Theorem 5.17. If $X_{n} \geq 0$ is a supermartingale then $X_{\infty}=\lim _{n \rightarrow \infty} X_{n}$ exists and $E X_{\infty} \leq E X_{0}$
The bad martingale in Example 5.9 shows that we can have $X_{0}=1$ and $X_{\infty}=0$. The key to the proof of this is the following maximal inequality.
Lemma 5.18. Let $X_{n} \geq 0$ be a supermartingale and $\lambda>0 .$
$$
P\left(\max _{n \geq 0} X_{n}>\lambda\right) \leq E X_{0} / \lambda
$$
Proof. Let $T=\min \left\{n \geq 0: X_{n}>\lambda\right\} .$ Theorem 5.13 implies that
$$
E X_{0} \geq E\left(X_{T \wedge n}\right) \geq \lambda P(T \leq n)
$$
i.e., $P(T \leq n) \leq E X_{0} / \lambda .$ Since this holds for all $n$ the desired result follows.
Proof of Theorem 5.17 Let $a<b,$ let $S_{0}=0$ and define stopping times for $k \geq$ 1 by
$$
\begin{array}{l}
R_{k}=\min \left\{n \geq S_{k-1}: X_{n} \leq a\right\} \\
S_{k}=\min \left\{n \geq R_{k}: X_{n} \geq b\right\}
\end{array}
$$
Using the reasoning that led to Lemma 5.18
$$
P\left(S_{k}<\infty \mid R_{k}<\infty\right) \leq a / b
$$
Iterating we see that $P\left(S_{k}<\infty\right) \leq(a / b)^{k} .$ Since this tends to 0 as $k \rightarrow \infty X_{n}$ crosses from below $a$ to above $b$ only finitely many times. To conclude from this that $\lim _{n \rightarrow \infty} X_{n}$ exists, let
$$
Y=\liminf _{n \rightarrow \infty} X_{n} \quad \text { and } \quad Z=\limsup _{n \rightarrow \infty} X_{n}
$$
If $P(Y<Z)>0$ then there are numbers $a<b$ so that $P(Y<a<b<Z)>0$ but in this case $X_{n}$ crosses from below $a$ to above $b$ infinitely many times with positive probability, a contradiction. To prove $E X_{\infty} \leq E X_{0}$ note that for any time $n$ and positive real number $M$
$$
E X_{0}=E X_{n} \geq E\left(X_{n} \wedge M\right) \rightarrow E\left(X_{\infty} \wedge M\right)
$$
where the last conclusion follows from the reasoning in the proof of Theorem $5.14 .$ The last conclusion implies $E X_{0} \geq E\left(X_{\infty} \wedge M\right) \uparrow E X_{\infty}$ as $M \uparrow \infty$
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This theorem 5.17 is from Richard Durrett's 《Essentials of Stochastic the Processes Second Edition》(Page 200-201).Now there are two problems with this theorem and I still can't prove it after trying.
1)How to use Lemma5.18 to derive$P\left(S_{k}<\infty \mid R_{k}<\infty\right) \leq a / b$.
2)How to derive $P\left(S_{k}<\infty\right) \leq(a / b)^{k}$
Could anyone show me detailed proof procedure of these two problem?Thank you very much!
Best Answer
Let $Y_n={X_{n+R_k}}$ for $n\ge 0$, so this is the process after $R_k$. $$P(S_k<\infty|R_k<\infty)=P(\max_n Y_n\ge b)\le E[Y_0]/b= E[X_{R_k}]/b \le a/b,$$ since $X_{R_k}\le a$ almost surely.
You can then prove $P(S_k<\infty)<(a/b)^k$ by induction on $k$. Indeed,
$$ P(S_k<\infty)=P(S_{k-1}<\infty)\cdot P(R_k<\infty |S_{k-1}<\infty)\cdot P(S_k<\infty |R_k<\infty, S_{k-1}<\infty) \\\le \underbrace{(a/b)^{k-1}}_{\text{induction hypothesis}}\cdot 1\cdot (a/b) $$