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Mar 4, 2018 at 14:26 vote accept QuantumLogarithm
Mar 4, 2018 at 14:25 comment added Calculon It is hard to say anything new based on the additional condition that $E[X_{\tau}] = 0$. Think of the following stopping time $\tau = \min\{n: X_n = b \text{ or } X_n = -b\}$ for some $b > 0$. Then $E[\tau] = b^2$, $E[X_{\tau}] = 0$ and $E[\lvert X_{\tau}\rvert] = b$. This satisfies the inequality $E[\lvert X_{\tau}\rvert] \leq \left(E[X_{\tau}]\right)^{\frac{1}{2}} $ with an equality. It is tempting to think that such an equality would hold for any stopping time with $E[X_\tau] =0$ but I doubt this is true. Maybe if $\tau$ is a stopping time wrt the natural filtration of $X$.
Mar 4, 2018 at 14:02 comment added QuantumLogarithm Thanks for the comment. I added an additional condition that I know, namely that $E[X(\tau)]=0$.
Mar 4, 2018 at 10:56 history answered Calculon CC BY-SA 3.0