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Dob's Inequality Proof 📂Probability Theory

Dob's Inequality Proof

Theorem

Given a probability space (Ω,F,P)( \Omega , \mathcal{F} , P) and a submartingale {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}.

If for some N∈NN \in \mathbb{N} and p>1p>1, Xn≥0(n≤N)X_{n} \ge 0 (n \le N), EXNp<∞E X_{N}^{p} < \infty then E(max⁡n≤NXnp)≤(pp−1)pEXNp a.s. E \left( \max_{n \le N} X_{n}^{p} \right) \le \left( {{ p } \over { p-1 }} \right)^{p} E X_{N}^{p} \text{ a.s.}

Explanation

The form of the equation can be seen as extracting the (pp−1)p\displaystyle \left( {{ p } \over { p-1 }} \right)^{p} resulting from max⁡n≤N⋅np\displaystyle \max_{n \le N} \cdot_{n} ^{p} and calculating its upper bound. Since (pp−1)>1\displaystyle \left( {{ p } \over { p-1 }} \right)>1, if pp is too large, the inequality’s value decreases, and it is crucial that the final number NN can be large. In practical applications to statistics beyond theory, this NN can be considered quite large, so there is no particular concern that NN is limited.

Proof

Strategy: First, for a constant LL, choose (L∧max⁡n≤NXnp)\displaystyle \left( L \land \max_{n \le N} X_{n} ^{p} \right) and cut off the value based on LL to develop the equation. Here, ∧\land means (f∧g)(x):=min⁡{f(x),g(x)}\displaystyle (f \land g) (x):= \min \left\{ f(x) , g(x) \right\} for two functions f,gf,g. After obtaining a satisfactory expression through various tricks, sending LL to infinity will eventually leave only ∞∧max⁡n≤NXnp=max⁡n≤NXnp\displaystyle \infty \land \max_{n \le N} X_{n} ^{p} = \max_{n \le N} X_{n} ^{p}.

Part 1.

Consider an increasing sequence {Ln}\left\{ L_{n} \right\} that satisfies Ln+1>LnL_{n+1} > L_{n} for all n∈Nn \in \mathbb{N}, where Y:=max⁡n≤NXn\displaystyle Y:= \max_{n \le N} X_{n}.

EX=∫0∞P(X>t)dt EX = \int_{0}^{\infty} P(X>t) dt

By considering t:=λpt:= \lambda^{p}, according to another expression of expectation, E[(Y∧Ln)p]=∫0∞P[(Y∧Ln)p>t]dt=∫0∞P[(Y∧Ln)p>λp]pλp−1dλ=∫0∞P(Y∧Ln>λ)pλp−1dλ=∫0LnP(Y∧Ln>λ)pλp−1dλ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] =& \int_{0}^{\infty} P \left[ \left( Y \land L_{n} \right)^{p} > t \right] dt \\ =& \int_{0}^{\infty} P \left[ \left( Y \land L_{n} \right)^{p} > \lambda^{p} \right] p \lambda^{p-1} d \lambda \\ =& \int_{0}^{\infty} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \\ =& \int_{0}^{L_{n}} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \end{align*} The reason why the integration interval can be reduced from [0,∞][0,\infty] to [0,Ln][0,L_{n}] is because if 0≤λ≤Ln0 \le \lambda \le L_{n}, then there is no need to consider up to ∞\infty since it deals with P(Y∧Ln>λ)P \left( Y \land L_{n} > \lambda \right) anyway. Meanwhile, since LnL_{n} will be sent to infinity at the end of the proof, cases where Ln<λL_{n} < \lambda is even less of a concern.

Part 2.

If {(Xn,Fn)}\left\{ (X_{n} , \mathcal{F}_{n}) \right\} is a submartingale, for all λ>0\lambda > 0 λP(max⁡n≤NXn≥λ)≤∫(max⁡n≤NXn≥λ)XNdP \lambda P \left( \max_{n \le N} X_{n} \ge \lambda \right) \le \int_{(\max_{n \le N} X_{n} \ge \lambda)} X_{N} dP

If 0≤λ≤Ln0 \le \lambda \le L_{n}, since it’s essentially Y≥LnY \ge L_{n} and according to the inequality for submartingales [3], P(Y≥λ)≤∫(Y≥λ)XNdP\displaystyle P \left( Y \ge \lambda \right) \le \int_{(Y \ge \lambda)} X_{N} dP, thus E[(Y∧Ln)p]≤∫0LnP(Y∧Ln>λ)pλp−1dλ≤∫0LnP(Y≥λ)pλp−1dλ≤∫0Ln1λ∫(Y≥λ)XNdPpλp−1dλ≤p∫0Ln∫(Y≥λ)XNλp−2dPdλ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & \int_{0}^{L_{n}} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \\ \le & \int_{0}^{L_{n}} P \left( Y \ge \lambda \right) p \lambda^{p-1} d \lambda \\ \le & \int_{0}^{L_{n}} {{ 1 } \over { \lambda }} \int_{(Y \ge \lambda)} X_{N} dP p \lambda^{p-1} d \lambda \\ \le & p \int_{0}^{L_{n}} \int_{(Y \ge \lambda)} X_{N} \lambda^{p-2} dP d \lambda \end{align*}

Part 3.

