Question

Let $X$ be a random variable with the following PDF:

$f_X\left(x\right)=\{\begin{array}{c}C{x}^2;|x|\le 1\\ 0;\text{otherwise}\end{array}$

Find Var(X).

• A. $\frac{2}{5}$
• B. $\frac{4}{5}$
• C. $\frac{3}{5}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3}{5}$

Given, $X$ is a continuous random variable with PDF $f_X\left(x\right)=\{\begin{array}{c}C{x}^2,|x|\le 1\\ 0,otherwise\end{array}$

We have to find the positive constant $c$.

We know that a PDF is valid if it satisfies the following

$\left(1\right)$ $f\left(x\right)$ is positive everywhere.

$\left(2\right)$ The area under the curve $f\left(x\right)$ in the range $(a,b)$is 1, that is: ${\int }_a^{b}f\left(x\right)dx=1$.

Thus we have ${\int }_{-1}^1{f}_X\left(x\right)dx=1$

$\Rightarrow {\int }_{-1}^{1}C{x}^{2}dx=1$

$\Rightarrow C{\int }_{-1}^1{x}^{2}dx=1$

$\Rightarrow C{\left[\frac{x^3}{3}\right]}_{-1}^1=1$

$\Rightarrow C[\frac{1}{3}-\left(\frac{-1}{3}\right)]=1$

$\Rightarrow C\left(\frac{2}{3}\right)=1$

$\Rightarrow C=\frac{3}{2}$

$\therefore f_X\left(x\right)=\{\begin{array}{c}\frac{3}{2}{x}^2,|x|\le 1\\ 0,otherwise\end{array}$

Now let us find $E\left(X\right)$

We know that $E\left(X\right)={\int }_a^{b}xP\left(x\right)dx$ where $P\left(x\right)$ is probability density function.

Thus $E\left(X\right)={\int }_{-1}^{1}x\cdot \frac{3}{2}{x}^{2}dx$

$={\int }_{-1}^1\frac{3}{2}{x}^{3}dx$

$=\frac{3}{2}{\int }_{-1}^1{x}^{3}dx$

$=\frac{3}{2}{\left[\frac{x^4}{4}\right]}_{-1}^1$

$=\frac{3}{2}[\frac{1}{4}-\frac{1}{4}]$

$=0$

$\therefore E\left(X\right)=0$

We have to find $Var\left(X\right)$

We know that $Var\left(X\right)={\int }_a^b{x}^{2}P\left(x\right)dx-{\mu }^2$ where $P\left(x\right)$ is probability density function and $\mu$ is mean. Also $\mu =E\left(X\right)$

Thus $Var\left(X\right)={\int }_{-1}^1{x}^2\cdot \frac{3}{2}{x}^{2}dx-\left[E\right(X){]}^2$

$={\int }_{-1}^1\frac{3}{2}{x}^{4}dx-[0{]}^2$

$=\frac{3}{2}{\int }_{-1}^1{x}^{4}dx-0$

$=\frac{3}{2}{\int }_{-1}^1{x}^{4}dx$

$=\frac{3}{2}{\left[\frac{x^5}{5}\right]}_{-1}^1$

$=\frac{3}{2}[\frac{1}{5}-\left(\frac{-1}{5}\right)]$

$=\frac{3}{2}\left[\frac{2}{5}\right]$

$=\frac{3}{5}$

$\therefore Var\left(X\right)=\frac{3}{5}$