Simple Random Walk

Question

How to find:

$\lim\limits_{N \to \infty}\sum\limits_{m=0}^Nu_m$ where $u_m$=${2m \choose m}p^mq^m$

I know there are two cases to consider depending if $p$ and $q$ are equal or not. I should probably mention p and q are probabilities related to a simple random walk. I suspect the infinite sum is bounded for an asymmetric walk and unbounded for a symmetric walk but I'm struggling to come up with a closed form, ive tried relating it to the $(1+pq)^a$ expansion for some $a$ but I can't seem to progress.

Thanks

You've got the same binomial coefficient both in the sum and in definition of $u_m$; Did was asking whether it is intentional or a typo. — Peter Košinár, Dec 15 '13 at 14:50
This should help: http://math.stackexchange.com/questions/69270/show-sum-limits-n-0-infty2n-choose-nxn-1-4x-1-2 — , Dec 15 '13 at 16:22

score 1 · Accepted Answer · answered Dec 15 '13 at 16:28

1

As @Byron indicated, for every $p\ne\frac12$ in $[0,1]$, $$ \sum_{m=0}^\infty{2m\choose m}(pq)^m=\frac1{\sqrt{1-4pq}}=\frac1{|p-q|}.$$

answered Dec 15 '13 at 16:28

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Simple Random Walk

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