TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Conditional jensen inequality

Statement

Assume AC. If $ϕ : R \to R$ is finite convex and both $X$ and $ϕ (X)$ are integrable, then $ϕ (E [X ∣ G]) \leq E [ϕ (X) ∣ G]$ almost surely; the left side is measurable and integrable.

Facts & Assumptions

Given: AC, finite convex $ϕ : R \to R$ , and integrable real X such that $ϕ (X)$ is integrable.

[F1]

Integrable inputs have conditional classes under AC. (Conditional expectation as an ae class)

[F2]

Conditional expectation is linear, fixes constants and preserves order. (Basic algebra and order properties of conditional expectation)

[F3]

Finite convex functions are Borel and are suprema of their rational-contact supporting lines. (Convex functions have countable supporting line representations)

Proof

technique · direct

1.1

Set $U = E [X ∣ G]$ and $V = E [ϕ (X) ∣ G]$ . For each supporting line $ℓ_{q} (t) = m_{q} t + b_{q}$ of [F3], the variable $m_{q} X + b_{q}$ is integrable and bounded above by $ϕ (X)$ . Linearity and order give $m_{q} U + b_{q} \leq V$ almost surely. There are countably many $q$ , so remove one measurable null union to make all inequalities hold together.

F1F2F3

2.1

On the resulting conull set take the supremum over $q$ . By [F3], $ϕ (U) = \sup_{q} (m_{q} U + b_{q}) \leq V$ . Measurability follows either from that countable supremum or composition with the Borel function $ϕ$ . The fixed supporting line at $q = 0$ also gives $m_{0} U + b_{0} \leq ϕ (U)$ . Hence $(ϕ (U))^{+} \leq V^{+}$ and $(ϕ (U))^{-} \leq (m_{0} U + b_{0})^{-}$ almost surely. Both upper bounds are integrable, proving integrability as well as the inequality.

step 1.1F3∎

Source notes

Durrett Theorem 4.1.10 and following remark, printed p.211; van der Vaart Lemma 1.9(vi), printed p.4. The integrability of the left side is checked using one lower supporting line and the conditional upper bound.

Depends on

Used by

Conditional lp contraction Corollary
Convex functions of martingales are submartingales Theorem

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Durrett, Probability: Theory and Examples, 5th ed. (standard reference, not scraped)