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5.1.6 Solved Problems

Problem
Consider two random variables $X$ and $Y$ with joint PMF given in Table 5.3.
  1. Find $P(X \leq 2, Y \leq 4)$.
  2. Find the marginal PMFs of $X$ and $Y$.
  3. Find $P(Y=2 | X=1)$.
  4. Are $X$ and $Y$ independent?
  $Y = 2$ $Y = 4$ $Y = 5$
$X = 1$ $\frac{1}{12}$ $\frac{1}{24}$ $\frac{1}{24}$
$X = 2$ $\frac{1}{6}$ $\frac{1}{12}$ $\frac{1}{8}$
$X = 3$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{1}{12}$
  • Solution
      1. To find $P(X \leq 2, Y \leq 4)$, we can write \begin{align}%\label{} \nonumber P(X \leq 2, Y \leq 4) &=P_{XY}(1,2)+ P_{XY}(1,4)+P_{XY}(2,2)+ P_{XY}(2,4)\\ \nonumber &=\frac{1}{12}+\frac{1}{24}+\frac{1}{6}+\frac{1}{12}=\frac{3}{8}. \end{align}
      2. Note from the table that \begin{align}%\label{} \nonumber R_X=\{1,2,3\} \textrm{ and } R_Y=\{2,4,5\}. \end{align} Now we can use Equation 5.1 to find the marginal PMFs: \begin{equation} \nonumber P_X(x) = \left\{ \begin{array}{l l} \frac{1}{6} & \quad x=1 \\ & \quad \\ \frac{3}{8} & \quad x=2 \\ & \quad \\ \frac{11}{24} & \quad x=3 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} \begin{equation} \nonumber P_Y(y) = \left\{ \begin{array}{l l} \frac{1}{2} & \quad y=2 \\ & \quad \\ \frac{1}{4} & \quad y=4 \\ & \quad \\ \frac{1}{4} & \quad y=5 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
      3. Using the formula for conditional probability, we have \begin{align}%\label{} \nonumber P(Y=2 | X=1)&=\frac{P(X=1, Y=2)}{P(X=1)}\\ \nonumber &=\frac{P_{XY}(1,2)}{P_X(1)}\\ \nonumber &=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}. \end{align}
      4. Are $X$ and $Y$ independent? To check whether $X$ and $Y$ are independent, we need to check that $P(X=x_i,Y=y_j)=P(X=x_i)P(Y=y_j)$, for all $x_i \in R_X$ and all $y_j \in R_Y$. Looking at the table and the results from previous parts, we find \begin{align}%\label{} \nonumber P(X=2,Y=2)=\frac{1}{6} \neq P(X=2)P(Y=2)=\frac{3}{16}. \end{align} Thus, we conclude that $X$ and $Y$ are not independent.

Problem
I have a bag containing $40$ blue marbles and $60$ red marbles. I choose $10$ marbles (without replacement) at random. Let $X$ be the number of blue marbles and $y$ be the number of red marbles. Find the joint PMF of $X$ and $Y$.
  • Solution
    • This is, in fact, a hypergeometric distribution. First, note that we must have $X+Y=10$, so \begin{align}%\label{} \nonumber R_{XY}&=\{(i,j) | i+j=10, i,j \in \mathbb{Z}, i,j \geq 0\}\\ \nonumber &=\{(0,10),(1,9),(2,8),...,(10,0)\}. \end{align} Then, we can write \begin{equation} \nonumber P_{XY}(i,j)= \left\{ \begin{array}{l l} \frac{{40 \choose i} {60 \choose j}}{{100 \choose 10}} & \quad i+j=10, i,j \in \mathbb{Z}, i,j \geq 0 \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

Problem
Let $X$ and $Y$ be two independent discrete random variables with the same CDFs $F_{X}$ and $F_Y$ . Define \begin{align}%\label{} \nonumber Z=\max (X,Y), \\ \nonumber W=\min(X,Y). \end{align} Find the CDFs of $Z$ and $W$.
  • Solution
    • To find the CDF of $Z$, we can write \begin{align}%\label{} \nonumber F_Z(z)&=P(Z \leq z) \\ \nonumber &=P(\max(X,Y) \leq z)\\ \nonumber &=P\bigg((X \leq z) \textrm{ and } (Y \leq z)\bigg)\\ \nonumber &=P(X \leq z) P(Y \leq z) &(\textrm{ since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=F_X(z)F_Y(z). \end{align} To find the CDF of $W$, we can write \begin{align}%\label{} \nonumber F_W(w)&=P(W \leq w) \\ \nonumber &=P(\min(X,Y) \leq w)\\ \nonumber &=1-P(\min(X,Y) > w)\\ \nonumber &=1-P\bigg((X > w) \textrm{ and } (Y > w)\bigg)\\ \nonumber &=1-P(X > w) P(Y > w) &(\textrm{ since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=1-(1-F_X(w))(1-F_Y(w))\\ \nonumber &=F_X(w)+F_Y(w)-F_X(w)F_Y(w). \end{align}

