most recent 30 from http://mathoverflow.net 2013-05-20T22:39:30Z http://mathoverflow.net/feeds/user/25060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101986/find-closed-form-for-comparison-of-two-binomial-random-variable-solve-inequalit greatel 2012-07-11T20:01:08Z 2012-07-12T07:32:22Z

<p>Hi, Dear All, </p> <p>I come up with this problem, which I think for a long time without a good answer.</p> <p>Suppose two independent random variables $X \sim \mathrm{Binomial}(n, p)$ and $Y \sim \mathrm{Binomial}(m, p)$, and n &lt; m .</p> <p>We know that: \begin{eqnarray} Pr(X>Y) = \sum_{k_1=1}^n\sum_{k_2=1}^{m}\mathrm{I}(k_1>k_2) {n \choose k_1}{m \choose k_2}p^{k_1+k_2}(1-p)^{m+n-k_1-k_2} \end{eqnarray} \begin{eqnarray} = \sum_{k_1=1}^n\sum_{k_2=1}^{k_1 -1}{n \choose k_1}{m \choose k_2}p^{k_1+k_2}(1-p)^{m+n-k_1-k_2} \end{eqnarray}</p> <p><strong>Question:</strong></p> <p>Do we have closed form for the above problem?</p> <p>Motivation: because I want to solve Pr(X>Y| n&lt; m) > a, and try to bound the p, so I have to get a closed form and try to solve this inequality. </p> <p>$$\sum_{k_1=1}^n\sum_{k_2=1}^{k_1 -1}{n \choose k_1}{m \choose k_2}p^{k_1+k_2}(1-p)^{m+n-k_1-k_2} > a$$</p> <p>$$ p > ???$$</p> <p>Any hints? Thanks.</p>