# Monotonicity of expectation of a concave function of a random variable wrt the variance of the random variable

This is a question motivated from utility function. (See here and here.) I have been trying to develop some common sense in Economics by the way.

Given a function $f: \mathbb{R} \to \mathbb{R}$ and a real-valued random variable $X$, I was wondering if there is some relation between the convexity of $f$ and the monotonicity of $\mathrm{E}(f(X))$ in $\mathrm{dev}(X)$, the deviation of $X$?

1. When $f$ is a linear function, $\mathrm{E}(f(X))$ is constant in $\mathrm{dev}(X)$. Is the reverse true?

2. Does $f$ being concave/convex imply $\mathrm{E}(f(X))$ is decreasing/increasing in $\mathrm{dev}(X)$? If not, some conditions for them to be true? Is the reverse true?

For example, when the utility function is $f(x) := x - x^2/2$, it is concave over $\mathbb{R}$. $$\mathrm{E}(f(X)) = \mathrm{E} X - (\mathrm{E}X)^2/2 - \mathrm{Var}X/2$$ which is decreasing in $\mathrm{dev}(X)$. By the decreasing, I guess this utility function is therefore risk-averse, based on my limited knowledge in economics ($\mathrm{dev}(X)$ represents the risk).

Jensen's inequality only gives an inequality relation between $\mathrm{E}(f(X))$ and $\mathrm{E} X$, where $\mathrm{dev}(X)$ is not involved.

Thanks and regards!

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Is dev(X) the standard deviation of X, $\sqrt{Var(X)}$? What do you mean for $E[f(X)]$ to be increasing in $dev(X)$? Is it that, for any $X$ and $Y$ such that $dev(X) < dev(Y)$, $E[f(X)] < E[f(Y)]$. This doesn't work in your example, if you let $E[Y] \neq E[X]$. What is the class of random variables you are considering? All of them? Only some kind of parametric famility (For example, normals with fixed mean)? –  madprob Jun 25 '12 at 22:59
@madprob: (1) $dev(X)$ is the standard deviation of $X$. (2) In economics, $X$ represents some random wealth. At the same expected value of $X$, the higher $dev(X)$ is, the riskier $X$ is, and for those risk-averse people, the lower the utility of $X$ is. (3) The statements in (2) is distribution-free, but if it is not true, it could be my translation of the questions from economics to mathematics is wrong, and I wonder what conditions can make it true? –  Tim Jun 25 '12 at 23:52
Take $f(x) = X^{4}$. $f$ is convex and $E[f(X)]$ is $E[X^{4}]$ which cannot be written in terms of $dev(X)$. It is possible for $dev(X) = dev(Y)$, $E[X] = E[Y]$ but $[X^{4}] \neq E[Y^{4}]$. So I think that stronger assumptions on $f$ or on the distributions are necessary. For example, you can consider distribution which are completely specified with mean and variance or $f$ just among the quadratic functions. –  madprob Jun 26 '12 at 0:23