Question

Let $X$ be a random variable with the following Probability Distribution Function:

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

Find the constant $C$.

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

Answer 1

The correct option is B $\frac{3}{2}$

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

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

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

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

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

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

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

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