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Proof of the Submartingale Convergence Theorem 📂Probability Theory

Proof of the Submartingale Convergence Theorem

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 we assume sup⁡n∈NEXn+<∞\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty, then XnX_{n} converges almost surely to some random variable X∞:Ω→RX_{\infty}: \Omega \to \mathbb{R}. EX∞<EX∞+<∞E X_{\infty} < E X_{\infty}^{+} < \infty

Proof

Strategy: Use the properties of limit supremum and limit infimum. X∗:=lim sup⁡n∈NXnX∗:=lim inf⁡n∈NXn X^{\ast}:= \limsup_{n \in \mathbb{N}} X_{n} \\ X_{\ast}:= \liminf_{n \in \mathbb{N}} X_{n} Then, (X∗>X∗)=⋃a<ba,b∈Q(X∗>b>a>X∗) \left( X^{\ast} > X_{\ast} \right) = \bigcup_{a < b \\ a, b \in \mathbb{Q}} \left( X^{\ast} > b > a > X_{\ast} \right) since a,b∈Qa,b \in \mathbb{Q}, we can divide (X∗>X∗)(X^{\ast} > X_{\ast}) into PP countably. Therefore, for all rational numbers aa, bb between X∗X^{\ast} and X∗X_{\ast}, if P(X∗>b>a>X∗)=0P \left( X^{\ast} >b >a> X_{\ast} \right) = 0 then P(X∗>X∗)=0P \left( X^{\ast} > X_{\ast} \right) = 0, meaning that XnX_{n} converges almost surely to X∞X_{\infty}.

Part 1. P(X∗>b>a>X∗)≤P(β∞(a,b)=∞)P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right)

Let’s denote the number of upcrossings within the closed interval formed by two rational numbers a,b∈Qa, b \in \mathbb{Q} and [a,b][a,b] as βN(a,b)\beta_{N} (a,b) , and its limit as β∞(a,b):=lim⁡N→∞βN(a,b)\displaystyle \beta_{\infty} (a,b):= \lim_{N \to \infty} \beta_{N} (a,b). If X∗>b>a>X∗X^{\ast} > b > a > X_{\ast}, it implies that XnX_{n} has descended below aa and ascended above bb for infinitely many n∈Nn \in \mathbb{N}. Hence, β∞(a,b)=∞\beta_{\infty} (a,b) = \infty, and in proposition, X∗>b>a>X∗  ⟹  β∞(a,b)=∞ X^{\ast} > b > a > X_{\ast} \implies \beta_{\infty} (a,b) = \infty In set notation, (X∗>b>a>X∗)⊂(β∞(a,b)=∞) \left( X^{\ast} > b > a > X_{\ast} \right) \subset \left( \beta_{\infty} (a,b) = \infty \right) Taking the probability of PP, P(X∗>b>a>X∗)≤P(β∞(a,b)=∞) P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right)

Part 2. P(β∞(a,b)<∞)=1P \left( \beta_{\infty} (a,b) < \infty \right) = 1

Since P(∣X∣=∞)≠0  ⟹  E(∣X∣)=∞ P \left( |X| = \infty \right) \ne 0 \implies E \left( |X| \right) = \infty by the contrapositive, almost surely Eβ∞(a,b)<∞  ⟹  β∞(a,b)<∞ E \beta_{\infty} (a,b) < \infty \implies \beta_{\infty} (a,b) < \infty

Upcrossing expected value upper bound: EβN(a,b)≤EXN++∣a∣b−a\displaystyle E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }}

Monotone convergence theorem: Let the sequence of non-negative measurable functions {fn}\left\{ f_{n} \right\} satisfy fn↗ff_{n} \nearrow f. Then, lim⁡n→∞∫Efndm=∫Efdm \lim_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm

By assumption, EβN(a,b)≤EXN++∣a∣b−a≤sup⁡N∈NEXN++∣a∣b−a<∞ E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }} \le {{ \sup_{N \in \mathbb{N}} E X_{N}^{+} + |a| } \over { b-a }} < \infty since by definition of βN(a,b)\beta_{N} (a,b), we have βN(a,b)↗β∞(a,b)\beta_{N} (a,b) \nearrow \beta_{\infty} (a,b), by the monotone convergence theorem, ∞>lim⁡N→∞EβN(a,b)=Elim⁡N→∞βN(a,b)=Eβ∞(a,b) \begin{align*} \infty &>& \lim_{N \to \infty} E \beta_{N} (a,b) \\ =& E \lim_{N \to \infty} \beta_{N} (a,b) \\ =& E \beta_{\infty} (a,b) \end{align*} Summing up, since Eβ∞(a,b)<∞E \beta_{\infty} (a,b) < \infty, almost surely β∞(a,b)<∞\beta_{\infty} (a,b) < \infty, meaning P(β∞(a,b)<∞)=1P \left( \beta_{\infty} (a,b) < \infty \right) = 1 holds.

