Question

4) (20 pts) Let X be a RV with the following PDF: fx(x) = že=fal for all x. Let Y = X?. (a) Compute E[X]. (b) Find the PDF of Y, fy(y). (c) Compute E[Y].

Answer 1

(a)

The expectation of X is given by:
$\begin{align*} E[X] &= \int_{-\infty}^{\infty} x*f_X(x) \ dx \\ &= \int_{-\infty}^{\infty} x*\frac{1}{2}e^{-|x|} \ dx \\ &= \int_{-\infty}^{0} x*\frac{1}{2}e^{-|x|} \ dx + \int_{0}^{\infty} x*\frac{1}{2}e^{-|x|} \ dx \\ &= \int_{-\infty}^{0} x*\frac{1}{2}e^{-(-x)} \ dx + \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &\text{[Since, when x is negative, }|x| = -x \text{ and when x is positive, }|x| = x] \\ &= \int_{-\infty}^{0} x*\frac{1}{2}e^{x} \ dx + \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &\text{In the first integral let: }x=-y \ \ \ \ \Rightarrow dx = -dy \\ &\text{when }x=-\infty, y = \infty \text{ and when }x=0 , y=0 \text{, thus we get:} \\ &= \int_{\infty}^{0} (-y)*\frac{1}{2}e^{-y} \ (-dy)+ \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &= \int_{\infty}^{0} y*\frac{1}{2}e^{-y} \ dy+ \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &= -\int_{0}^{\infty} y*\frac{1}{2}e^{-y} \ dy+ \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \end{align*}$

$\begin{align*} E[X] &= -\int_{0}^{\infty} y*\frac{1}{2}e^{-y} \ dy+ \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &\left[\text{Using: }\int_a^b g(t) \ dt = -\int_b^a g(t) \ dt \right ] \\ &\text{In the first integral let }y=x \ \ \Rightarrow dy = dx\text{, thus we get:} \\ &= -\int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dy+ \int_{0}^{\infty} x*\frac{1}{2}e^{-x} \ dx \\ &= \bf 0 \ \ \ \ \ \ \ \ \ \ [ANSWER] \end{align*}$

(b)

The support of X is given by:
$\\ -\infty < x< \infty \\ \Rightarrow 0\le |x| < \infty \\ \Rightarrow 0^2\le |x|^2 < \infty^2 \\ \Rightarrow 0\le x^2 < \infty \\ \Rightarrow 0\le y < \infty$

Thus, the support of Y is given by the interval: .

0,0)

Now, we find the cumulative distribution function of Y:
$\begin{align*} F_Y(y) &= P(Y\le y) \ \ \ \ \ \ \ \ \ \ ;0\le y < \infty \\ &= P(X^2 \le y) \\ &= P(|X| \le \sqrt{y}) \\ &= P(-\sqrt{y} \le X \le \sqrt{y}) \\ &= \int_{-\sqrt{y}}^{\sqrt{y}} f_X(x) \ dx \\ &= \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{2}e^{-|x|} \ dx \\ &= \int_{-\sqrt{y}}^{0} \frac{1}{2}e^{-|x|} \ dx + \int_{0}^{\sqrt{y}} \frac{1}{2}e^{-|x|} \ dx \\ &= \int_{-\sqrt{y}}^{0} \frac{1}{2}e^{-(-x)} \ dx + \int_{0}^{\sqrt{y}} \frac{1}{2}e^{-x} \ dx \\ &\text{[Since, when x is negative, }|x|=-x \text{ and when x is positive, }|x|=x] \\ &= \int_{-\sqrt{y}}^{0} \frac{1}{2}e^{x} \ dx + \int_{0}^{\sqrt{y}} \frac{1}{2}e^{-x} \ dx \end{align*}$

$\begin{align*} F_Y(y) &= \frac{1}{2}\int_{-\sqrt{y}}^{0} e^{x} \ dx + \frac{1}{2}\int_{0}^{\sqrt{y}} e^{-x} \ dx \\ &= \frac{1}{2}*\left[e^x \right ]_{x=-\sqrt{y}}^{x=0} + \frac{1}{2}*\left[-e^{-x} \right ]_{x=0}^{x=\sqrt{y}} \\ &= \frac{1}{2}*\left[e^0 - e^{-\sqrt{y}} \right ] + \frac{1}{2}*\left[-e^{-\sqrt{y}} - (-e^{0}) \right ] \\ &= \frac{1}{2}*\left[1 - e^{-\sqrt{y}} \right ] + \frac{1}{2}*\left[-e^{-\sqrt{y}} - (-1) \right ] \\ &= \frac{1}{2}*\left[1 - e^{-\sqrt{y}} \right ] + \frac{1}{2}*\left[1-e^{-\sqrt{y}} \right ] \\ &= 1-e^{-\sqrt{y}} \ \ \ \ \ \ \ ; 0\le y < \infty \end{align*}$

Thus, the PDF of Y is given by:
$\begin{align*} \boldsymbol{f_Y(y)} &= \frac{\mathrm{d} F_Y(y)}{\mathrm{d} y} \\ &= \frac{\mathrm{d} }{\mathrm{d} y}\left(1-e^{-\sqrt{y}} \right ) \\ &= 0-e^{-\sqrt{y}}*\frac{\mathrm{d} }{\mathrm{d} y}\left(-\sqrt{y} \right ) \ \ \ \ \ \ \ \ \ \ \ \ \text{[Using Chain Rule]} \\ &= -e^{-\sqrt{y}}*\left(-\frac{1}{2\sqrt{y}} \right ) \\ &= \boldsymbol{\frac{e^{-\sqrt{y}}}{2\sqrt{y}} \ \ \ \ \ \ \ ; 0\le y < \infty} \end{align*}$

(c)

The expectation of Y is given by:
$\begin{align*} \boldsymbol{E(Y)} &= \int_0^\infty y*f_Y(y) \ dy \\ &= \int_{0}^{\infty} y*\frac{e^{-\sqrt{y}}}{2\sqrt{y}} \ dy \\ &=\frac{1}{2} \int_{0}^{\infty} \sqrt{y}*e^{-\sqrt{y}} \ dy \\ &\text{Let: }t=\sqrt{y} \ \ \Rightarrow y = t^2 \ \ \Rightarrow dy = 2t \ dt \\ &=\frac{1}{2} \int_{0}^{\infty} t*e^{-t} *2t \ dt \\ &= \int_{0}^{\infty} t^2*e^{-t} \ dt \\ &= \Gamma(3) \ \ \ \ \ \ \ \ \ \ \left[\text{Using: }\int_{0}^{\infty}z^{n-1}e^{-z} \ dz = \Gamma(n) \right ] \\ &= 2! \ \ \ \ \ \ \ \ \ \ \ \left[\text{Using: When n is a natural number, }\Gamma = n! \right ] \\ &= \bf 2 \ \ \ \ \ \ \ \ \ \ \ [ANSWER] \end{align*}$

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