Question

If a random variable $X$ has probability distribution function $f\left(x\right)=\frac{c}{x},1<x<3,c>0$,
find $c,E\left(X\right)$ and $Var\left(X\right)$.

Answer 1

$fx=\frac{c}{x}:f\left(x\right)$ is probability density function

$\therefore {}_a^b\int f\left(x\right)\left(x\right)dx=1$

$\Rightarrow {}_1^3\int \frac{c}{x}dx=1$

$\Rightarrow c[\log 3-\log 1]=1\Rightarrow c=\frac{1}{\log 3}$

$E\left(x\right)={}_a^b\int xf\left(x\right)dx$

$\Rightarrow {}_1^3\int \frac{x}{\log 3\times x}dx$

$\Rightarrow \frac{1}{\log 3}[x{]}_1^3$

$\Rightarrow E\left(x\right)=\frac{2}{\log 3}$

$Var\left(x\right)={}_b^a\int (x-\mu {)}^{2}f(x)dx$

${\Rightarrow }_1^3\int {(x-\frac{2}{\log 3})}^2\left(\frac{1}{x\log 3}\right)dx$

$=\frac{1}{\log 3}{}_1^3\int (x^2+\frac{4}{{\log }^{2}3}-\frac{4x}{\log 3})\times \frac{1}{x}dx$

$\Rightarrow \frac{1}{\log 3}{}_1^3\int [x+\frac{4}{{\log }^{2}3x}-\frac{4}{\log 3}]dx$

$\Rightarrow \frac{1}{\log 3}{[\frac{x^2}{2}+\frac{4}{{\log }^{2}3}\log x-\frac{4x}{\log 3}]}_1^3$

$\Rightarrow \frac{1}{\log 3}[4+\frac{4}{{\log }^{2}3}(\log 3)-\frac{8}{\log 3}]$

$Var\left(x\right)=\frac{1}{\log 3}[4-\frac{4}{\log 3}]$