Part 3. P(X∗=X∗)=1\displaystyle P \left( X^{\ast} = X_{\ast} \right) = 1

According to Part 1~2, for all a,b∈Qa, b \in \mathbb{Q} where X∗>b>a>X∗X^{\ast} > b > a > X_{\ast}, P(X∗>b>a>X∗)≤P(β∞(a,b)=∞)=0 P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right) = 0 Since probability PP is a measure, P(X∗>X∗)=P[⋃a<ba,b∈Q(X∗>b>a>X∗)]=∑a<ba,b∈QP(X∗>b>a>X∗)≤∑a<ba,b∈Q0=0 \begin{align*} P \left( X^{\ast} > X_{\ast} \right) =& P \left[ \bigcup_{a < b \\ a, b \in \mathbb{Q}} \left( X^{\ast} > b > a > X_{\ast} \right) \right] \\ =& \sum_{a < b \\ a, b \in \mathbb{Q}} P \left( X^{\ast} > b > a > X_{\ast} \right) \\ \le & \sum_{a < b \\ a, b \in \mathbb{Q}} 0 \\ =& 0 \end{align*} Summing up, since P(X∗≤X∗)=1P \left( X^{\ast} \le X_{\ast} \right) = 1, the limit X∞X_{\infty} of XnX_{n} almost surely exists.

Part 4. EX∞<∞E X_{\infty} < \infty

By the decomposition of absolute values, ∣Xn∣=Xn++Xn−=2Xn+−Xn |X_{n}| = X_{n}^{+} + X_{n}^{-} = 2 X_{n}^{+} - X_{n} Since {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\} is a submartingale, hence EXn≥EX1E X_{n} \ge E X_{1}, E∣Xn∣=2EXn+−EXn≤2EXn+−EX1 E |X_{n}| = 2 E X_{n}^{+} - E X_{n} \le 2 E X_{n}^{+} - E X_{1} Given that we assumed sup⁡n∈NEXn+<∞\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty, sup⁡n∈NE∣Xn∣≤2sup⁡n∈NEXn+−EX1<∞ \sup_{n \in \mathbb{N}} E | X_{n} | \le 2 \sup_{n \in \mathbb{N}} E X_{n}^{+} - E X_{1} < \infty

Fatou’s lemma: For a sequence of non-negative measurable functions {fn}\left\{ f_{n} \right\}, ∫E(lim inf⁡n→∞fn)dm≤lim inf⁡n→∞∫Efndm\displaystyle \int_{E} \left( \liminf_{n \to \infty} f_{n} \right) dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm

sup⁡n∈NE∣Xn∣<∞\sup_{n \in \mathbb{N}} E | X_{n} | < \infty and by Fatou’s lemma, ∞>sup⁡n∈NE∣Xn∣≥lim inf⁡n→∞E∣Xn∣=lim inf⁡n→∞∫Ω∣Xn∣dP=∫Ωlim inf⁡n→∞∣Xn∣dP=∫Ω∣X∞∣dP=E∣X∞∣ \begin{align*} \infty &>& \sup_{n \in \mathbb{N}} E | X_{n} | \\ \ge& \liminf_{n \to \infty} E | X_{n} | \\ =& \liminf_{n \to \infty} \int_{\Omega} | X_{n} | d P \\ =& \int_{\Omega} \liminf_{n \to \infty} | X_{n} | d P \\ =& \int_{\Omega} | X_{\infty} | d P \\ =& E |X_{\infty}| \end{align*} Therefore, E∣X∞∣E | X_{\infty} | also exists.

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Corollary

Particularly, in the proof process of Part 4., if Xn<0X_{n} < 0, then from Xn=Xn+−Xn−X_{n} = X_{n}^{+} - X_{n}^{-} to Xn+=0X_{n}^{+} = 0, implying that even the condition sup⁡n∈NEXn+<∞\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty becomes unnecessary.