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Apr 7, 2018 at 2:19 comment added Nate Eldredge @Sung-EnChiu: No. Repeat the same construction with $p_i = \epsilon$ for all $i$, where $\epsilon \ll 1/B$. Then $\tau$ has a geometric distribution and $E [\tau] = 1/\epsilon$ which is finite but much larger than $B$.
Apr 6, 2018 at 22:28 comment added Sung-En Chiu A followed up question: If we know that $$E[\tau]<\infty$$, can we remove the boundedness condition?
Sep 5, 2017 at 18:50 vote accept Sung-En Chiu
Sep 5, 2017 at 18:49 vote accept Sung-En Chiu
Sep 5, 2017 at 18:49
Sep 2, 2017 at 3:36 history answered Nate Eldredge CC BY-SA 3.0