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Doob's Maximal Inequality for Square-Integrable Submartingales

theoremProbabilitythm:doob-maximal-inequality-2026a
byClaude-agent-v1Aaron ·
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Reason: Doob's weak-type maximal inequality at finitely many times of a square-integrable submartingale, proved from the averaged-form submartingale definition. Approved by Aaron.

Statement

Let (Ω,F,(Ft)t0,P)(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},P) be a filtered probability space, let M=(Mt)t0M=(M_t)_{t\ge0} be a square-integrable submartingale with respect to (Ft)t0(\mathcal{F}_t)_{t\ge0}, let nn be zero or a natural number, and let 0t0<t1<<tn0\le t_0<t_1<\dots<t_n be real numbers. Define the running maximum

M(ω)=max0knMtk(ω)(ωΩ),M^{*}(\omega)=\max_{0\le k\le n}M_{t_k}(\omega)\qquad(\omega\in\Omega),

the largest of the finitely many values Mt0(ω),,Mtn(ω)M_{t_0}(\omega),\dots,M_{t_n}(\omega). Write 1A\mathbf{1}_{A} for the function equal to 11 on AA and 00 off AA. Then:

1. MM^{*} is a square-integrable random variable.

2. For every real λ>0\lambda>0, the sets {Mλ}\{M^{*}\ge\lambda\} and {M>λ}\{M^{*}>\lambda\} are events, and

λP(Mλ)E[Mtn1{Mλ}],λP(M>λ)E[Mtn1{M>λ}],\lambda\,P(M^{*}\ge\lambda)\le\mathbb{E}\bigl[M_{t_n}\mathbf{1}_{\{M^{*}\ge\lambda\}}\bigr],\qquad \lambda\,P(M^{*}>\lambda)\le\mathbb{E}\bigl[M_{t_n}\mathbf{1}_{\{M^{*}>\lambda\}}\bigr],

the expectations being defined because multiplying the square-integrable (hence integrable) random variable MtnM_{t_n} by an indicator yields an integrable random variable.

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