Problem
Let $X$ and $Y$ be two discrete random variables, with range \begin{align}%\label{} \nonumber R_{XY}=\{(i,j) \in \mathbb{Z}^2|i,j \geq 0, |i-j|\leq 1\}, \end{align} and joint PMF given by \begin{align}%\label{} \nonumber P_{XY}(i,j)=\frac{1}{6 \cdot 2^{\min(i,j)}}, \hspace{10pt} \textrm{ for }(i,j) \in R_{XY}. \end{align}
  1. Pictorially show $R_{XY}$ in the $x-y$ plane.
  2. Find the marginal PMFs $P_X(i)$, $P_Y(j)$.
  3. Find $P(X=Y|X<2)$.
  4. Find $P(1 \leq X^2+Y^2 \leq 5)$.
  5. Find $P(X=Y)$.
  6. Find $E[X|Y=2]$.
  7. Find Var$(X|Y=2)$.
  • Solution
      1. Figure 5.5 shows the $R_{XY}$ in the $x-y$ plane.

        Figure 5.5: Figure shows $R_{XY}$ for $X$ and $Y$ in problem 4.

      2. First, by symmetry we note that $X$ and $Y$ have the same PMF. Next, we can write \begin{align}\label{} \nonumber &P_X(0)=P_{XY}(0,0)+P_{XY}(0,1)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3},\\ \nonumber &P_X(1)=P_{XY}(1,0)+P_{XY}(1,1)+P_{XY}(1,2)=\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{2}\right)=\frac{1}{3},\\ \nonumber &P_X(2)=P_{XY}(2,1)+P_{XY}(2,2)+P_{XY}(2,3)=\frac{1}{6}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{6},\\ \nonumber &P_X(3)=P_{XY}(3,2)+P_{XY}(3,3)+P_{XY}(3,4)=\frac{1}{6}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{12}. \end{align} In general, we obtain \begin{equation} \nonumber P_X(k) = P_Y(k)=\left\{ \begin{array}{l l} \frac{1}{3} & \quad k=0 \\ & \quad \\ \frac{1}{3 \cdot 2^{k-1}} & \quad k=1,2,3,... \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
      3. Find $P(X=Y|X<2)$: We have \begin{align}\label{} \nonumber P(X=Y|X<2)&=\frac{P(X=Y,X<2)}{P(X<2)}\\ \nonumber &=\frac{P_{XY}(0,0)+P_{XY}(1,1)}{P_X(0)+P_X(1)}\\ \nonumber &=\frac{\frac{1}{6}+\frac{1}{12}}{\frac{1}{3}+\frac{1}{3}}\\ \nonumber &=\frac{3}{8}. \end{align}
      4. Find $P(1 \leq X^2+Y^2 \leq 5)$: We have \begin{align}\label{} \nonumber P(1 \leq X^2+Y^2 \leq 5)&=P_{XY}(0,1)+P_{XY}(1,0)+P_{XY}(1,1)+P_{XY}(1,2)+P_{XY}(2,1)\\ \nonumber &=\frac{1}{6}+\frac{1}{6}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\\ \nonumber &=\frac{7}{12}. \end{align}
      5. By symmetry, we can argue that $P(X=Y)=\frac{1}{3}$. The reason is that $R_{XY}$ consists of three lines with points with the same probabilities. We can also find $P(X=Y)$ by \begin{align}\label{} \nonumber P(X=Y)&=\sum_{i=0}^{\infty} P_{XY}(i,i)\\ \nonumber &=\sum_{i=0}^{\infty} \frac{1}{6. 2^i}\\ \nonumber &=\frac{1}{3}. \end{align}
      6. To find $E[X|Y=2]$, we first need the conditional PMF of $X$ given $Y=2$. We have \begin{align}\label{} \nonumber P_{X|Y}(k|2) &=\frac{P_{XY}(k,2)}{P(Y=2)}\\ \nonumber &=6P_{XY}(k,2), \end{align} so we obtain \begin{equation} \nonumber P_{X|Y}(k|2)=\left\{ \begin{array}{l l} \frac{1}{2} & \quad k=1 \\ & \quad \\ \frac{1}{4} & \quad k=2,3 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Thus, \begin{align}\label{} \nonumber E[X|Y=2]&=1 \cdot \frac{1}{2}+2 \cdot \frac{1}{4}+3 \cdot \frac{1}{4}\\ \nonumber &=\frac{7}{4}. \end{align}
      7. Find Var$(X|Y=2)$: we have \begin{align}\label{} \nonumber E[X^2|Y=2]&=1 \cdot \frac{1}{2}+4 \cdot \frac{1}{4}+9 \cdot \frac{1}{4}\\ \nonumber &=\frac{15}{4}. \end{align} Thus, \begin{align}\label{} \nonumber \textrm{Var}(X)&=E[X^2|Y=2]-\big(E[X|Y=2]\big)^2\\ \nonumber &=\frac{15}{4}-\frac{49}{16}\\ \nonumber &=\frac{11}{16}. \end{align}

