Question

If the probability density function of a random variable is given by,
$f\left(x\right)=\{\begin{array}{cc}k(1-x^2),& 0<x<1\\ 0,& \text{elsewhere}\end{array}$ find $k$ and the distribution function of the random variable.

Answer 1

(i) Since $f\left(x\right)$ is a p.d.f. ${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$
$\Rightarrow {\int }_0^{1}k(1-x^2)dx=1$

$\Rightarrow k{[x-\frac{x^3}{3}]}_0^1=1$
$\Rightarrow k(1-\frac{1}{3})=1$

$\Rightarrow \frac{2k}{3}=1$

$\Rightarrow k=\frac{3}{2}$
(ii) The distribution function $F\left(x\right)={\int }_{-\infty }^{x}f\left(t\right)dt$
(a) When $x\in (-\infty ,0]$
$F\left(x\right)={\int }_{-\infty }^{x}f\left(t\right)dt=0$
(b) When $x\epsilon (0,1]$
$F\left(x\right)={\int }_{-\infty }^{x}f\left(t\right)dt={\int }_{-\infty }^{0}f\left(t\right)dt+{\int }_0^{x}f\left(t\right)dt$
$=0+\frac{3}{2}{\int }_0^x(1-t^2)dt$
$F\left(x\right)=\frac{3}{2}(x-\frac{x^3}{3})$
(c) When $x\in [1,\infty )$
$F\left(x\right)={\int }_{-\infty }^{x}f\left(t\right)dt$
$={\int }_{-\infty }^{0}f\left(t\right)dt+{\int }_0^{1}f\left(t\right)dt+{\int }_1^{x}f\left(t\right)dt$
$=0+{\int }_0^1\frac{3}{2}(1-t^2)dt+0$
$=\frac{3}{2}{\left[t\frac{t^3}{3}\right]}_0^1=1$
$\therefore F\left(x\right)=\{\begin{array}{cc}0& -\infty <x\le 0\\ \frac{3}{2}(x-\frac{x^3}{3}),& 0<x<1\\ 1& 1\le x\infty \end{array}$