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$\begingroup$ Yes, Omer, I would greatly appreciate if you could kindly elaborate a little more, anyway thank you so much. $\endgroup$– andyCommented Nov 1, 2013 at 3:00
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2$\begingroup$ For two positive real numbers, this gives the geometric mean, not the logarithmic mean. Have you looked at en.wikipedia.org/wiki/Logarithmic_mean ? $\endgroup$– S. Carnahan ♦Commented Nov 1, 2013 at 9:42
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Once we know how to define a geometric mean of two quantities, the logarithmic mean is just an integral away.
Indeed, say $g(t) := e^{E[(1-t)\log X]+E[t\log Y]}$, then we could define a logarithmic mean as $$L(X,Y) := \int_0^1 g(t)dt.$$
Reasoning: The above idea is inspired by noting that the ordinary logarithmic mean between two positive scalars, $x$ and $y$ may be viewed as $L(x,y) = \int_0^1 x^{1-t}y^tdt$, where the integrand is nothing but the (weighted) geometric mean.
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$\begingroup$ As I understood it, the question is not asking “give the logarithmic mean of two random variables”, but rather “define the logarithmic expectation of a single random variable, generalising logarithmic mean analogously to how ordinary expectation generalises the arithmetic mean of a finite multiset”. $\endgroup$ Commented Nov 2, 2013 at 23:25
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$\begingroup$ @Peter: so maybe I've misunderstood the question....hopefully the OP clarifies; but since the name "logarithmic mean" has a well-defined meaning, I interpreted it as above.... $\endgroup$– SuvritCommented Nov 3, 2013 at 0:01