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Given a sequence of random variables $X_0, X_1, \cdots$. Assume that $X_0=0$ and $X_{n+1}= X_n+U[0,1]$, where $U[0,1]$ denotes a uniform distribution over $[0,1]$. Define $N=\min\{n: X_n\geq 2\}$.

Q: how to obtain a "tight" bound on $Pr[X_N\leq 2.1]$?

Obviously, $X_i$ is a supermartingale, and $E[N]=4$, and we have that $E[X_N]=2$, so one might use Markov inequality. However, I am wondering whether powerful martingale results can give a tighter bound?

maomao
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  • Sorry if I'm misunderstanding your question, but why is it that $\mathbb{E}[N]=4$? Shouldn't it be something like $e^2-e$? [Eq. (15)] – Clement C. Jun 17 '16 at 12:15
  • @ClementC. I might be wrong: on average, $X_n$ increases by 0.5 each time, from 0 to 4 it takes on average 4 times. But I admit I did not think carefully; will check the reference you give. – maomao Jun 17 '16 at 12:29
  • At first glance, you can use (12) from the link above to compute what you want, by conditioning on N, getting the conditional probability, then taking the expectation over N using the probability given in (12). – Clement C. Jun 17 '16 at 12:43
  • @ClementC. you are right, here by the simple argument, one can only have $E(N)\geq 4$, which is consistent to the link. The point here is that I am actually considering a general $X_{n+1}=X_n + N$, where $N$ could be Gaussian, uniform, ... (in general any distribution), so specific result on uniform distributions, being preciser, is probably not quite useful in the general setting. – maomao Jun 17 '16 at 12:46
  • See also e.g. this approach. – joriki Jun 17 '16 at 13:07

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