Jensen’s Inequality and Equality Conditions for Convex Functions

Summary

The document asks when the expectation of a function of a random variable equals the function evaluated at that variable’s expectation. It considers a convex function and whether equality at a particular parameter value implies that the function is locally linear there, with a zero second derivative.

The answer invokes Jensen’s inequality: for a convex function, the expected function value is at least the function evaluated at the expected input. Strict inequality follows under additional conditions, such as a nonconstant random variable and strict convexity. However, the response corrects its initial overstatement: equality alone does not establish that the second derivative vanishes. The example discussion notes that the function’s behavior may be unspecified at the parameter of interest, and derivatives may not even exist. The conclusion is therefore limited: stronger assumptions about the function and random variable are needed to infer local linearity.

Key ideas

• Jensen’s inequality gives an expected function value no smaller than the function at the expected input for a convex function.
• Strict convexity and a nonconstant random variable yield strict inequality under the stated conditions.
• Equality alone does not prove that the function has zero second derivative at the point of interest.
• Any conclusion about local linearity depends on additional assumptions, including differentiability and the function’s behavior.

Tags

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


# 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.

## Answer by Charles Fox (score -1)

https://quant.stackexchange.com/a/45130

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) $f''(X)>0$

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.

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