Question

Show that the median of a sample of size $2n+1$from a uniform distribution on $(0,1)$ has a beta distribution with parameters $(n+1,n+1)$.

Short answer

$\frac{(2n+1)!}{n!n!}{\left(x\right)}^n{(1-x)}^n$is a Beta function with parameters$l=n+1\&m=n+1$.

Step-by-step solution

Uniform distribution :

A continuous probability distribution is a Uniform distribution that describes occurrences that are equally likely to happen.

Explanation :

Sample size$2n+1$.

Uniform distribution on$(0,1)$.

Beta distribution with parameters$(n+1,n+1)$.

Sample size$2n+1$

Uniform distribution over$(0,1)$

For median$j=n+1$

Applying equation $(6.2)$

localid="1647349001706" $$\begin{gathered} f_{x_{ij}}\left(x\right)=\frac{n!}{(n-j)!(j-1)!}{\left[F\left(x\right)\right]}^{-1}{\left[1-F\left(x\right)\right]}^{n-j}f\left(x\right) \\ f\left(x\right)=1\Rightarrow F\left(x\right)=x \\ \Rightarrow f_{x_{(n+!)}}\left(x\right)=\frac{(2n+1)!}{n!n!}{\left(x\right)}^n{(1-x)}^n\times 1 \\ =\frac{(2n+1)!}{n!n!}{\left(x\right)}^n{(1-x)}^n.........\left(1\right) \end{gathered}$$

A form of Beta distribution is

localid="1648088016166" $f\left(x\right)=\frac{1}{\beta (l,m)}{x}^{l-1}{(1-x)}^{m-1}0\le x\le 1;l,m>0$

Thus, $\left(1\right)$is a Beta function with parameters $l=n+1\&m=n+1$

Hence proved.