# When $E[f(\alpha,X)] = f(\alpha, E[X])$

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.

• Do you mean for any random variable $X$ or just a particular one. Apr 15, 2019 at 16:41
• @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.
– user34971
Apr 15, 2019 at 16:50
• I think you can ignore the first variable, and then show that $f''(x)=0$ everywhere. Apr 15, 2019 at 19:12
• 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.
– g g
Apr 16, 2019 at 19:29
• -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 $f$.
– Hans
Apr 18, 2019 at 18:32

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

1) $$X$$ is not a constant

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

then

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

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.

• -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 $X$.
– Hans
Apr 16, 2019 at 2:53
• @Hans, thank you - I have corrected it in an edit. Apr 18, 2019 at 17:18
• 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.
– Hans
Apr 18, 2019 at 18:56