When $E[f(\alpha,X)] = f(\alpha,E[X])$, where $f$ is some convex function of the first and second variables, except when the first variable takes the value $\alpha$ in which case the equality holds, then intuitively f is a (locally) linear function of the second variable. But how do you prove this, i.e. prove that $f''(\alpha, E[X]) = 0$ where the prime denotes differentiation to the second variable? It's maybe simple to prove but I can't figure it out.

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    $\begingroup$ Do you mean for any random variable $X$ or just a particular one. $\endgroup$ – Gordon Apr 15 at 16:41
  • $\begingroup$ @Gordon thanks, $X$ is a particular random variable, but its distribution is not specified. $f$ is a bounded function and convex in both variables, $f$ is in fact the price of a claim. $\endgroup$ – ilovevolatility Apr 15 at 16:50
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    $\begingroup$ I think you can ignore the first variable, and then show that $f''(x)=0$ everywhere. $\endgroup$ – Gordon Apr 15 at 19:12
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    $\begingroup$ This is answered here:math.stackexchange.com/a/1160128/249524. You need to be careful about 1) the range of $X$ and 2) sets of measure zero. This is why the answer has the caveat about the essential range. $\endgroup$ – g g Apr 16 at 19:29
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    $\begingroup$ -1. In addition to what I have pointed out regarding the ill-posed-ness of your problem, it is ill-posed in that, you do not specify the domain of $f$. $\endgroup$ – Hans Apr 18 at 18:32

By Jenson's Inequality, $E[f(X)] >= f(E[X])$ if $f''(X) >= 0$.

When two additional constraints apply:

1) $X$ is not a constant

2) $f''(X)>0$


$E[f(X)] > f(E[X])$.

By contradiction, if $E[f(X)] = f(E[X])$, then either $X$ is a constant or $f''(X) = 0$.

Edit: @Hans is correct. I had assumed f''(x) is a constant, but that was not stated in your question. You claim is not true. You could have a discreet random variable X and and some arbitrary f(x) like the below. $E[f(X)] = 0 = f(E[X])$ but the derivatives of $f$ are undefined. Although you have told us the shape of $f$ for other values of $\alpha$ we know nothing about the shape for the $\alpha$ value of interest.

arbitrary pmf

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    $\begingroup$ -1. This "proof" is ambiguous and wrong. $f"(X)>0$ means $f"(X)>0, \forall X$ values with at least two distinct values. First you have not proved the strict inequality $E[f(X)] > f(E[X])$ under these conditions. Second, the contradiction can only lead to the existence of some $X_0$ such that $f"(X_0)=0$ but not for all $X$. $\endgroup$ – Hans Apr 16 at 2:53
  • $\begingroup$ @Hans, thank you - I have corrected it in an edit. $\endgroup$ – Charles Fox Apr 18 at 17:18
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    $\begingroup$ Your $f(\alpha=1,\cdot)$ does not seem to be convex. The problem is ill-posed as I have pointed out in the comment right below the question. Yes, if $f$ is defined on a discrete domain, $f''$ is undefined. However, if we let $f$ be defined on an interval containing the range of $X$ and if $X$ has strictly positive measure on no less than $2$ values, $f''$ exists at $\mathbf E[X]$ and $f''(\mathbf E[X])=0$. Try to prove this. $\endgroup$ – Hans Apr 18 at 18:56

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