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

\(\begin{array}{l} f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{x}{a} + 1}&{: - a \le x < 0}\\ { - \frac{x}{a} + 1}&{0 \le x \le a}\\ 0&{otherwise} \end{array}} \right.\\ g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - \frac{x}{a}}&{:-a \le x \le 0}\\ {\frac{x}{a}}&{:0 \le x \le a}\\ 0&{:otherewise} \end{array}} \right. \end{array}\)

Which one of the following statements is true?

\(E_1(x)=\int^{\infty}_{-\infty}xf(x)dx\)

\(⇒ \int^0_{-a}xf(x)dx+\int^a_0af(x)dx\)

\(⇒ \int^0_{-a}x\left(\frac{x}{a}+1\right)+\int^a_{0}x\left(-\frac{x}{a}+1\right)\)

\(⇒ \int^0_{-a}\frac{x^2}{a}dx+\int^0_{-a}xds+\int^a_{0}\frac{-x^2}{a}dx+\int^a_{0}xdx\)

⇒ 0

∴ \(\left[\int^0_{-a}\frac{x^2}{a}=-\int^a_0-\frac{x^2}{a}\right]\)

∴ \(\left[\int^0_{-a}xdx=-\int^a_0xdx\right]\)

\(E_2(x)=\int^{\infty}_{-\infty}xg(x)dx\)

⇒ \(\int^0_{-a}x\left(-\frac{x}{a}\right)dx+\int^0_{a}x\left(\frac{x}{a}\right)dx\)

⇒ \(\int^0_{-a}-\frac{x^2}{a}dx+\int^0_a\frac{x^2}{a}dx\)

⇒ 0

∴ \(\left[\int^0_{-a}-\frac{x^2}{a}dx=-\int^0_a\frac{x^2}{a}dx\right]\)

Variance

E(x²) - {E(x)}²

\(E_1(x^2)=\int^{\infty}_{-\infty}x^2f(x)\)

⇒ \(\int^0_{-a}x^2\left(\frac{x}{a}+1\right)dx+\int^a_{0}x^2\left(-\frac{x}{a}+1\right)dx\)

\(\Rightarrow \int^0_{-a}\frac{x^3}{a}dx+\int^0_{-a}x^2dx+\int^a_0-\frac{x^3}{a}dx+\int^a_0x^2dx\)

\(\Rightarrow -\frac{a^3}{4}+\frac{a^3}{3}-\frac{a^3}{4}+\frac{a^3}{3}=\frac{a^3}{6}\)

\(E_2(x^2)=\int^{\infty}_{-\infty}x^2g(x)\)

\(\int^0_{-a}x^2\left(-\frac{x}{a}\right)dx+\int^a_0x^2\left(\frac{x}{a}\right)dx\)

\(\int^0_{-a}\frac{-x^3}{a}dx+\int^a_0\frac{x^3}{a}dx\)

\(\left[-\frac{x^4}{4a}\right]^0_{-a}+\left[\frac{x^4}{4a}\right]^a_{0}\)

\(\left\{0-\left[\frac{-(a)^4}{4a}\right]\right\}+\left\{\frac{a^4}{4a}-0\right\}\)

\(\frac{a^3}{4}+\frac{a^3}{4}=\frac{a^3}{2}\)

Mean of f(x) is E(x):

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^0 X \left( {\frac{X}{a} + 1} \right)dx + \mathop \smallint \limits_0^a X \left( {\frac{{ - X}}{a} + 1} \right)dx\\ = \left( {\frac{{{X^3}}}{{3a}} + \frac{{{X^2}}}{2}} \right)_{ - a}^0 + \left( {\frac{{ - {X^3}}}{{3a}} + \frac{{{X^3}}}{3}} \right)_0^a = 0\end{array}\)

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^0 X^2 \left( {\frac{X}{a} + 1} \right)dx + \mathop \smallint \limits_0^a X^2 \left( {\frac{{ - X}}{a} + 1} \right)dx\\ \end {array}\)

\(\left( {\frac{{{X^4}}}{{4a}} + \frac{{{X^3}}}{3}} \right)_{ - a}^0 + \left( {\frac{{ - {X^4}}}{{4a}} + \frac{{{X^3}}}{3}} \right)_0^a = \frac{a^3}{6}\)

⇒ Variance is \(\frac{{{a^3}}}{6}\)

Next, mean of g(x) is E(x)

\(= \mathop \smallint \limits_{-a}^0 x\left( {\frac{{ - x}}{a}} \right)dx + \mathop \smallint \limits_0^a x\times \left( {\frac{X}{a}} \right)dx = 0\)

Variance of g(x) is E(x²) – {E(X)}², where

\(E\left( {{X^2}} \right) = \mathop \smallint \limits_{ - a}^0 {X^2}\left( {\frac{{ - X}}{a}} \right)dX + \mathop \smallint \limits_0^a {X^2}\left( {\frac{X}{a}} \right)dx = \frac{{{a^3}}}{2}\)

⇒ Variance is \(\frac{{{a^3}}}{2}\)

∴ Mean of f(x) and g(x) are same but variance of f(x) and g(x) are different