Quiz 7 Solution

The random variable $X$ has range ${\mathcal{S}}_X=[0,\mathrm{\infty })$ with probability density function

$f_X(x)=\{\begin{array}{ll}\exp (-x),& x\ge 0\\ 0,& x<0.\end{array}$

Let $Y=1-2\exp (-X)$.  Then $Y$ is itself a random variable.  Note that, since $0<\exp (-x)\le 1$ when $x\ge 0$, the random variable $Y$ has range ${\mathcal{S}}_Y=[-1,1)$.

The first three questions pertain to the random variable $X$.

1.  True or false?  $X$ is an exponential random variable.

Answer:  true.  By definition an exponential random variable $Z$ has a pdf of the form

$f_Z(z)=\{\begin{array}{ll}\lambda \exp (-\lambda z),& z\ge 0,\\ 0,& z<0.\end{array}$

for some parameter $\lambda >0$.  The pdf of $X$ has this form for $\lambda =1$.

2.  True or false?  $X$ is a Gaussian random variable.

Answer: false.  By definition a Gaussian random variable $Z$ has a range ${\mathcal{S}}_Z=\mathbb{R}$, and the pdf of $Z$, given as

$f_Z(z)=\frac{1}{\sigma \sqrt{2\pi }}\exp (-\frac{(z-m{)}^2}{2{\sigma }^2})$

is nonzero for all $z$.   This does not match the pdf of $X$.

3.  True or false?  $X$ is a gamma random variable.

Answer: true.  By definition a gamma random variable $Z$ has a pdf of the form

$f_Z(z)=\{\begin{array}{ll}\frac{\lambda }{\mathrm{\Gamma }(\alpha )}(\lambda x{)}^{\alpha -1}\exp (-\lambda x),& x\ge 0\\ 0,& x<0\end{array}$

for parameters $\alpha >0$ and $\lambda >0$, where $\mathrm{\Gamma }(x)$ is the gamma function.  Setting $\alpha =1$ and $\lambda =1$ gives the pdf for $X$, since $\mathrm{\Gamma }(1)=0!=1$.  (In general, the exponential pdf is the special case of the gamma pdf obtained when $\alpha =1$.)

The last four questions pertain to the random variable $Y$.

4.  Determine $E(Y)$.

Answer:  $E(Y)=E(1-2\exp (-X))$

$=1-2E(\exp (-X))$

$=1-2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp (-x)f_X(x)\text{d}x$

$=1-2{\int }_0^{\mathrm{\infty }}\exp (-x)\exp (-x)\text{d}x$

$=1-2{\int }_0^{\mathrm{\infty }}\exp (-2x)\text{d}x$

$=1-2\left(\frac{1}{2}\right)$

$=0$.

5.  Determine $E(Y^2)$.

Answer:  $E(Y^2)=E((1-2\exp (-X){)}^2)$

$={\int }_0^{\mathrm{\infty }}(1-4\exp (-x)+4\exp (-2x))\exp (-x)\text{d}x$

$={\int }_0^{\mathrm{\infty }}(\exp (-x)-4\exp (-2x)+4\exp (-3x))\text{d}x$

$=1-\frac{4}{2}+\frac{4}{3}$

$=\frac{1}{3}$.

6.  Determine $Pr(Y>\frac{1}{2})$.

Answer:  Note the $F_X(x)=Pr(X\le x)=1-\exp (-x)$ when $x\ge 0$.  We have

$Pr(Y>\frac{1}{2})=Pr(1-2\exp (-X)\ge \frac{1}{2})$

$=Pr(-2\exp (-X)\ge -\frac{1}{2})$

$=Pr(\exp (-X)\le \frac{1}{4})$

$=Pr(-X\le \ln (1/4))$

$=Pr(X\ge \ln (1/4))$

$=1-Pr(X<\ln (1/4))$

$=1-F_X(\ln (1/4))$

$=1-(1-\exp (\ln (1/4)))$

$=\frac{1}{4}$.

7.  True or false?  $Y$ is uniformly distributed over ${\mathcal{S}}_Y$.

Answer:  true.

For $y\in {\mathcal{S}}_Y$, we have

$Pr(Y\le y)=Pr(1-2\exp (-X)\le y)$

$=Pr(2\exp (-X)\ge 1-y)$

$=Pr(-X\ge \ln \left(\frac{1-y}{2}\right))$

$=Pr(X\le -\ln \left(\frac{1-y}{2}\right))$

$=1-\exp (\ln \left(\frac{1-y}{2}\right))$

$=1-\frac{1-y}{2}$

$=\frac{y+1}{2}$

It follows that $f_Y(y)=\frac{\text{d}}{\text{d}y}{F}_Y(y)=\frac{1}{2}$ for all $y\in {\mathcal{S}}_Y$.   Since the pdf is a constant, $Y$ is indeed uniformly distributed.