183 views

Que:-
The variance of the random variable X with P.D.F  f(x)=0.5|x| e-|x| is ___?

answer $6$ ???
yes 6, i need detailed solution please
writing it...long sol
updated !

$$\large f(x) = \begin{cases} \\ {-} \; 0.5.{\color{red}{x}}.\mathbf e^{\color{red}{x}} & , x \leq 0 \\ \\ {+} \; 0.5.{\color{red}{x}}.\mathbf e^{\color{red}{-x}} & , x \geq 0 \\ \\ \end{cases}$$

\begin{align*} E(x) & = \int_{- \infty}^{+\infty}x.f(x).dx \\ & =\int_{- \infty}^{0}x.f(x).dx + \int_{0}^{+\infty}x.f(x).dx \\ & =\int_{- \infty}^{0}x.\left [ -0.5.x.e^x \right ].dx+ \int_{0}^{+\infty}x.\left [ 0.5.x.e^{-x} \right ].dx \\ & =\int_{0}^{- \infty}\left [ 0.5.x^2.e^x \right ].dx + {\color{red}{\int_{0}^{-\infty}\left [ 0.5.y^2.e^{y} \right ].\left ( -dy \right )}} \;\; \left \{ \text{putting x = -y} \right \} \\ &= 0 \\ \end{align*}

\begin{align*} E(X^{2}) & = \int_{- \infty}^{+\infty}x^2.f(x).dx \\ & =\int_{- \infty}^{0}x^2.f(x).dx + \int_{0}^{+\infty}x^2.f(x).dx \\ & =\int_{- \infty}^{0}x^2.\left [ -0.5.x.e^x \right ].dx+ \int_{0}^{+\infty}x^2.\left [ 0.5.x.e^{-x} \right ].dx \\ & =\int_{0}^{- \infty}\left [ 0.5.x^3.e^x \right ].dx + {\color{red}{\int_{0}^{-\infty}\left [ 0.5.(-y)^3.e^{y} \right ].\left ( -dy \right )}} \;\; \left \{ \text{putting x = -y} \right \} \\ &= \int_{0}^{- \infty}\left [x^3.e^x \right ].dx \;\; \left \{ \text{y is dummy variable only} \right \} \\ &= \left [ (x^3 - 3x^2 + 6x - 6)e^x \right ]_0^{-\infty} \\ &=6 \\ \end{align*}

\begin{align*} \text{VAR(X)} & = E(X^2) - \left [ E(X) \right ]^2 \\ & = 6 - 0 \\ &= 6\\ \end{align*}

NOTE

$$\int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C.$$

\color{maroon}{\begin{align*} &\text{Using the idea of ODD / EVEN function} \\ &\Rightarrow f(x) \;\; \text{is EVEN} \\ &\Rightarrow x.f(x) \;\; \text{is ODD} \\ &\Rightarrow \int_{-\infty}^{+\infty}xf(x)dx = 0 \\ &\Rightarrow E(X) = 0 \\ \\ \text{Now} \\ &\Rightarrow x^2.f(x) \;\; \text{is EVEN} \\ &\Rightarrow E(X^2) = \int_{-\infty}^{+\infty}x^2f(x)dx = 2.\int_{0}^{+\infty}x^2f(x)dx \\ &\Rightarrow E(X^2) = 2.\int_{0}^{+\infty}0.5.x^3.e^{-x}dx \\ &\Rightarrow E(X^2) = \int_{0}^{+\infty}x^3.e^{-x}dx \\ &\Rightarrow E(X^2) = 6 \\ \end{align*}}

edited
I solved this question using the definitions of ODD and Even functions :) :)
I wanted to get it solved using other way. you solved it using both ways thanks a lot!!
QS was meant to be done in ODD/EVEN way