Suppose $(X, \mathscr{F})$ is a martingale. Show that $(X_{\tau \wedge n}, \mathscr{F})$ is a uniformly integrable for any finite stopping time $\tau$ such that $\{X_n\}$ is uniformly integrable.
My attempt: In one of my textbooks (Resnick- probability path- section 6.5.1) it says that if a family of random variables $\{X_n\}$ is dominated by a uniformly integrable family $\{Y_n\}$ then the $\{X_n\}$ are also U.I.
Therefore, for this question I simply said that $|X_{\tau \wedge n}| \le |X_n|$ $\forall n$ and the result holds.
However, this is the solution to the problem.
Is this another way to solve this question? Or was there a mistake in my attempt.
Thanks.
Best Answer
Just because $m:= \tau \wedge n \le n$ we cannot conclude that $X_{m} \le X_{n}$ pointwise... take for example the simple random walk, the stopping time $\tau = \inf_{n}\{|X_{n}|=a\}$ and some $n > \tau$ such that $X_{n}=0$