# median of a uniform distribution [0,1]

I need to find the distribution of the median from the given distribution, where n is known to be odd.

The formula given in class for this is:

$n=2m+1$ where $m\in\mathbb{N}$

$f_{x(m+1)}(x)=\frac{(2m+1)!}{m!m!}[F_x(x)]^m[f_x(x)][1-F_x(x)]^m$

Noting that $f_x(x)=1$ and $F_x(x)=x$ we get:

$f_{x(m+1)}(x)=\frac{(2m+1)!}{m!m!}[x]^m[1][1-x]^m$ $=\frac{(2m+1)(2m)(2m-1)!}{m(m-1)!m(m-1)!}[x]^m[1][1-x]^m$ $=\frac{(2m+1)(2\not{m})\Gamma(2m)}{\not{m}\Gamma(m)m\Gamma(m)}[x]^m[1-x]^m$ $=\frac{(2m+1)(2)\Gamma(2m)}{m\Gamma(m)\Gamma(m)}[x]^m[1-x]^m$

recalling that the beta function is: B(x,y)=$\Gamma(x)\Gamma(y)\over{\Gamma(x+y)}$ its easy to see that $\Gamma2\over{\Gamma(m)\Gamma(m)}$ is $1\over{B(m,m)}$

Thus $f_{x(m+1)}(x)=\frac{(2m+1)(2)}{mB(m,m)}[x]^m[1-x]^m$

Which is where I'm stuck. I'm thinking at this point the distribution of the median will be a gamma distribution, but I'm unsure how to simplify this any further to achieve this. Any thoughts?

-

If you calculations are correct, what you've shown is that the sample median follows a Beta distribution with parameters $(m+1,m+1)$. By the way, the distribution of the sample median cannot be Gamma because the domain of Gamma distribution is unbounded.
What if its not a standard uniform, say $U[0,\theta]$. O cant seem to get rid of $\frac{1}{\theta}$ to show its a beta distribution. –  user191919 Jan 31 '14 at 12:28