Let $h:[0,\infty)\to [0,1)$ be a strictly increasing and strictly concave function. Let the argument of this function be a random variable $C$ with probability density function (pdf) $f_{C}(c)$ with support on some subset of $[0,\infty)$. By an application of Jensen's inequality, we have

$$\mathrm{E}(h(C)) = \int_{0}^{\infty}h(c)f_{C}(c)dc < h\left(\int_{0}^{\infty}cf_{C}(c)dc\right) = h(\mathrm{E}(C))$$ where the expectations are assumed to exist. I wonder if this can be generalized as outlined below when the function $h$ is defined as above.

Assume that we have two random variables $C_{1}$ and $C_{2}$ for which $\mathrm{E}(C_{1}) = \mathrm{E}(C_{2})$, but the higher moments of their distributions $f_{C_{1}}$ and $f_{C_{2}}$ differ in general. In particular, we may take $\mathrm{Var}(C_{2})>\mathrm{Var}(C_{1})$.

  • Can anything be said about which quantity is larger in this case, $\mathrm{E}(h(C_{1}))$ or $\mathrm{E}(h(C_{2}))$?

Intuitively, since $h$ is concave and increasing, I expect $\mathrm{E}(h(C_{2})) < \mathrm{E}(h(C_{1}))$ (the variable with the largest variance produces the smallest expectation), and certainly if $f_{C_{1}}$ and $f_{C_{2}}$ are both symmetric. But does this hold in general? I would be a little surprised if it does, but I haven't been able to find a simple counterexample.

  • $\begingroup$ can you not use the fact that a the negative of a strictly concave function is strictly convex (at least from my limited knowledge in this area) and apply the same tricks? $\endgroup$
    – Chinny84
    Apr 7 '15 at 8:38
  • $\begingroup$ @Chinny84 It is correct that the negative of a concave function is convex, but I can't see a way to exploit that fact here... $\endgroup$
    – vgnils
    Apr 8 '15 at 12:33

To answer my own question: It does not hold in general, i.e. $\mathrm{Var}(C_{2}) > \mathrm{Var}(C_{1})$ does not guarantee that $\mathrm{E}(h(C_{2})) < \mathrm{E}(h(C_{1}))$. Simple numerical testing produced counterexamples.

However, the concept of stochastic dominance is what I was looking for. It provides neccesary and sufficient conditions on $h$, $C_{1}$ and $C_{2}$ for an inequality like $\mathrm{E}(h(C_{2})) < \mathrm{E}(h(C_{1}))$ to hold. For example, with $h$ concave and increasing, the inequality holds whenever $C_{2}$ is a mean-preserving spread of $C_{1}$ (2nd order stochastic dominance). If $C_{2}$ is a mean-preserving spread of $C_{1}$, then $\mathrm{Var}(C_{2}) > \mathrm{Var}(C_{1})$, but the converse is not true in general.


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