Question

The p.m.f. of a r.v. X is 

P(x) = ${\displaystyle \{\begin{array}{l}\frac{2x}{n(n+1)}\text{,}x=1\text{,}2\text{,}....\text{,}\text{n}\\ 0\text{,}\text{otherwise}\end{array}}$ Then, E(X) = ______.

Options

• ${\displaystyle \frac{n+1}{3}}$

• ${\displaystyle \frac{2n+1}{3}}$

• ${\displaystyle \frac{n+2}{3}}$

• ${\displaystyle \frac{2n-1}{3}}$

Answer 1

The p.m.f. of a r.v. X is 

P(x) = ${\displaystyle \{\begin{array}{l}\frac{2x}{n(n+1)}\text{,}x=1\text{,}2\text{,}....\text{,}\text{n}\\ 0\text{,}\text{otherwise}\end{array}}$ Then, E(X) = ${\displaystyle \underline{\frac{2n+1}{3}}}$.

Explanation:

X	1	2	3	....	n
P(X)	${\displaystyle \frac{2}{n(n+1)}}$	${\displaystyle \frac{4}{n(n+1)}}$	${\displaystyle \frac{6}{n(n+1)}}$	....	${\displaystyle \frac{2n}{n(n+1)}}$

${\displaystyle E\left(x\right)=\sum x_i\cdot p\left(x_i\right)}$

${\displaystyle =1\cdot \frac{2}{n(n+1)}+2\cdot \frac{4}{n(n+1)}+3\cdot \frac{6}{n(n+1)}+...+n\cdot \frac{2n}{n(n+1)}}$

${\displaystyle =\frac{2}{n(n+1)}(1+4+9+...+n^2)}$

${\displaystyle =\frac{2}{n(n+1)}(1^2+2^2+3^2+...+n^2)}$

${\displaystyle =\frac{2}{n(n+1)}\cdot \frac{n(n+1)(2n+1)}{6}}$

${\displaystyle =\frac{2n+1}{3}}$