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Suppose we have two random variables $X$ and $Y$ with the same mean but different variances, say $\operatorname{Var}(X) > \operatorname{Var}(Y)$, and $f$ is a convex function. Is it possible to compare the expectations $E(f(X))$ and $E(f(Y))$? In the case $\operatorname{Var}(Y)=0$, it reduces to Jensen's inequality.

This is motivated by the following thought: if $f$ is convex increasing, we can interpret it as the utility function of a risk-loving individual, in which case $E(f(X)) > f(E(X))$, i.e. facing a gamble $X$ the individual would prefer to play the gamble instead of taking the expected value. I'm wondering whether we can generalise it, so that if a gamble $X$ is riskier than a gamble $Y$ (in the sense that $\operatorname{Var}(X)>\operatorname{Var}(Y)$, the risk-loving individual would prefer the riskier gamble (i.e. $E(f(X)) > E(f(Y))$)?

Thanks!

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Taking $f$ as the identiy function (which is convex and increasing) would yield $E(X)>E(Y)$, so there's a very fundamental contradiction in this statement.

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No, variance is not strong enough to ensure such a property. Think of random variables $X$ and $Y$ with

$$ P(X=-1) = P(X=1) = 1/2 $$

and

$$ P(Y = -10)=P(Y = 10)= 1/1000, \quad P(Y = 0)= 1 - 2/1000.$$

You can check that the variance of $X$ is larger than that of $Y$, but $E[X^{10}] < E[Y^{10}]$ for instance.

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  • $\begingroup$ That was such a simple counterexample. Thank you!. Do you think an inequality like that would hold if we strengthen $f$ to be convex increasing? $\endgroup$ – Viet Dang Jul 1 '19 at 21:41
  • $\begingroup$ @Viet Dang: No, see my answer below. $\endgroup$ – Mars Plastic Jul 1 '19 at 22:49

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