Let $X=(X_1,\dots,X_n)$ be a $\mathbb R^n$-valued random vector such that $E(|X_i|)<\infty$ for all $i$. Let $f: \mathbb R^n \to \mathbb R$ be a convex function.

Jensen's inequality tells us that $E(f(X_1,\dots,X_n))$ exists (in $]-\infty,\infty]$) and that $$E(f(X_1,\dots,X_n)) \ge f(E(X_1),\dots,E(X_n)).$$

So if we replace each $X_i$ by its expectation $E(X_i)$ we get something smaller. Does this still hold if we substitute only some of the $X_i$ by their expectations?

Question: Does it hold that $E(f(X_1,\dots,X_n)) \ge E(f(E(X_1),X_2\dots,X_n))$?

Here are my thoughts:

Using the conditional Jensen's inequality we get that \begin{align*} E(f(X_1,\dots,X_n)) &= E(E(f(X_1,\dots,X_n)|X_2,\dots,X_n))\\ &\ge E(f(E(X_1|X_2,\dots,X_n),X_2\dots,X_n)) \end{align*} holds whenever $E(|X_1||X_2,\dots,X_n)$ is a.s. finite.

If $X_1,\dots,X_n$ are independent it follows that $$E(f(X_1,\dots,X_n)) \ge E(f(E(X_1),X_2\dots,X_n))$$ and we can iterate this to get $$E(f(E(X_1),X_2\dots,X_n)) \ge E(f(E(X_1),E(X_2),X_3\dots,X_n)),$$ etc.

But what if $X_1, \dots, X_n$ are not independent?

  • 1
    $\begingroup$ convex functions are continuous and thus measurable $\endgroup$ Jun 26, 2020 at 19:12
  • 1
    $\begingroup$ @mathworker21 I'm not sure if this is true. Consider a function defined on the unit circle in $\mathbb R^2$. Take a nonmeasurable set $B$ from the boundary of the circle (I am guessing this exists). Define a function to be 1 on $B$ and 0 on the rest of the circle. Then the function is convex but neither continuous nor measurable. $\endgroup$
    – Epiousios
    Jun 26, 2020 at 19:20
  • 1
    $\begingroup$ convex functions are continuous on the interior of the domain on which they are convex. you gave domain $\mathbb{R}^n$ $\endgroup$ Jun 26, 2020 at 19:22
  • 1
    $\begingroup$ Ok, fair enough. $\endgroup$
    – Epiousios
    Jun 26, 2020 at 19:23
  • $\begingroup$ Can't you just restrict to $n=2$? (i.e. if you prove it for $n=2$, you get it for $n \ge 3$) $\endgroup$ Jun 26, 2020 at 19:29

1 Answer 1


Let $X_1$ be any non-constant random variable, and let $X_2=-X_1$.

For $f(x,y)=(x+y)^2$, we have





You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .