I am having trouble understanding the concept of Jensen's inequality in a probabilistic setting with multiple variables. Specifically, I am interested in cases where equality holds. I understand that in the univariate case
\begin{equation}
f(\text{E}[X])\leq \text{E}[f(X)]
\end{equation}
for an integrable, real-valued random variable $X$ and a convex function $f$, and that equality holds if $X$ is constant almost surely or $f$ is linear on some set $A$ such that $\text{P}(X \in A) = 1$ (according to wikipedia and some answered questions on this site).
I am trying to find a similar condition for the multivariate case with independent random variables. For example, consider the function $g(X, Y) = XY$ for two independent, integrable, real-valued random variables $X$ and $Y$ (e.g. normal distributed). In that case
\begin{equation}
\text{E}[g(X, Y)] = \text{E}[XY] = \text{E}[X]\text{E}[Y] = g(\text{E}[X],\text{E}[Y])
\end{equation}
which suggests to me that $g$ has some property that causes equality. However, if my understanding is correct, $g$ is not linear (it is for example not homogeneous of degree 1). I assume that the paragraph about general inequality in a probabilistic setting on wikipedia is relevant to my question, but it is far above my level of understanding.
Is there a condition for equality in the multivariate case besides almost surely constant random variables (and specifically how does $g$ above meet this condition) or do I have a serious misunderstanding about the applicability of Jensen's inequality on this case?
Best Answer
Your function $\ g\ $ is neither convex nor concave, so Jensen's inequality isn't applicable. If $\ x=\Big(\frac{1}{2},\frac{-1}{2}\Big)\ $ and $\ y=\Big(\frac{-1}{2},\frac{1}{2}\Big)\ $, for instance, then \begin{align} g\Big(\frac{1}{2}x+\frac{1}{2}y\Big)&=g((0,0))\\ &=0\\ &> \frac{-1}{4}\\ &= \frac{1}{2}g(x)+\frac{1}{2}g(y)\ , \end{align} so $\ g\ $ isn't convex. On the other hand, if $\ u=\Big(\frac{1}{2},\frac{1}{2}\Big)\ $ and $\ v=\Big(\frac{-1}{2},\frac{-1}{2}\Big)\ $, then \begin{align} g\Big(\frac{1}{2}u+\frac{1}{2}v\Big)&=g((0,0))\\ &=0\\ &< \frac{1}{4}\\ &= \frac{1}{2}g(u)+\frac{1}{2}g(v)\ , \end{align} so $\ g\ $ isn't concave either.