Problem
Suppose that the number of customers visiting a fast food restaurant in a given day is $N \sim Poisson(\lambda)$. Assume that each customer purchases a drink with probability $p$, independently from other customers, and independently from the value of $N$. Let $X$ be the number of customers who purchase drinks. Let $Y$ be the number of customers that do not purchase drinks; so $X+Y=N$.
  1. Find the marginal PMFs of $X$ and $Y$.
  2. Find the joint PMF of $X$ and $Y$.
  3. Are $X$ and $Y$ independent?
  4. Find $E[X^2Y^2]$.
  • Solution
      1. First note that $R_X=R_Y=\{0,1,2,...\}$. Also, given $N=n$, $X$ is a sum of $n$ independent $Bernoulli(p)$ random variables. Thus, given $N=n$, $X$ has a binomial distribution with parameters $n$ and $p$, so \begin{align}\label{} \nonumber &X|N=n \hspace{10pt} \sim \hspace{10pt} Binomial(n,p),\\ \nonumber &Y|N=n \hspace{10pt} \sim \hspace{10pt} Binomial(n,q=1-p). \end{align} We have \begin{align}\label{} \nonumber P_X(k)&=\sum_{n=0}^{\infty} P(X=k|N=n)P_N(n) & (\textrm{law of total probability})\\ \nonumber &=\sum_{n=k}^{\infty} {n \choose k} p^k q^{n-k} exp(-\lambda) \frac{\lambda^n}{n!}\\ \nonumber &=\sum_{n=k}^{\infty} \frac{p^k q^{n-k} exp(-\lambda) \lambda^n}{k! (n-k)!} \\ \nonumber &=\frac{exp(-\lambda) (\lambda p)^k}{k!} \sum_{n=k}^{\infty} \frac{(\lambda q)^{n-k}}{(n-k)!} \\ \nonumber &=\frac{exp(-\lambda) (\lambda p)^k}{k!} exp(\lambda q) & (\textrm{Taylor series for } e^x)\\ \nonumber &=\frac{exp(-\lambda p) (\lambda p)^k}{k!}, \hspace{40pt} \textrm{ for }k=0,1,2,... \end{align} Thus, we conclude that \begin{align}\label{} \nonumber &X \hspace{10pt} \sim \hspace{10pt} Poisson(\lambda p),\\ \nonumber &Y \hspace{10pt} \sim \hspace{10pt} Poisson(\lambda q). \end{align}
      2. To find the joint PMF of $X$ and $Y$, we can also use the law of total probability: \begin{align}\label{} \nonumber P_{XY}(i,j)&=\sum_{n=0}^{\infty} P(X=i, Y=j|N=n)P_N(n) & (\textrm{law of total probability}). \end{align} But note that $P(X=i, Y=j|N=n)=0$ if $N \neq i+j$, thus \begin{align}\label{} \nonumber P_{XY}(i,j)&=P(X=i, Y=j|N=i+j)P_N(i+j)\\ \nonumber &=P(X=i|N=i+j)P_N(i+j)\\ \nonumber &={i+j \choose i} p^i q^j exp(-\lambda) \frac{\lambda^{i+j}}{(i+j)!}\\ \nonumber &=\frac{exp(-\lambda) (\lambda p)^i (\lambda q)^j}{i! j!}\\ \nonumber &=\frac{exp(-\lambda p) (\lambda p)^i}{i!}. \frac{exp(-\lambda q) (\lambda q)^j}{j!}\\ \nonumber &=P_X(i)P_Y(j). \end{align}
      3. $X$ and $Y$ are independent, since as we saw above \begin{align}\label{} \nonumber P_{XY}(i,j)=P_X(i)P_Y(j). \end{align}
      4. Since $X$ and $Y$ are independent, we have \begin{align}\label{} \nonumber E[X^2Y^2]=E[X^2]E[Y^2]. \end{align} Also, note that for a Poisson random variable $W$ with parameter $\lambda$, \begin{align}\label{} \nonumber E[W^2]=\textrm{Var}(W)+(EW)^2=\lambda+\lambda^2. \end{align} Thus, \begin{align}\label{} \nonumber E[X^2Y^2]&=E[X^2]E[Y^2]\\ \nonumber &=(\lambda p+ \lambda^2 p^2)(\lambda q+ \lambda^2 q^2)\\ \nonumber &=\lambda^2 pq(\lambda^2 pq+ \lambda+1). \end{align}

