This is the Jensen's inequality I saw in my textbook:

$$E{ f(X) } \geq f( E(X) ),$$

where $f$ is a convex function.

Is this also applicable for two random variables--independent or otherwise--like this:

$$E{ f(X,Y) } \geq f( E(X,Y) )?$$

  • $\begingroup$ Yes this is the same than saying X = (X,Y) and applying it $\endgroup$
    – statquant
    May 15, 2013 at 10:16
  • $\begingroup$ @statquant And I should be summing over the joint probability distribution? $\endgroup$ May 15, 2013 at 10:21
  • $\begingroup$ What do you mean summing ? $\endgroup$
    – statquant
    May 15, 2013 at 10:27
  • $\begingroup$ I mean, the integration. $E(x,y) = \int\int{g(x,y)p(x,y)dxdy}$ where $g(x,y)$ is an arbitrary function of $X,Y$. $\endgroup$ May 15, 2013 at 13:47
  • $\begingroup$ @statquant, that right? And why is this expected value dependent on an arbitrary function while for single variables, it's not? $\endgroup$ May 15, 2013 at 13:48

1 Answer 1


Yes Jensen inequality holds for multiple variable.

We can find a general formulation in the mesure theoretic article in Wikipedia.

"Let (Ω, A, μ) be a measure space, such that μ(Ω) = 1. If g is a real-valued function that is μ-integrable, and if g is a convex function on the real line. $$\varphi\left(\int_\Omega g\, d\mu\right) \le \int_\Omega \varphi \circ g\, d\mu$$"

This generalizable for convex function on vectorial spaces. (To be more precise, the real case is the restriction to one dimension). See article general inequality in probabilistic setting.

More intuitively: pose Z = (X,Y), Z is a random variable, apply Jensen inequality to Z.

To compute it: (from wikipedia again...)

If one considers the joint probability density function of ''X'' and ''Y'', say ''j(x,y)'', then the expectation of ''XY'' is

$$\operatorname{E}[(X,Y)] = \int\int xy \, j(x,y)\,dx\,dy$$

For g:

$$\operatorname{E}[g(X,Y)] = \int\int g(x,y) \, j(x,y)\,dx\,dy$$

Note that with didn't not discussed this issue before, but you have to make sure that X is integrable, g measurable.

  • 3
    $\begingroup$ @Were_cat how does the jensen inequality apply for more variables? $E[f(X,Y)]=f(E[X,Y])$ does not make sense, as $f$ is a function of two arguments in the first case, and a function of one argument in the second. $\endgroup$
    – spurra
    Dec 24, 2016 at 9:36
  • 3
    $\begingroup$ @suppra see whuber's comment posted below the question: E[X,Y] is a vector itself (a two-element vector), so it makes sense to apply f() to E[X,Y]. $\endgroup$ May 30, 2021 at 2:06

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