Quiz 4 Solution

The random variable $X$ has cumulative distribution function (cdf) given by

$F_X(x)=\{\begin{array}{ll}0,& x<0,\\ \frac{1}{8}(x+u(x-1)+u(x-2)+u(x-3)+u(x-4)),& 0\le x\le 4,\\ 1,& x>4,\end{array}$

where $u(x)$ denotes the unit step function.  The cdf $F_X(x)$ is sketched below.

1.  The random variable $X$ would be classified as:

• discrete,
• continuous,
• mixed-type

Answer:  The cdf has jump discontinuities, so $X$ is not continuous.  The cdf is not piece-wise flat, so $X$ is not discrete.  Therefore $X$ is a mixed-type random variable.

2.  Find $Pr(2\le X\le 4)$.

Answer:  $Pr(2\le X\le 4)=F_X(4)-F_X(2^{-})=1-\frac{3}{8}=\frac{5}{8}=0.625$.

Remark:  Recall that the notation $F_X(a^{-})$ denotes $\underset{h\to 0^{+}}{lim}{F}_X(a-h)$, i.e., the limit of $F_X(x)$ as $x$ approaches $a$ from below.  We need this limit because $X=2$ is included in the event $[2,4]$.  If, instead, we were interested in $Pr(2<X\le 4)$, this would simply be $F_X(4)-F_X(2)$, which excludes $Pr(X=2)$.

3.  Find $Pr(\frac{3}{2}\le X\le \frac{5}{2})$.

Answer:  $Pr(\frac{3}{2}\le X\le \frac{5}{2})=F_X(\frac{5}{2})-F_X({\frac{3}{2}}^{-})=\frac{9}{16}-\frac{5}{16}=\frac{1}{4}=0.25$.

4.  Find $Pr(X=3)$.

Answer:  $Pr(X=3)=F_X(3)-F_X(3^{-})=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}=0.125$.

Remark:  this is the height of the jump-discontinuity in $F_X(x)$ at $x=3$.

5.  Find $Pr(X=\frac{5}{2})$.

Answer:  $Pr(X=\frac{5}{2})=F_X(\frac{5}{2})-F_X({\frac{5}{2}}^{-})=\frac{9}{16}-\frac{9}{16}=0$.

Remark:  this is the height of the jump-discontinuity in $F_X(x)$ at $x=\frac{5}{2}$.  However, $F_X(x)$ is continuous at $x=\frac{5}{2}$, i.e., there is no jump discontinuity, so the height of the jump-discontinuity is zero.

6.  Find $Pr(X\text{ is not an integer})$.

Answer:  Let us first find the probability of the complementary event.   We have $Pr(X\text{ is an integer})=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}Pr(X=k)$.    Now $Pr(X=k)$ is the height of the jump-discontinuity in $F_X(x)$ at $x=k$.  Only for $k\in \{1,2,3,4\}$ is this height nonzero.   Thus $Pr(X\text{ is an integer})=Pr(X=1)+Pr(X=2)+Pr(X=3)+Pr(X=4)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$.  Finally, $Pr(X\text{is not an integer})=1-Pr(X\text{ is an integer})=1-\frac{1}{2}=\frac{1}{2}=0.5$.

7.  A discrete random variable $Y$ has range ${\mathcal{S}}_Y=\{0,2,4,6,\dots \}$ (the nonnegative even integers).  The probability mass function for $Y$ is given as $p_Y(k)=C\cdot 5^{-k}$ for $k\in {\mathcal{S}}_Y$, where $C$ is a constant.  Determine the value of $C$.

Answer:  The probabilities must sum to one.  Even integers take the form $2m$ where $m$ is an integer.  Thus we require $1=\sum _{m=0}^{\mathrm{\infty }}C\cdot 5^{-2m}=C\sum _{m=0}^{\mathrm{\infty }}(5^{-2}{)}^m=\frac{C}{1-5^{-2}}$.  We find that $C=1-5^{-2}=1-\frac{1}{25}=\frac{24}{25}=0.96$.

8.   An urn contains two $5 bills, two $10 bills, and one $20 bill.  Let the random variable  $M$ denote the total amount that results when two bills are drawn one at a time at random from the urn, without replacement.    The range of $M$ is the set ${\mathcal{S}}_M=\{10,15,20,25,30\}$.  Determine the probability mass function for $M$ .

Hint:  let $b_1$ denote the value of the first bill drawn, and let $b_2$ denote the value of the second bill drawn.  There are then eight possible $(b_1,b_2)$ pairs comprising the set

$\mathcal{S}=\{(5,5),(5,10),(5,20),(10,5),(10,10),(10,20),(20,5),(20,10)\}$.

Start by determining the probability of each these elementary outcomes, and then use these values to determine the probability mass function for $M$.

Answer:  We are sampling without replacement, so $b_1$ and $b_2$ are not independent.  We have

$Pr(5,5)=\frac{2}{5}\cdot \frac{1}{4}=\frac{1}{10}$

$Pr(5,10)=\frac{2}{5}\cdot \frac{2}{4}=\frac{2}{10}$

$Pr(5,20)=\frac{2}{5}\cdot \frac{1}{4}=\frac{1}{10}$

$Pr(10,5)=\frac{2}{5}\cdot \frac{2}{4}=\frac{2}{10}$

$Pr(10,10)=\frac{2}{5}\cdot \frac{1}{4}=\frac{1}{10}$

$Pr(10,20)=\frac{2}{5}\cdot \frac{1}{4}=\frac{1}{10}$

$Pr(20,5)=\frac{1}{5}\cdot \frac{2}{4}=\frac{1}{10}$

$Pr(20,10)=\frac{1}{5}\cdot \frac{2}{4}=\frac{1}{10}$

We can verify that these probabilities do indeed sum to one.  Finally, we can compute the probability mass function for $M$ by considering which ordered pairs $(b_1,b_2)$ sum to a particular value. We get

$p_M(10)=Pr(5,5)=\frac{1}{10}=0.1$

$p_M(15)=Pr(5,10)+Pr(10,5)=\frac{4}{10}=0.4$

$p_M(20)=Pr(10,10)=\frac{1}{10}=0.1$

$p_M(25)=Pr(5,20)+Pr(20,5)=\frac{2}{10}=0.2$

$p_M(30)=Pr(10,20)+Pr(20,10)=\frac{2}{10}=0.2$

Again, as a check, we can verify that these probabilities do indeed sum to one.