According to Fubini’s theorem, E[(Y∧Ln)p]≤p∫0Ln∫(Y≥λ)XNλp−2dPdλ≤p∫Ω∫0Y∧LnXNλp−2dλdP=p∫ΩXN∫0Y∧Lnλp−2dλdP=p∫ΩXN1p−1(Y∧Ln)p−1dP=pp−1∫ΩXN(Y∧Ln)p−1dP=pp−1E[XN(Y∧Ln)p−1] \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & p \int_{0}^{L_{n}} \int_{(Y \ge \lambda)} X_{N} \lambda^{p-2} dP d \lambda \\ \le & p \int_{\Omega} \int_{0}^{ Y \land L_{n}} X_{N} \lambda^{p-2} d \lambda dP \\ =& p \int_{\Omega} X_{N} \int_{0}^{ Y \land L_{n}} \lambda^{p-2} d \lambda dP \\ =& p \int_{\Omega} X_{N} {{ 1 } \over { p-1 }} \left( Y \land L_{n} \right)^{p-1} dP \\ =& {{ p } \over { p-1 }} \int_{\Omega} X_{N} \left( Y \land L_{n} \right)^{p-1} dP \\ =& {{ p } \over { p-1 }} E \left[ X_{N} \left( Y \land L_{n} \right)^{p-1} \right] \end{align*}

Hölder’s inequality: For p>1p>1, if 1p+1q=1\displaystyle {{1} \over {p}} + {{1} \over {q}} = 1 and f∈Lp(E)f \in \mathcal{L}^{p} (E) , g∈Lq(E)g \in \mathcal{L}^{q} (E) then fg∈L1(E)fg \in \mathcal{L}^{1} (E) and ∣fg∣1≤∣f∣p∣g∣q| fg |_{1} \le | f |_{p} | g |_{q}

If we set q:=pp−1\displaystyle q: = {{ p } \over { p-1 }}, then q(p−1)=pq (p-1) = p and, according to Hölder’s inequality, E[(Y∧Ln)p]≤pp−1E[XN(Y∧Ln)p−1]≤pp−1∥XN(Y∧Ln)p−1∥1≤q∥XN∥p∥(Y∧Ln)p−1∥q≤q[EXNp]1/pE[(Y∧Ln)(p−1)⋅q]1/q≤q[EXNp]1/pE[(Y∧Ln)p]1/q \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & {{ p } \over { p-1 }} E \left[ X_{N} \left( Y \land L_{n} \right)^{p-1} \right] \\ \le & {{ p } \over { p-1 }} \left\| X_{N} \left( Y \land L_{n} \right)^{p-1} \right\|_{1} \\ \le & q \left\| X_{N} \right\|_{p} \left\| \left( Y \land L_{n} \right)^{p-1} \right\|_{q} \\ \le & q \left[ E X_{N}^{p} \right]^{1/p} E \left[ \left( Y \land L_{n} \right)^{(p-1) \cdot q} \right]^{1/q} \\ \le & q \left[ E X_{N}^{p} \right]^{1/p} E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1/q} \end{align*} Dividing both sides by E[(Y∧Ln)p]1/q\displaystyle E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1/q}, E[(Y∧Ln)p]1−1/q≤q[EXNp]1/p E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1-1/q} \le q \left[ E X_{N}^{p} \right]^{1/p} Taking the pp power to both sides, 1−1/q=1−(p−1)/p=1/p1 - 1/q = 1 - (p-1)/p = 1/p, therefore E[(Y∧Ln)p]≤qp[EXNp] E \left[ \left( Y \land L_{n} \right)^{p} \right] \le q^{p} \left[ E X_{N}^{p} \right] Taking the limit to both sides, lim⁡n→∞E[(Y∧Ln)p]≤lim⁡n→∞qp[EXNp] \lim_{n \to \infty} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le \lim_{n \to \infty} q^{p} \left[ E X_{N}^{p} \right] When n→∞n \to \infty, since Y∧Ln↗YY \land L_{n} \nearrow Y, according to the Monotone Convergence Theorem, Elim⁡n→∞[(Y∧Ln)p]≤qp[EXNp] E \lim_{n \to \infty} \left[ \left( Y \land L_{n} \right)^{p} \right] \le q^{p} \left[ E X_{N}^{p} \right] Organizing the equation, EYp≤qp[EXNp] E Y^{p} \le q^{p} \left[ E X_{N}^{p} \right]

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