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Hi, Dear All,

I come up with this problem, which I think for a long time without a good answer.

Suppose two independent random variables $X \sim \mathrm{Binomial}(n, p)$ and $Y \sim \mathrm{Binomial}(m, p)$, and n < m .

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}

Question:

Do we have closed form for the above problem?

Motivation: because I want to solve Pr(X>Y| n< m) > a, and try to bound the p, so I have to get a closed form and try to solve this inequality.

$$\sum_{k_1=1}^n\sum_{k_2=1}^{k_1 -1}{n \choose k_1}{m \choose k_2}p^{k_1+k_2}(1-p)^{m+n-k_1-k_2} > a$$

$$ p > ???$$

Any hints? Thanks.

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  • $\begingroup$ I doubt that there is a closed form for this probability; even if there is a closed form for that, it's extremely doubtful that you could solve for $p$ in closed form. However, if $n$ and $m$ are large you can approximate $X-Y$ with a normal distribution. $\endgroup$
    – Robert Israel
    Commented Jul 11, 2012 at 21:32
  • $\begingroup$ If the normal approximation is good enough, then this is a simple exercise and not research level. If it is not good enough for your purposes, then please explain why it isn't. By the way, binomial random variables can take the value $0$, so you have the wrong lower limits. $\endgroup$
    – Douglas Zare
    Commented Jul 11, 2012 at 22:45
  • $\begingroup$ No chance of an exact formula except in some special cases like $p=\frac12$. $\endgroup$
    – Brendan McKay
    Commented Jul 12, 2012 at 1:01
  • $\begingroup$ One special case is $a=1/2, m=n+1$. There is equality when $p=1/2$: $P(X < Y) = P((n-X) + Y >= n+1)$ and $n-X$ also has a binomial distribution, so $(n-X)+Y$ is distributed as $\text{Binomial}(2n+1,1/2).$ $\endgroup$
    – Douglas Zare
    Commented Jul 12, 2012 at 7:33
  • $\begingroup$ ?Why do you need a closed form? The summation can be easily programmed (say, in R) so a numerical solution of your inequality is easy! For the cases I tried, I get that the probability sum as a function of p is concave, so the solution for prob_sum > a is a open interval or empty. $\endgroup$
    – kjetil b halvorsen
    Commented Aug 9, 2012 at 5:37

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