Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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 at 12:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.