For a random variable X, Jensen's inequality says, $$E[g(x)]\geq g(E[x])$$ for a convex function $g(x)$. Is there a generalization of this to two variables? That is, for two random variables, does $$ E[g(x,y)]\geq g(E[x],E[y]) $$ hold for a convex g?

In my specific case, $g(x,y) = (x^2+y^2)^{\alpha/2}$ and $X \sim \mathcal{N}(\mu_x,\sigma_x)$ and $Y \sim \mathcal{N}(\mu_y,\sigma_y)$ and $X$ and $Y$ are independent.

Any help would be appreciated.

  • $\begingroup$ The "multivariate version" with $x$ a vector variable for the scalar function $g(x)$, it is really the same theorem with the same proof. No new ideas needed $\endgroup$ Sep 17, 2017 at 12:20

1 Answer 1


Suppose $g$ is a convex function and suppose that a subgradient exists at $(E(X), E(Y))$, then we can write

$$g(X, Y) \geq a+b^T(X, Y)^T\tag{1}$$


$$g(\mathbb{E}(X), \mathbb{E}[Y]) =a+b^T(\mathbb{E}[X],\mathbb{E}[Y])^T\tag{2}$$

for some $a\in \mathbb{R}$ and $b \in \mathbb{R}^2$.

Hence, taking expectation over equation $(1)$ and using linearity of expectation, I obtain

$$\mathbb{E}[g(X, Y) ]\geq a+b^T(\mathbb{E}[X], \mathbb{E}[Y])^T=g(\mathbb{E}(X), \mathbb{E}[Y])$$

which is the Jensen's inequality in two variables.

Note that the proof holds for any finite dimensions as long as we know that the subgradient exists. Note that if a function is differentiable, the subgradient exists.

  • $\begingroup$ Thanks for the answer. Could you please point out how you got (2)? Thanks. $\endgroup$ Sep 17, 2017 at 20:47
  • $\begingroup$ For the case when it is smooth, you can think of it as tangent at the point $(E[x], E[y])$, equality holds at that point. $\endgroup$ Sep 18, 2017 at 0:44

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