Question

If f(x) and g(x) are two probability density functions,

$f\left(x\right)=(\begin{array}{cc}\frac{x}{a}+1:& -a\le x<0\\ -\frac{x}{a}+1:& 0\le x\le a\end{array}$

$g\left(x\right)=(\begin{array}{cc}-\frac{x}{a}:& -a\le x<0\\ \frac{x}{a}:& 0\le x\le a\\ 0:& \mathrm{otherwise}\end{array}$

Which of the following statements is true?

• A. Mean of f(x) and g(x) are same; Variance of f(x) and g(x) are same.
• B. Mean of f(x) and g(x) are same; Variance of f(x) and g(x) are different.
• C. Mean of f(x) and g(x) are different; Variance of f(x) and g(x) are same.
• D. Mean of f(x) and g(x) are different; Variance of f(x) and g(x) are different.

Answer 1

The correct option is B Mean of f(x) and g(x) are same; Variance of f(x) and g(x) are different.

$\because f(-x)=f\left(x\right)\text{and}g(-x)=g\left(x\right)\text{so both}f\left(x\right)\text{and}g\left(x\right)$ are even functions.

So, $\text{Mean (f)}=E\left(x\right)={\int }_{-\infty }^{\infty }xf\left(x\right)dx=0$

$(\because$ integrand is odd function)

& $\text{Mean (g)}=E\left(x\right)={\int }_{-\infty }^{\infty }xg\left(x\right)dx=0$

$(\because$ integrand is odd function)

Now, $Varf=E\left(x^2\right)-\left(E\right(x){)}^2$

$={\int }_{-\infty }^{\infty }{x}^{2}f\left(x\right)dx-(0{)}^2$

$=2{\int }_0^a{x}^{2}f\left(x\right)dx=\frac{a^3}{6}$

& $Var\left(g\right)=E\left(x^2\right)-\left(E\right(x){)}^2$

$={\int }_{-\infty }^{\infty }{x}^{2}g\left(x\right)dx-(0{)}^2$

$=2{\int }_0^a{x}^{2}g\left(x\right)dx=\frac{a^3}{2}$

Hence, variance of f(x) and g(x) are different.