Problem
I have a coin with $P(H)=p$. I toss the coin repeatedly until I observe two consecutive heads. Let $X$ be the total number of coin tosses. Find $EX$.
  • Solution
    • We solve this problem using a similar approach as in Example 5.6. Let $\mu=EX$. We first condition on the result of the first coin toss. Specifically, \begin{align} \nonumber \mu=EX&=E[X|H]P(H)+E[X|T]P(T)\\ \nonumber &=E[X|H]p+(1+\mu)(1-p). \end{align} In this equation, $E[X|T]=1+EX$, because the tosses are independent, so if the first toss is tails, it is like starting over on the second toss. Thus, \begin{align}\label{alhh} p\mu=pE[X|H]+(1-p) \hspace{100pt} (5.14) \end{align} We still need to find $E[X|H]$ so we condition on the second coin toss \begin{align} \nonumber E[X|H]&=E[X|HH]p+E[X|HT](1-p)\\ \nonumber &=2p+(2+\mu)(1-p)\\ \nonumber &=2+(1-p)\mu. \end{align} Here, $E[X|HT]=2+EX$ because, if the first two tosses are $HT$, we have wasted two coin tosses and we start over at the third toss. By letting $E[X|H]=2+(1-p)\mu$ in Equation 5.14, we obtain \begin{align} \nonumber \mu=EX=\frac{1+p}{p^2}. \end{align}

Problem
Let $X,Y \sim Geometric(p)$ be independent, and let $Z=\frac{X}{Y}$.
  1. Find the range of $Z$.
  2. Find the PMF of $Z$.
  3. Find $EZ$.
  • Solution
      1. The range of $Z$ is given by \begin{align}%\label{} \nonumber R_Z=\left\{\frac{m}{n}|m,n \in \mathbb{N}\right\}, \end{align} which is the set of all positive rational numbers.
      2. To find PMF of $Z$, let $m,n \in \mathbb{N}$ such that $(m,n)=1$, where $(m,n)$ is the largest divisor of $m$ and $n$. Then \begin{align}%\label{} \nonumber P_Z\left(\frac{m}{n}\right) &= \sum_{k=1}^{\infty} P(X=mk, Y=nk)\\ \nonumber &= \sum_{k=1}^{\infty} P(X=mk)P(Y=nk) \hspace{30pt} (\textrm{since }X \textrm{ and } Y \textrm{are independent})\\ \nonumber &= \sum_{k=1}^{\infty} pq^{mk-1}pq^{nk-1} \hspace{40pt} (\textrm{where }q=1-p)\\ \nonumber &= p^2q^{-2}\sum_{k=1}^{\infty} q^{(m+n)k}\\ \nonumber &= \frac{p^2q^{m+n-2}}{1-q^{m+n}}\\ \nonumber &=\frac{p^2(1-p)^{m+n-2}}{1-(1-p)^{m+n}}. \end{align}
      3. Find $EZ$: We can use LOTUS to find $EZ$. Let us first remember the following useful identities: \begin{align}%\label{} \nonumber \sum_{k=1}^{\infty} kx^{k-1}= \frac{1}{(1-x)^2}, \hspace{40pt} \textrm{ for }|x|<1, \end{align} \begin{align}%\label{} \nonumber -\ln (1-x)=\sum_{k=1}^{\infty} \frac{x^{k}}{k}, \hspace{40pt} \textrm{ for }|x|<1. \end{align} The first one is obtained by taking derivative of the geometric sum formula, and the second one is a Taylor series. Now, let's apply LOTUS. \begin{align}%\label{} \nonumber E\bigg[\frac{X}{Y}\bigg]&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{m}{n}P(X=m, Y=n)\\ \nonumber &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{m}{n}p^2q^{m-1}q^{n-1}\\ \nonumber &= \sum_{n=1}^{\infty} \frac{1}{n} p^2q^{n-1}\sum_{m=1}^{\infty} mq^{m-1}\\ \nonumber &= \sum_{n=1}^{\infty} \frac{1}{n} p^2q^{n-1}\frac{1}{(1-q)^2}\\ \nonumber &= \sum_{n=1}^{\infty} \frac{1}{n} q^{n-1}\\ \nonumber &=\frac{1}{q}\sum_{n=1}^{\infty} \frac{ q^{n}}{n}\\ \nonumber &=\frac{1}{1-p}\ln \frac{1}{p}. \end{align}

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