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