Question

A random variable X has the density function $f\left(x\right)=K\frac{1}{1+x^2},\text{where}-\infty <x<\infty .$ Then the value of K is

• A. $\pi$
• B. $\frac{1}{\pi }$
• C. $2\pi$
• D. $\frac{1}{2\pi }$

Answer 1

The correct option is B $\frac{1}{\pi }$

$\because f\left(x\right)=\frac{K}{1+x^2}$ is p.d.f.

so we have,
${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

$\Rightarrow {\int }_{-\infty }^{\infty }\frac{K}{1+x^2}dx=1$

$\Rightarrow 2K{\left[ta{n}^{-1}x\right]}_0^{\infty }=1$

$\Rightarrow 2K(\frac{\pi }{2}-0)=1$

$\Rightarrow K=\frac{1}{\pi }$