In his probability book Bauer proves the following version of Jensen's inequality:

Proposition. Let $X$ be an integrable random variable taking values in an open interval $I\subset\mathbb{R}$, and let $q$ be a convex function on $I$. If $q\circ X$ is integrable, then

$$q(E(X))\leq E(q\circ X).$$

Now am asked to prove that the result holds for an arbitrary interval, e.g. $I=[a,b]$. As a hint Bauer suggests to show that $q$ is lower semicontinuous on $I$, i.e. that $\{x\in I:q(x)>\alpha\}$ is relatively open in $I$ for every $\alpha \in \mathbb{R}$.

How can I do this? I tried an answer below. Any comment is greatly appreciated.

EDIT: I believe Bauer means to show that $q$ is upper semicontinuous, not lower. Indeed by considering the indicator of $\{0,1\}$ on the interval $[0,1]$ it is clear that a convex function need not be lower semicontinuous. To show this, suppose $I=[a,b]$. By convexity we know that $q$ is continuous on $(a,b)$, and so in particular upper semicontinuous on $(a,b)$. Also by convexity, we know that $q'_+(a)$ exists, but might be $-\infty$ (convexity implies that $q'_+(x)$ exists and is nondecreasing on $I$, and is real-valued on $I^\mathrm{o}$). If $q'_+(a)\in\mathbb{R}$, then $q$ is continuous at $a$ and so in particular upper semicontinuous at $a$. If $q'_+(a)=-\infty$, then we see that that $q$ must be decreasing in a neighborhood of $a$, which implies upper semicontinuity of $q$ at $a$. A similar argument using the left derivative $q'_-(x)$ applies for the endpoint $b$.

  • $\begingroup$ Is $p$ real valued or extended real valued? Is $p$ proper convex? Presumably it is assumed that $p$ is real valued, in which case is well known that $p$ is continuous in the one interval $(a,b)$. Since $p(a)\in\mathbb{R}$, then $p$ most be, as you pointed out, upper semicontinuous for $\liminf_{x\rightarrow a+}p(x)\leq p(a)$. That takes care of measurability. The rest is pretty much standard. $\endgroup$
    – Mittens
    Sep 2, 2020 at 17:49
  • $\begingroup$ @OliverDiaz Yes $q$ is real-valued and convex. You disagree with my answer below? $\endgroup$
    – Alain
    Sep 2, 2020 at 18:51
  • $\begingroup$ I don't disagree. It is pretty much textbook answer once you realize $p$ is u.s.c. to take care of measurability and then use the well know properties of left and right derivatives univariate convex functions. It was not clear from the way you wrote the problem whether $\phi$ is a proper convex function or not. Also, I did not down-vote you. $\endgroup$
    – Mittens
    Sep 2, 2020 at 18:58
  • $\begingroup$ @OliverDiaz OK but do you think what I wrote is correct? I just need to show that the inequality holds for an arbitrary $x\in I$ and not only for $x\in I^\mathrm{o}$. $\endgroup$
    – Alain
    Sep 2, 2020 at 19:00
  • $\begingroup$ I would have only focused on what happens at $a$. At other points is trivial. To me that does the trick is that one cannot have $p(a)\leq\liminf_{x\rightarrow a}p(x)$. The issue with the right derivative of $p$ at $a$ become then clear. $\endgroup$
    – Mittens
    Sep 2, 2020 at 19:10

1 Answer 1


Suppose $I=[a,b]$. If $E(X)=a$ or $E(X)=b$, then $X=a$ almost surely or $X=b$ almost surely, and in this case we see that Jensen's inequality is in fact an equality. Hence we may suppose $E(X)\in I^\mathrm{o}$. As noted, we have the inequality

$$ q(y)\geq q(x) + q'_+(x)(y-x) \hspace{0.5cm} x\in I^\mathrm{o}, y\in I $$

and so

$$ q(X)\geq q(x) + q'_+(x)(X-x) \hspace{0.5cm} x\in I^\mathrm{o} .$$

Taking expectations on both sides yields

$$ E(q(X))\geq q(x) + q'_+(x)(E(X)-x) \hspace{0.5cm} x\in I^\mathrm{o} .$$

In particular, for $x=E(X)$ we get

$$ E(q(X))\geq q(E(X)) $$

which is Jensen.

The upper semicontinuity part is to ensure that the function $q:I\to \mathbb{R}$ is Borel measurable, so that the composite function $q\circ X$ is measurable if $X$ is a random variable taking values in $I$.


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