# Generalization of Jensen's inequality

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?

• convex functions are continuous and thus measurable Jun 26, 2020 at 19:12
• @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. Jun 26, 2020 at 19:20
• convex functions are continuous on the interior of the domain on which they are convex. you gave domain $\mathbb{R}^n$ Jun 26, 2020 at 19:22
• Ok, fair enough. Jun 26, 2020 at 19:23
• Can't you just restrict to $n=2$? (i.e. if you prove it for $n=2$, you get it for $n \ge 3$) Jun 26, 2020 at 19:29

Let $$X_1$$ be any non-constant random variable, and let $$X_2=-X_1$$.
For $$f(x,y)=(x+y)^2$$, we have
$$Ef(X_1,X_2)=E((X_1+X_2)^2)=0$$
$$E(f(EX_1,X2))=E((EX_1+X_2)^2)=E((X_2-EX_2)^2)=Var(X_2)>0$$.