Let $Z,Y_0,Y_1,\cdots$ be joint random variables with $\Bbb E[|Z|]<\infty$. Doob's martingale process is defined by $X_n=\Bbb E[Z\mid Y_0,\cdots,Y_n]$ for $n\geq 0$.

In Karlin and Taylor's A First Course in Stochastic Processes (p. 296), the authors prove that the martingale is uniformly integrable as follows:

\begin{align*} |\Bbb E[X_nI\{X_n>c\}]|&\leq \Bbb E[I\{X_n>c\}\Bbb E[|Z|\mid Y_0,\cdots,Y_n]]\\ & \leq \Bbb E[|Z|I\{|X_n|>c\}]\\ & \leq \Bbb E[|Z|I\{U>c\}]\rightarrow 0 \end{align*} as $c\rightarrow \infty$. (Here, $U=\sup_{k\geq 0} |X_k|$ and $I\{\cdot\}$ is the indicator function.)

I understand all the steps except the second inequality; why does it hold?


\begin{align} &\mathsf{E}[I\{|X_n|>c\}\mathsf{E}[|Z|\mid Y_0,\cdots,Y_n]] \\ =\,&\mathsf{E}[\mathsf{E}[|Z|I\{|X_n|>c\}\mid Y_0,\cdots,Y_n]] \\ =\,&\mathsf{E}[|Z|I\{|X_n|>c\}] \le \mathsf{E}[|Z|I\{U>c\}] \end{align}

  • $\begingroup$ Thanks for your answer. Could you explain the first equality? $\endgroup$ – Empty Set Jun 12 '17 at 1:33
  • $\begingroup$ @EmptySet $X_n$ is $\sigma(Y_0,\ldots,Y_n)$-measurable. $\endgroup$ – d.k.o. Jun 12 '17 at 1:46

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