Question

Suppose the random variable X has the probability density function (p.d.f) given as
f(x) = c(1 − x^2), for − 1 < x < 1
0, elsewhere
for some constant c.
(i) What is the value of c?
(ii) Derive the cumulative distribution function of X, F(X).
(iii) Hence or otherwise, find P(X > 0.7).

Answer 1

$f(x)=\begin{cases} c(1-x^2), \ \ -1<x<1 \\ 0, \ \ \text{ elsewhere } \end{cases}$

(i) To find the value of $c$ , we can use the fact that $\int \limits _{-\infin}^{\infin}f(x)dx=1$

$\int \limits _{-\infin}^{\infin}f(x)dx= \int \limits _{-1}^{1}c(1-x^2)dx =c(x-\frac{x^3}{3}) \bigg| _{x=-1}^{x=1}= \frac{4c}{3}=1 \ \Rightarrow c=3/4$

Answer: $c=3/4$

(ii) $F(x)=\int \limits _{-\infin}^{x}f(t)dt$

if $x\leq -1:$ $F(x)= \int \limits _{-\infin}^{x}f(t)dt= \int \limits _{-\infin}^{x}0dt=0$

if $-1<x<1: F(x)= \int \limits _{-\infin}^{x}f(t)dt= \int \limits _{-1}^{x}\frac{3}{4}(1-t^2)dt=\frac{3}{4}(t-\frac{t^3}{3})-\frac{3}{4}(-1-\frac{(-1)^3}{3})=$

$= \frac{3}{4}t-\frac{1}{4}t^3+\frac{1}{2}$

if $x\geq 1: F(x)=\int \limits _{-\infin}^{x}f(t)dt= \int \limits _{-1}^{1}f(t)dt+ \int \limits _{1}^{x}f(t)dt=\ 1+0=1$

Answer: $F(x)=\begin{cases} 0, \ \ x\leq -1 \\ \frac{3}{4}t-\frac{1}{4}t^3+\frac{1}{2} , \ \ -1<x<1 \\ 1, \ \ 1\leq x \end{cases}$

(iii) $P(X>0.7)=1-P(X\leq 0.7)=1-F(0.7)=1-(\frac{3}{4}\times 0.7-\frac{1}{4}\times 0.7^3+\frac{1}2)=$

$=0.06075$

Answer: $P(X>0.7)=0.06075$