Consider some differentiable function $f(X,Y)$, where $X$ and $Y$ are scalar RVs. Assume that $\frac{\partial^2 f}{\partial X^2}>0$ but that nothing is known about the rest of the Hessian of $f$. Can we nonetheless make the following claim, which is effectively Jensen's inequality in "one direction"? $$ E[f(X,Y)]>E[f(E[X],Y)] $$


1 Answer 1


Yes, that is true. For any fixed $y$, define $g_y(X) = f(X, y)$. Note that $g_y$ is a function only of $X$, and is strictly convex in it. Therefore, for any $y$,

$$ E_X[f(X, y)] = E_X[g_y(X)] > g_y(E[x]) = f(E_X[X], y). $$

So, for any $y$,

$$ E_X[f(X, y)] > f(E_X[X], y). $$

However, the LHS and RHS are RVs of $Y$, and you can write

$$ E_Y\left[ E_X[f(X, y)] \right] > E_Y \left[ f(E_X[X], y) \right]. $$

  • $\begingroup$ Just to be clear, to complete the argument, we'd apply the law of iterated expectations as follows: $E[f(X,y)] > f(E[X],y)\Rightarrow E[E[f(X,Y)\vert Y=y]]>E[E[f(E[X],Y)\vert Y=y]],$ which yields the expression given in the original post? $\endgroup$
    – sunga
    Jul 1, 2017 at 2:52
  • $\begingroup$ @sunga Exactly. The inequality holds for any $y$, but that means the LHS and RHS are random variables in $Y$. The law of iterated expectations completes it. $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 3:02

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