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Grimmett and Stirzaker's "Probability and Random Processes" gives a nice discussion about random walk, for example, it considers

$M_n=\max\{S_i,0\le i \le n\}$ where as usual $$S_i=\sum_{k=1}^i{X_k},$$ here $S_0=0$, $\mathbb{P}(X_k=1)=p$ and $\mathbb{P}(X_k=-1)=q=1-p$.

One of the results about $M_n$ is $\mathbb{P}(M_n\ge r, S_n=b)$ or even $\mathbb{P}(M_n\ge r)$. I am just wondering how to relate to the minimum $m_n=\min\{S_i,0\le i \le n\}$ given this result? Is there any simple way to do this?

I just found sometimes considering the minimum is most convenient for some problems.

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1 Answer 1

Let $m_n^p = \min\{S_i, 0\leq i\leq n, P(X_i=1)=p\}$, with a similar definition for $M_n^p$.

Then $P(m_n^p < r) = P(M_n^{1-p} > -r)$.

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nice. So essential you devise a new random walk with inverted probability? Initially I was thinking in the line of a new random walk with inverted $X_k$'s. –  Qiang Li Jun 9 '11 at 15:39
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The two are trivially equivalent (think about the distribution of $-X_k$). –  Did Jun 9 '11 at 16:38

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