logo

Proof of Conditional Jensen's Inequality 📂Probability Theory

Proof of Conditional Jensen's Inequality

Theorem

Given a probability space (Ω,F,P)( \Omega , \mathcal{F} , P) and a sub-sigma field G⊂F\mathcal{G} \subset \mathcal{F}, let’s assume XX is a random variable.

Regarding the convex function ϕ:R→R\phi : \mathbb{R} \to \mathbb{R} and ϕ(X)∈L1(Ω)\phi (X) \in \mathcal{L}^{1} ( \Omega ) , ϕ(E(X∣G))≤E(ϕ(X)∣G) \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∈Rx,y \in \mathbb{R} and α∈[0,1]\alpha \in [0,1], it satisfies the following: ϕ(αx+(1−α)y)≤αϕ(x)+(1−α)ϕ(y) \phi ( \alpha x + (1 - \alpha ) y ) \le \alpha \phi (x) + (1 - \alpha ) \phi (y)
  • G\mathcal{G} being a sub-sigma field of F\mathcal{F} means they are both sigma fields of Ω\Omega, but G⊂F\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 XX is F\mathcal{F}-measurable, then E(X∣F)=X a.s.E(X|\mathcal{F}) =X \text{ a.s.}
  • For constants aa and bb, E(aX+b∣G)=aE(X∣G)+b a.s.E(aX + b | \mathcal{G}) = a E(X | \mathcal{G}) + b \text{ a.s.}

Since ϕ\phi is convex, for all μ∈R\mu \in \mathbb{R}, ϕ(x)≥m(x−μ)+ϕ(μ) \begin{align} \phi ( x ) \ge m ( x - \mu ) + \phi ( \mu) \end{align} there exists a slope mm that satisfies it. Now setting μ:=E(X∣G)\mu := E \left( X | \mathcal{G} \right) and taking the conditional expectation E(⋅∣G)E \left( \cdot | \mathcal{G} \right) of (1)(1), both μ\mu and ϕ(E(X∣G))\phi \left( E \left( X | \mathcal{G} \right) \right) are G\mathcal{G}-measurable, thus E(ϕ(X)∣G)≥mE(X−μ∣G)+E(ϕ(μ)∣G)=mE(X∣G)−mE(E(X∣G)∣G)+E(ϕ(μ)∣G)=mE(X∣G)−mE(X∣G)+E(ϕ(μ)∣G)=E(ϕ(μ)∣G)=E(ϕ(E(X∣G))∣G)=ϕ(E(X∣G)) \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*}

■

See Also