Quiz 6 Solution

The random variable $X$ is uniformly distributed over the interval $[-1,5]$.

1.  What is $E(X)$?

Answer:  for a random variable uniformly distributed over $[a,b]$, we showed in class that the expected value is $(a+b)/2$.  In this case $a=-1$, $b=5$, so $E(X)=\frac{1}{2}(-1+5)=2$.

More laboriously, since $f_X(x)=\{\begin{array}{ll}\frac{1}{6},& -1\le x\le 5,\\ 0,& \text{otherwise,}\end{array}$, we get

$E(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{f}_X(x)\text{d}x$

$={\int }_{-1}^{5}x\cdot \frac{1}{6}\text{d}x$

$=\frac{1}{6}\cdot {\frac{x^2}{2}|}_{-1}^5$

$=\frac{5^2-(-1{)}^2}{12}$

$=2$.

2.  What is the variance, $\text{VAR}(X)$, of $X$?

Answer, for a random variable uniformly distributed over $[a,b]$, we showed in class that the variance is $\frac{1}{12}(b-a{)}^2$.  In this case $a=-1$, $b=5$, so $\text{VAR}(X)=\frac{(5-(-1){)}^2}{12}=\frac{6^2}{12}=3$.

More laboriously, using the pdf as above, and noting the $E(X)=2$, we get

$\text{VAR}(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x-E(X){)}^2{f}_X(x)\text{d}x$

$={\int }_{-1}^5(x-2{)}^2\cdot \frac{1}{6}\text{d}x$

$={\int }_{-3}^3{u}^2\cdot \frac{1}{6}\text{d}u$ (substituting $u=x-2$)

$=\frac{1}{6}\cdot {\frac{u^3}{3}|}_{-3}^3$

$=\frac{3^3+3^3}{18}$

$=3$.

3.  What is the second moment of $X$?

Answer:  We know that $\text{VAR}(X)=E(X^2)-(E(X){)}^2$, thus $E(X^2)=\text{VAR}(X)+(E(X){)}^2=3+2^2=7$.

More laboriously, we have

$E(X^2)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2{f}_X(x)\text{d}x$

$={\int }_{-1}^5{x}^2\cdot \frac{1}{6}\text{d}x$

$=\frac{1}{6}\cdot {\frac{x^3}{3}|}_{-1}^5$

$=\frac{5^3+1^3}{18}$

$=7$.

4.  What is $Pr(|X-2|<\frac{3}{4})$?

Answer: To have $|X-2|<\frac{3}{4}$ we need $-\frac{3}{4}<X-2<\frac{3}{4}$ or $2-\frac{3}{4}<X<2+\frac{3}{4}$.  We then get

$Pr(\frac{5}{4}<X<\frac{11}{4})={\int }_{\frac{5}{4}}^{\frac{11}{4}}{f}_X(x)\text{d}x$

$={\int }_{\frac{5}{4}}^{\frac{11}{4}}\frac{1}{6}\text{d}x$

$=\frac{1}{6}(\frac{11}{4}-\frac{5}{4})$

$=\frac{1}{4}=0.25$

5.  How should the constant $c$ be chosen so that the random variable $c\cdot (X-2)$ has unit variance?

Answer:  in general, if $Y=aX+b$, then $\text{VAR}(Y)=a^2\text{VAR}(X)$.  The shift value, $b$, does not influence the variance of $Y$.  Here $a=c$ and $b=-2c$.  We choose $c$ so that $c^2\text{VAR}(X)=1$, thus $c=\pm \frac{1}{\sqrt{\text{VAR}(X)}}$.  Since $\text{VAR}(X)=3$, we can choose $c=\pm \frac{1}{\sqrt{3}}\approx \pm 0.57735$.

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The random variable $Y$ is exponentially distributed with probability density function

$f_Y(y)=\{\begin{array}{ll}\frac{1}{2}\exp (-y/2),& y\ge 0;\\ 0,& y<0.\end{array}$

6.  What is $E(Y)$?

Answer:  $Y$ is exponentially distributed with parameter $\lambda =\frac{1}{2}$.   From the lecture we know that $E(Y)=\frac{1}{\lambda }=2$.

We may also compute

$E(Y)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}y{f}_Y(y)\text{d}y$

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

(now integrate by parts with $u=y$, $\text{d}u=\text{d}y$, $\text{d}v=-\frac{1}{2}exp(-y/2)\text{d}y$, $v=-\exp (-y/2)$, to obtain)

$E(Y)=\underset{0}{\underbrace{{[-y\exp (-y/2)]|}_0^{\mathrm{\infty }}}}+{\int }_0^{\mathrm{\infty }}\exp (-y/2)\text{d}y$

$=-{2\exp (-y/2)|}_0^{\mathrm{\infty }}$

$=2$.

We could also express this using the gamma function.  We have

$E(Y)={\int }_0^{\mathrm{\infty }}(y/2)\exp (-y/2)\text{d}y=2{\int }_0^{\mathrm{\infty }}u\exp (-u)\text{d}u=2\mathrm{\Gamma }(2)=\mathrm{\Gamma }(3)=2!=2$.

We could also compute this using the cdf of $Y$.  (See @126).  Since $Y$ is a positive random variable with cdf $F_Y(y)=1-\exp (-y/2)$, we have $E(Y)={\int }_0^{\mathrm{\infty }}1-F_Y(y)\text{d}y={\int }_0^{\mathrm{\infty }}\exp (-y/2)\text{d}y=2$.

7.  What is $Pr(1\le Y\le 3)$?

Answer:  $Pr(1\le Y\le 3)={\int }_1^3{f}_Y(y)\text{d}y={\int }_1^3\frac{1}{2}\exp (-y/2)\text{d}y=\exp (-1/2)-\exp (-3/2)\approx 0.3834$

8.  What is $Pr(Y>2)$?

Answer: $Pr(Y>2)={\int }_2^{\mathrm{\infty }}{f}_Y(y)\text{d}y={\int }_2^{\mathrm{\infty }}\frac{1}{2}\exp (-y/2)\text{d}y=\exp (-1)\approx 0.367879$.

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The random variable $Z$ is normally distributed with probability density function $f_Z(z)=\exp (-\pi (z-1{)}^2)$.

9.  What is $E(Z)$?

Answer:  a normal random variable with mean $m$ and variance ${\sigma }^2$ has pdf $f(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}\exp (-\frac{(x-m{)}^2}{2{\sigma }^2})$.  We see that $Z$ has this form with $m=1$ and ${\sigma }^2=\frac{1}{2\pi }$.

Thus $E(Z)=1$.

10.  What is $\text{VAR}(Z)$.

Answer: as noted, we have $\text{VAR}(Z)={\sigma }^2=\frac{1}{2\pi }\approx 0.1591549$.