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This question inspires a further question: For a probability distribution supported on the positive half-line, is the difference between the arithmetic and geometric means bounded by some known function of either the coefficient of variation ($\sigma/\mu$) or some other measure of relative dispersion? $$ \begin{gather} \text{arithmetic mean minus geometric mean} \le g\left( \frac \sigma \mu \right) \text{?} \\ (\text{What function might $g$ be?}) \\[6pt] (\text{Clearly $g(0)$ would be $0$.}) \end{gather} $$[etc. etc. etc.]

It was pointed out in comments below, first implicitly by Nick Cox, then explicitly by me, then by whuber, that the above won't work. So here's a revised version of the question. Is there some function $h$ for which $$ \frac{\text{arithmetic mean minus geometric mean}} \sigma \le h\left( \frac \sigma \mu \right) \text{?} $$ I briefly thought that $h=$ the identity function, might work, so that $\sigma^2/\mu$ would serve as an upper bound. But the following seems to be a counterexample: $$ X = \begin{cases} 0.01 & \text{with probability } 0.01, \\ 1.01 & \text{with probability } 0.99. \end{cases} $$ Here we have $$ \frac{\sigma^2} \mu = \frac{0.0099} 1 $$ whereas $$ \text{arithmetic mean minus geometric mean} \approx 0.03555346. $$

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Michael Hardy
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    I don't know, but dimensional analysis alone implies that the difference between the arithmetic and geometric means has the same dimensions and units as the variable in question; as the coefficient of variation is dimensionless and unit-free, any such function must also, I conjecture, contain such a term. – Nick Cox Jun 28 '16 at 22:01
  • @NickCox : Good point. So I'd expect $g(x) = \sigma h(x)$, for some function $h$ whose value is dimensionless. (Or else maybe some other measure of dispersion would be where I put $\sigma$.) $\qquad$ – Michael Hardy Jun 28 '16 at 22:04
  • Maybe one should express it as $$ \frac{\text{arithmetic mean minus geometric mean}} \sigma \le h\left(\frac\sigma\mu\right). $$ – Michael Hardy Jun 28 '16 at 22:05
  • ok, So this bound cannot depend ONLY on $\sigma/\mu$, since, by rescaling, one can make the difference in means as large as desired without changing $\sigma/\mu$. But maybe $\sigma^2/\mu$ could be a bound? $\qquad$ – Michael Hardy Jun 28 '16 at 22:30
  • The answer is "obviously not," because by rescaling all the data, $\sigma/\mu$ will not change while the difference between the AM and GM can be made as large or small as you wish (provided the data are not all equal). The right question to pose concerns the *ratio* of the AM to the GM. For a partial analysis from a similar point of view, please see my answer in the referenced thread. – whuber Jun 28 '16 at 22:58
  • @whuber : You point out what I had already said in an earlier comment, and in light of that point, I've revised the question. $\qquad$ – Michael Hardy Jun 28 '16 at 23:47

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Call the probability distribution in question $p(x)$.

I think you're basically asking what the error term in Jensen's inequality is. Specifically, you are comparing $E[X]$ to $\exp (E[\log(X)])$, or equivalently $\log E[X]$ to $E[\log(X)]$. In this case the convex function is $f(x)=-\log(x)$. Lets assume for a moment that 0 is not in the support of $p(x)$, maybe even that $p(x)=0$ for $x\leq a$ for some $a>0$. This will ensure $f(x)$ is Lipschitz on $[a,\infty)$. Set $\mu:=E[X]$ and $\sigma:=\sqrt{\mbox{Var}(X)}$ Now start Taylor expanding $f(x)$ around $\mu$, i.e. like the top answer here. We also have $f''(x)=1/x^2\leq \lambda:=1/a^2.$ Then we get:

$$E[f(X)]-f(E[X])\leq \frac{\lambda\sigma^2}{2}=\frac{\sigma^2}{2a^2}$$

I'm guessing if you do a second order expansion, you'll get a bound that also involves the mean.

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