Q.6.28 - A First Course in Probability

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 $(n + 1, n + 1)$. 

Step-by-Step Solution

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 1 : Uniform distribution :

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

Step 2 : 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 $(6.2)$

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

A form of Beta distribution is 

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

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

Hence proved.