Proof of Conditional Jensen's Inequality
Theorem
Given a probability space $( \Omega , \mathcal{F} , P)$ and a sub-sigma field $\mathcal{G} \subset \mathcal{F}$, let’s assume $X$ is a random variable.
Regarding the convex function $\phi : \mathbb{R} \to \mathbb{R}$ and $\phi (X) \in \mathcal{L}^{1} ( \Omega ) $, $$ \phi \left( E \left( X | \mathcal{G} \right) \right) \le E \left( \phi (X) | \mathcal{G} \right) $$
- A function is said to be convex if, for all $x,y \in \mathbb{R}$ and $\alpha \in [0,1]$, it satisfies the following: $$ \phi ( \alpha x + (1 - \alpha ) y ) \le \alpha \phi (x) + (1 - \alpha ) \phi (y) $$
- $\mathcal{G}$ being a sub-sigma field of $\mathcal{F}$ means they are both sigma fields of $\Omega$, but $\mathcal{G} \subset \mathcal{F}$.
Description
The conditional Jensen’s inequality guarantees that the expected value form of Jensen’s inequality applies the same way conditionally, as the name implies.
Proof
Properties of conditional expectation:
- If $X$ is $\mathcal{F}$-measurable, then $E(X|\mathcal{F}) =X \text{ a.s.}$
- For constants $a$ and $b$, $E(aX + b | \mathcal{G}) = a E(X | \mathcal{G}) + b \text{ a.s.}$
Since $\phi$ is convex, for all $\mu \in \mathbb{R}$, $$ \begin{align} \phi ( x ) \ge m ( x - \mu ) + \phi ( \mu) \end{align} $$ there exists a slope $m$ that satisfies it. Now setting $\mu := E \left( X | \mathcal{G} \right)$ and taking the conditional expectation $E \left( \cdot | \mathcal{G} \right)$ of $(1)$, both $\mu$ and $\phi \left( E \left( X | \mathcal{G} \right) \right)$ are $\mathcal{G}$-measurable, thus $$ \begin{align*} E \left( \phi (X) | \mathcal{G} \right) \ge& m E \left( X - \mu | \mathcal{G} \right) + E \left( \phi ( \mu ) | \mathcal{G} \right) \\ =& m E \left( X | \mathcal{G} \right) - m E \left( E \left( X | \mathcal{G} \right) | \mathcal{G} \right) + E \left( \phi ( \mu ) | \mathcal{G} \right) \\ =& m E \left( X | \mathcal{G} \right) - m E \left( X | \mathcal{G} \right) + E \left( \phi ( \mu ) | \mathcal{G} \right) \\ =& E \left( \phi ( \mu ) | \mathcal{G} \right) \\ =& E \left( \phi \left( E \left( X | \mathcal{G} \right) \right) | \mathcal{G} \right) \\ =& \phi \left( E \left( X | \mathcal{G} \right) \right) \end{align*} $$
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