# Suppose $E(X)=E(Y)$ and $\operatorname{Var}(X)>\operatorname{Var}(Y)$, do we have $E(f(X))>E(f(Y))$ for convex functions $f$?

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!

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.

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.

• 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? – Viet Dang Jul 1 '19 at 21:41
• @Viet Dang: No, see my answer below. – Mars Plastic Jul 1 '19 at 22:49
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