Jensen's Inequality

Jensen’s inequality extends from ordinary expectation to conditional expectation with no change in form: applying a convex function to a conditional mean never exceeds the conditional mean of the convex function. The proof rests on the supporting-line characterization of convex functions, together with linearity and monotonicity of conditional expectation.

Statement

Theorem: Conditional Jensen's Inequality

Statement

Let $\varphi : \R \to \R$ be a convex function. Let $X$ be a random variable with $\E|X| < \infty$ and $\E|\varphi(X)| < \infty$. Then

$$\varphi\!\big(\E(X \mid \cG)\big) \;\le\; \E\!\big(\varphi(X) \mid \cG\big) \quad \text{a.s.}$$

Proof

Every convex function $\varphi : \R \to \R$ is the supremum of its affine minorants. With the supporting lines parametrized by rational slope and intercept, the supremum can be taken over a countable set:

$$S \;:=\; \{ (a, b) \in \mathbb{Q}^2 \;:\; a x + b \le \varphi(x) \text{ for every } x \in \R \},$$

and

$$\varphi(x) \;=\; \sup_{(a, b) \in S} (a x + b) \qquad \text{for every } x \in \R.$$

(Restricting to rational $(a, b)$ is what makes $S$ countable; convexity ensures the supremum still recovers $\varphi$ exactly at every point.)

For any fixed $(a, b) \in S$, $\varphi(X) \ge aX + b$ pointwise. Take conditional expectation, and apply monotonicity, linearity, and the fact that constants are unchanged by conditioning:

$$\E\!\big(\varphi(X) \mid \cG\big) \;\ge\; a \, \E(X \mid \cG) + b \quad \text{a.s.}$$

This holds a.s., with the null exceptional set depending on the chosen $(a, b)$. Since $S$ is countable, the union of these exceptional sets over all $(a, b) \in S$ is still a null set, and outside this union the inequality holds simultaneously for every $(a, b) \in S$. Take the supremum over $(a, b) \in S$ on both sides:

$$\E\!\big(\varphi(X) \mid \cG\big) \;\ge\; \sup_{(a, b) \in S} \big( a \, \E(X \mid \cG) + b \big) \;=\; \varphi\!\big(\E(X \mid \cG)\big) \quad \text{a.s.}$$

The last equality is the supporting-line characterization applied at the point $\E(X \mid \cG)(\omega)$.

The countability of $S$ is what lets us pass from “a.s. for each $(a, b)$” to “a.s. for all $(a, b)$ simultaneously”. Without it, the union of exceptional null sets could fail to be null.

$L^p$-norm contraction

A clean specialization: conditional expectation does not increase $L^p$ norms.

Corollary: Conditional Expectation Contracts L^p

Statement

For any $p \ge 1$ and any $X \in L^p$,

$$\| \E(X \mid \cG) \|_p \;\le\; \| X \|_p.$$

Equivalently, $\E(\cdot \mid \cG) : L^p \to L^p$ is a contraction.

Proof

The function $\varphi(x) = |x|^p$ is convex for $p \ge 1$. Apply Jensen’s inequality:

$$\big| \E(X \mid \cG) \big|^p \;\le\; \E\!\big( |X|^p \,\big|\, \cG \big) \quad \text{a.s.}$$

Take unconditional expectations on both sides. The tower property collapses the right side, $\E[\E(|X|^p \mid \cG)] = \E[|X|^p]$, so

$$\E\!\big[ \big| \E(X \mid \cG) \big|^p \big] \;\le\; \E[|X|^p].$$

Both sides are $p$-th powers of $L^p$ norms, $\| \cdot \|_p^p = \E[|\cdot|^p]$. Taking $p$-th roots gives $\| \E(X \mid \cG) \|_p \le \| X \|_p$.

The contraction property is the basis for the $L^2$ projection picture of conditional expectation: the map $X \mapsto \E(X \mid \cG)$ is the orthogonal projection from $L^2(\Omega, \cF, \Pr)$ onto the closed subspace $L^2(\Omega, \cG, \Pr)$. The norm cannot grow under projection, and conditional Jensen with $p = 2$ recovers exactly that. The geometric viewpoint is the subject of the projection perspective.