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.

  • 1
    $\begingroup$ Do you mean for any random variable $X$ or just a particular one. $\endgroup$ – Gordon Apr 15 '19 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 '19 at 16:50
  • 1
    $\begingroup$ I think you can ignore the first variable, and then show that $f''(x)=0$ everywhere. $\endgroup$ – Gordon Apr 15 '19 at 19:12
  • 1
    $\begingroup$ 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 '19 at 19:29
  • 1
    $\begingroup$ -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 '19 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

| improve this answer | |
  • 3
    $\begingroup$ -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 '19 at 2:53
  • $\begingroup$ @Hans, thank you - I have corrected it in an edit. $\endgroup$ – Charles Fox Apr 18 '19 at 17:18
  • 1
    $\begingroup$ 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 '19 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.