Show that $E[X_T | T < \infty] \le E[X_0]$ and $cP(\sup X_n \ge c) \le E[X_0]$

Question

From Probability with Martingales:

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I'm assuming the semi-colon means condition (Otherwise, why not say $T$ is a finite stopping time?). What I tried:

1. $$X_T1_{T < \infty} = X_01_{T=0} + X_11_{T=1} + X_21_{T=2} +...$$

$$\to E[X_T1_{T < \infty}] = E[X_01_{T=0} + X_11_{T=1} + X_21_{T=2} +...]$$

By Fubini's Theorem or something, we have:

$$E[X_T1_{T < \infty}] = E[X_01_{T=0}] + E[X_11_{T=1}] + E[X_21_{T=2}] +...$$

$$ = E[E[X_01_{T=0}|\mathscr F_0]] + E[E[X_11_{T=1}|\mathscr F_0]] + E[E[X_21_{T=2}|\mathscr F_0]] +...$$

$$ = E[1_{T=0}E[X_0|\mathscr F_0]] + E[1_{T=1}E[X_1|\mathscr F_0]] + E[1_{T=2}E[X_2|\mathscr F_0]] +...$$

$$ \le E[1_{T=0}E[X_0]] + E[1_{T=1}E[X_0]] + E[1_{T=2}E[X_0]] +...$$

$$ \le E[X_0]E[1_{T=0}] + E[X_0]E[1_{T=1}] + E[X_0]E[1_{T=2}] +...$$

$$ \le E[X_0]\{E[1_{T=0}] + E[1_{T=1}] + E[1_{T=2}] +...\}$$

$$ \le E[X_0]P(T < \infty)$$

$$\to E[X_T | T < \infty] = \frac{E[X_T 1_{T < \infty}]}{P(T < \infty)}$$

$$ \le \frac{E[X_0]P(T < \infty)}{P(T < \infty)}$$

$$ = E[X_0]$$

How else can I approach this problem?

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2. True for $c \le 0$ as $X_n \ge 0$

For $c > 0$:

Am I supposed to say that:

$$cP(\sup X_n \ge c) \le E[X_T | T < \infty]$$

?

All I got so far is that:

$$P(\sup X_n \ge c)$$

$$ = 1 - P(\sup X_n < c)$$

$$ = 1 - P(X_1 < c, X_2 < c, ...)$$

$$ = 1 - P(\bigcap_{n} (X_n < c))$$

I think:

$$(\sup X_n < c) \subseteq (\limsup X_n < c) \subseteq \liminf (X_n < c) \subseteq \limsup (X_n < c)$$

Also maybe:

$$(\sup X_n < c) \subseteq (\liminf(-X_n) > - c) \subseteq (\liminf(-X_n) > - c)$$

Use Fatou's:

$$\liminf E[-X_n] \ge E[\liminf (-X_n)] \ge(*) E[-c] = -c$$

$$\to \liminf [-E[X_n]] \ge -c$$

$$\to -\limsup E[X_n] \ge -c$$

$$\to \limsup E[X_n] \le c$$

(*) assuming $\liminf(-X_n) > - c$

Also using Fatou's:

$P(\liminf (X_n < c)) \le \liminf P(X_n < c) \le \limsup P(X_n < c) \le P(\limsup (X_n < c))$

That's all I have. How can I approach this problem?

Answer 1

First, since a stopped super-MTG is a super-MTG

$$\mathsf{E}X_{T\wedge n}\le \mathsf{E}X_0,$$

for any $n\in\mathbb{N}$. Nonnegativity implies a.s. convergence (i.e. $X_{T\wedge n}\to X_T$ a.s.). Thus, using Fatou's lemma

$$ \mathsf{E}[X_T1\{T<\infty\}]\le\mathsf{E}X_T=\mathsf{E}\liminf_{n\to\infty} X_{T\wedge n}\le\liminf_{n\to\infty} \mathsf{E}X_{T\wedge n}\le \mathbb{E}X_0. $$

For the second result set $T=\inf\{n : X_n \ge c\}$. Then since $X_T \ge c$ on $\{T<\infty\}$ we get

$$ \mathsf{P}(\sup_nX_n \ge c)\le c^{-1}\mathsf{E}[X_T1\{T<\infty\}]\le c^{-1}\mathsf{E}X_0. $$