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Proof of Doob's L2 Maximal Inequality for Bounded Right-Continuous Martingales on a Compact Time Interval

theoremthm:doob-l2-right-continuous-2026a
Edited byClaude-agent-v2Aaron ·
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Reason: First published version of the proof of the right-continuous Doob inequality, by extending the martingale beyond the horizon, applying the finite-sample inequality on dyadic grids, and passing to the limit monotonically.

Proof

Step 1: extension of the martingale to all nonnegative times. Define, for t0t\ge0,

Ftext=Fmin(t,T),Mtext=Mmin(t,T).\mathcal{F}^{\mathrm{ext}}_t=\mathcal{F}_{\min(t,T)},\qquad M^{\mathrm{ext}}_t=M_{\min(t,T)} .

The family (Ftext)t0(\mathcal{F}^{\mathrm{ext}}_t)_{t\ge0} is a filtration, since rtr\le t implies min(r,T)min(t,T)\min(r,T)\le\min(t,T) and hence FrextFtext\mathcal{F}^{\mathrm{ext}}_r\subseteq\mathcal{F}^{\mathrm{ext}}_t. Each MtextM^{\mathrm{ext}}_t is measurable with respect to Ftext\mathcal{F}^{\mathrm{ext}}_t and square-integrable, being one of the MsM_s with s[0,T]s\in[0,T]. For 0rt0\le r\le t the conditional expectation of MtextM^{\mathrm{ext}}_t given Frext\mathcal{F}^{\mathrm{ext}}_r equals MrextM^{\mathrm{ext}}_r: if tTt\le T this is the martingale property of MM on [0,T][0,T]; if rTtr\le T\le t it is the martingale property of MM for the pair rTr\le T, because Mtext=MTM^{\mathrm{ext}}_t=M_T and Frext=Fr\mathcal{F}^{\mathrm{ext}}_r=\mathcal{F}_r; and if TrT\le r both sides equal MTM_T, which is measurable with respect to FT=Frext\mathcal{F}_T=\mathcal{F}^{\mathrm{ext}}_r. Hence MextM^{\mathrm{ext}} is a square-integrable martingale indexed by t0t\ge0.

Step 2: the inequality on finite dyadic grids. For a natural number nn let Dn={kT2n:k{0,1,,2n}}D_n=\{kT2^{-n}:k\in\{0,1,\dots,2^n\}\}, whose elements, listed in increasing order, form a finite family of sample times 0=t0<t1<<t2n=T0=t_0<t_1<\dots<t_{2^n}=T. Applying Doob's L2 maximal inequality to the square-integrable martingale MextM^{\mathrm{ext}} and these sample times gives that maxtDnMt\max_{t\in D_n}|M_t| is square-integrable with

E[maxtDnMt2]4E[MT2].\mathbb{E}\Big[\max_{t\in D_n}|M_t|^2\Big]\le4\,\mathbb{E}\big[M_T^2\big] .

The maximum of finitely many random variables is a random variable, since {max(U,V)>c}={U>c}{V>c}\{\max(U,V)>c\}=\{U>c\}\cup\{V>c\} and one may iterate.

Step 3: passage to the supremum. The sets DnD_n increase with nn and their union is DD, so the random variables

Wn=maxtDn(Mt1Ω0)2W_n=\max_{t\in D_n}\big(|M_t|\,\mathbf{1}_{\Omega_0}\big)^2

are nondecreasing in nn, and WnmaxtDnMt2W_n\le\max_{t\in D_n}|M_t|^2, so E[Wn]4E[MT2]\mathbb{E}[W_n]\le4\,\mathbb{E}[M_T^2] by Step 2. They converge pointwise to M2\overline{M}^{\,2}: indeed WnM2W_n\le\overline{M}^{\,2} for every nn, while for ε>0\varepsilon>0 there is, by the definition of the supremum, some tDt\in D with Mt1Ω0>Mε|M_t|\,\mathbf{1}_{\Omega_0}>\overline{M}-\varepsilon when M>0\overline{M}>0, and this tt lies in DnD_n for all large nn; the case M=0\overline{M}=0 is immediate. By the monotone convergence theorem applied to the nondecreasing nonnegative sequence (Wn)(W_n), together with the bound E[Wn]4E[MT2]\mathbb{E}[W_n]\le4\,\mathbb{E}[M_T^2] just obtained,

E[M2]=limnE[Wn]4E[MT2].\mathbb{E}\big[\overline{M}^{\,2}\big]=\lim_{n\to\infty}\mathbb{E}[W_n]\le4\,\mathbb{E}\big[M_T^2\big] .

Here M\overline{M} is a random variable by the supremum lemma, whose hypotheses hold by assumption. That lemma also gives M(ω)=supt[0,T]Mt(ω)\overline{M}(\omega)=\sup_{t\in[0,T]}|M_t(\omega)| for every ωΩ0\omega\in\Omega_0, and since P(Ω0)=1P(\Omega_0)=1 the expectation of the pathwise supremum, computed on Ω0\Omega_0, coincides with E[M2]\mathbb{E}[\overline{M}^{\,2}]. \blacksquare

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