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I have a set of N samples from an infinite set of values which are known to be logarithmically distributed (lots of small values, a few much larger large values).

I have a hunch that a geometric mean will converge more quickly to the actual mean of the infinite set than an arithmetic mean would.

Am I right?

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    $\begingroup$ what do you mean by "the actual mean"? $\endgroup$
    – Suvrit
    Dec 1, 2010 at 21:10

1 Answer 1

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As a particular case of the following proposition, the geometric mean will converge (almost surely) to a number smaller than the actual mean.

Proposition. Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. non-constant random variables taking values in $[1,\infty)$ and having a finite mean ${\rm E}(X_1) < \infty$ (this is the case for the logarithmic distribution). Then, almost surely, $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{X_1 \cdots X_n }} < {\rm E}(X_1).$

Proof. Define $Y_i = \ln X_i$. Thus, $(Y_i)$ is a sequence of i.i.d. rv's taking values in $[0,\infty)$ and having a finite mean ${\rm E}(Y_1) (< {\rm E}(X_1))$. Then, by the strong law of large numbers, $\bar Y_n \to {\rm E}(Y_1)$ almost surely, where $\bar Y_n = \frac{1}{n}\sum\nolimits_{i = 1}^n {Y_i } $. Hence, since $\sqrt[n]{{X_1 \cdots X_n }} = \exp (\bar Y_n )$, almost surely $\sqrt[n]{{X_1 \cdots X_n }}$ converges to $\exp({\rm E}(Y_1))$. By Jensen's inequality, since the function $e^x$ is strictly convex and $Y_1$ is non-constant, $\exp({\rm E}(Y_1)) < {\rm E}[\exp(Y_1)]$. Noting that the right-hand side is ${\rm E}(X_1)$, the proposition is proved.

EDIT: In fact, the statement of this proposition continues to hold even if we replace $[1,\infty)$ with $(0,\infty)$. I note this since the OP might be interested in a distribution other than the logarithmic distribution (which is supported on the positive integers).

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  • $\begingroup$ Assuming the rv's are non-degenerate... $\endgroup$ Dec 2, 2010 at 16:43
  • $\begingroup$ Right! I'll edit my answer, $\endgroup$
    – Shai Covo
    Dec 2, 2010 at 18:02

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