most recent 30 from http://mathoverflow.net 2013-05-20T07:57:06Z http://mathoverflow.net/feeds/question/9615 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9615/probability-distribution-of-the-median Nick 2009-12-23T13:38:19Z 2009-12-24T12:47:18Z

<p>Suppose we have $2k + 1$ points $a_1, ..., a_{2k+1}$. Each point is uniformly distributed between 0 and N. What is the distribution of the median (i.e. of the k+1-th point) ?</p> <p>What happens if $a_1, ..., a_{2k+1}$ is a random set of $2k+1$ distinct points from $1, 2, ..., N$ ?</p>

http://mathoverflow.net/questions/9615/probability-distribution-of-the-median/9621#9621 Alekk 2009-12-23T16:23:43Z 2009-12-23T16:23:43Z

<p>have you checked the "order statistics" article on Wikipedia.</p>

http://mathoverflow.net/questions/9615/probability-distribution-of-the-median/9634#9634 David Speyer 2009-12-23T19:14:16Z 2009-12-24T12:47:18Z

<p>I assume the points are meant to take integer values between $0$ and $N$.</p> <p>The second question is easier. The probability that the median is $j$ is $$\binom{j}{k} \binom{N-j}{k} / \binom{N+1}{2k+1}.$$ (I hope I'm not answering homework here. If I am, remember that you need to justify this statement to get full marks.)</p> <p>The probability that $j/N$ is between $x$ and $x+dx$ is roughly $$\frac{(2k+1)!}{(k!)^2} x^k (1-x)^k dx ,$$ as $N$ goes to infinity with $k$ fixed.</p> <p>When you allow repetition, you should have the same asymptopics, but the problem is just messy enough that I can't get a simple answer. My best formula is</p> <p>$$\frac{(2k+1)!}{(k!)^2} \frac{1}{(N+1)^{2k+1}} \int_{x=0}^1 (x+j)^k (N-j+x)^k dx.$$</p> <p>Bounding the integral by the minimum and maximum values of the integrand gives reasonable bounds. I'll provide a proof if needed.</p>