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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?

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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.

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  • $\begingroup$ Thank you for the answer, this clears up my problem in this specific case. To confirm my understanding, would I be correct in saying the following then: If Jensen's inequality is applicable, then equality also holds in the multivariate case for linear functions. If Jensen's inequality is not applicable on a function, then I have to proof equality (or inequality) for that function using some other method. $\endgroup$
    – Fneuge
    Apr 29, 2021 at 15:08
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    $\begingroup$ Equality holds for all linear functions (for all affine functions, in fact) by the linearity of expectations, a stronger result than Jensen's inequality, although it can be deduced by applying Jensen's inequality to both the function and its negative. And, yes if a function $\ g\ $ is neither convex nor concave then you can't invoke Jensen's inequality, so it tells you nothing about the value of $\ \text{E}(g(X))-g(\text{E}(X)) \ $. $\endgroup$ May 3, 2021 at 16:19

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