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Given that ${X_n}$ is a submartingale with $\left|X_{k}-X_{k-1}\right|\leq M<\infty$, and defining stopping times $\tau_{1}<\tau_{2}$ with $E(\tau_{2})<\infty$, eventually I want to show that $E(X_{\tau_{2}}) \geq E(X_{\tau_{1}})$. In the proof, in order to take the expectation into the infinite sum I need to check that $\sum_{k=1}^{\infty}E(\left|X_{k}-X_{k-1}\right|\mathbf{1}_{\tau_{1}<k\leq\tau_{2}})<\infty$, but I am not sure how to do it since I can't write the expectation as the product of the two term due to their dependence. I'd appreciate any help or suggestions.

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up vote 1 down vote accepted

I don't understand where is you problem (your post is not very clear). If you want to prove that $E[\tau_2]\geq E[\tau_1]$, this follows by monotonicity of the Lebesgue integral.

In the summation you can try: $$E\sum_{k=1}^{\infty}|X_k-X_{k-1}|1_{\tau_1<k\leq \tau_2}=E\sum_{k=\tau_1+1}^{\tau_2}|X_k-X_{k-1}|\leq$$

$$\leq E M(\tau_2-\tau_1) < + \infty.$$

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Sorry, I had made a mistake in the statement of the problem; it is corrected now. Thanks for your help on the summation. – V.A. Mar 26 '12 at 15:48

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