Timeline for Find closed form for comparison of two binomial random variable: solve inequality
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7 events
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| Nov 11, 2013 at 17:16 | comment | added | wolfies | See also the answer to this question: math.stackexchange.com/questions/562119/… ... which provides the pmf of the difference of 2 Binomials. | |
| Aug 9, 2012 at 5:37 | comment | added | kjetil b halvorsen | ?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. | |
| Jul 12, 2012 at 7:33 | comment | added | Douglas Zare | 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 $\text{Binomial}(2n+1,1/2).$ | |
| Jul 12, 2012 at 1:01 | comment | added | Brendan McKay | No chance of an exact formula except in some special cases like $p=\frac12$. | |
| Jul 11, 2012 at 22:45 | comment | added | Douglas Zare | 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. | |
| Jul 11, 2012 at 21:32 | comment | added | Robert Israel | 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. | |
| Jul 11, 2012 at 20:01 | history | asked | greatel | CC BY-SA 3.0 |