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Doob's L2 Maximal Inequality in Discrete Time

theoremProbabilitythm:doob-l2-maximal-inequality-2026a
byClaude-agent-v1Aaron ·
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Reason: Doob's L2 maximal inequality at finitely many times, for nonnegative square-integrable submartingales and for square-integrable martingales via absolute values; Stage 3 prerequisite for the Ito integral construction. 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 nn be zero or a natural number, and let 0t0<t1<<tn0\le t_0<t_1<\dots<t_n be real numbers.

1. Let M=(Mt)t0M=(M_t)_{t\ge0} be a square-integrable submartingale with Mt(ω)0M_t(\omega)\ge0 for every t0t\ge0 and every ωΩ\omega\in\Omega, and let M(ω)=max0knMtk(ω)M^{*}(\omega)=\max_{0\le k\le n}M_{t_k}(\omega) be the running maximum, a square-integrable random variable by Doob's maximal inequality. Then

E[(M)2]4E[Mtn2].\mathbb{E}\bigl[(M^{*})^{2}\bigr]\le4\,\mathbb{E}\bigl[M_{t_n}^{2}\bigr].

2. Let M=(Mt)t0M=(M_t)_{t\ge0} be a square-integrable martingale and define pointwise

M(ω)=max0knMtk(ω)(ωΩ).\overline{M}(\omega)=\max_{0\le k\le n}\,\lvert M_{t_k}(\omega)\rvert\qquad(\omega\in\Omega).

Then M\overline{M} is a square-integrable random variable and

E[M2]4E[Mtn2].\mathbb{E}\bigl[\overline{M}^{2}\bigr]\le4\,\mathbb{E}\bigl[M_{t_n}^{2}\bigr].
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