Question

If $f(x)$ and $g(x)$ are two probability density functions,$f(x)=\{\begin{array}{cc}\frac{x}{a}+1& :-a\le x<0\\ -\frac{x}{a}+1& 0\le x\le a\\ 0& \text{ otherwise }\end{array}$$g(x)=\{\begin{array}{cc}-\frac{x}{a}& :-a\le x\le 0\\ \frac{x}{a}& :0\le x\le a\\ 0& :\text{ otherewise }\end{array}$Which one 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

Explanation:
Mean of $f(x)$:$E(x)={\int }_{-a}^{0}X(\frac{X}{a}+1)dx+{\int }_0^{a}X(\frac{-X}{a}+1)dx$$={(\frac{X^3}{3a}+\frac{X^2}{2})}_{-a}^0+{(\frac{-X^3}{3a}+\frac{X^3}{3})}_0^a=0$$={\int }_{-a}^0{X}^2(\frac{X}{a}+1)dx+{\int }_0^a{X}^2(\frac{-X}{a}+1)dx$${(\frac{X^4}{4a}+\frac{X^3}{3})}_{-a}^0+{(\frac{-X^4}{4a}+\frac{X^3}{3})}_0^a=\frac{a^3}{6}$$\Rightarrow$ Variance $=\frac{a^3}{6}$Mean of $g(x)$:$E(x)={\int }_{-a}^{0}x\left(\frac{-x}{a}\right)dx+{\int }_0^{a}x\times \left(\frac{X}{a}\right)dx=0$Variance of $g(x)$ is $E\left(x^2\right)-\{E(X){\}}^2$, Where $E\left(X^2\right)={\int }_{-a}^0{X}^2\left(\frac{-X}{a}\right)dX+{\int }_0^a{X}^2\left(\frac{X}{a}\right)dx=\frac{a^3}{2}$$\Rightarrow$ Variance $=\frac{a^3}{2}$$\therefore$ Mean of $f(x)$ and $g(x)$ are same but variance of $f(x)$ and $g(x)$ are different.Hence, the correct option is (B).