MTH 202 Homework

HW 13, due May 1

Hints

Problem 1

Customers arrive at a service facility according to a Poisson process of intensity λ. The service times \(Y_1, Y_2 , \ldots\) of the arriving customers are independent random variables having the common probability distribution function \(G(y) = \Prob\{Y_k ≤ y\}\). Assume that there is no limit to the number of customers that can be serviced simultaneously; i.e., there is an infinite number of servers available. Let \(M(t)\) count the number of customers in the system at time t. Argue that \(M(t)\) has a Poisson distribution with mean \(λpt\), where

Problem 2

Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let \(X(t)\) be the number of customers that have arrived up to time t. Let \(W_1 , W_2 , \ldots\) be the successive arrival times of the customers.

1. Determine the conditional mean $E[W_1|X(t) = 2]$.

2. Determine the conditional mean $E[W_3|X(t) = 5]$.

3. Determine the conditional probability density function for \(W_2\), given that \(X(t) = 5\).

Problem 3

Suppose that N points are uniformly distributed over the surface of a circular disk of radius r. Determine the probability distribution for the number of points within a distance of one of the origin as \(N → ∞, r → ∞, N/(π r^2) = λ\).

Problem 4

Let \(X(t)\) be a Poisson process of rate \(\lambda\). Validate the identity

Use this to determine the joint upper tail probability

Finally, differentiate once in $w_1$ and once in $w_2$ to obtain the joint density function

Problem 5

The joint probability density function for the waiting times \(W_1\) and \(W_2\) is given by

Determine the conditional probability density function for \(W_1\), given that \(W_2 = w_2\). How does this result differ from that in Theorem 5.7 (I know the 4th edition of the textbook says Theorem 5.6, which is probably a typo)?

Problem 6

A critical component on a submarine has an operating lifetime that is exponentially distributed with mean \(0.50\) years. As soon as a component fails, it is replaced by a new one having statistically identical properties. What is the smallest number of spare components that the submarine should stock if it is leaving for a one-year tour and wishes the probability of having an inoperable unit caused by failures exceeding the spare inventory to be less than \(0.02\)?

HW 12 due Apr 24

Problem 1

For each value of \(h > 0\), let \(X(h)\) have a Poisson distribution with parameter \(λh\). Let \(p_k (h) = P\left( X(h) = k \right)\) for \(k = 0, 1, \ldots\) Verify that

Here \(o(h)\) stands for any remainder term of order less than \(h\) as \(h → 0\).

Problem 2

Shocks occur to a system according to a Poisson process of rate \(λ\). Suppose that the system survives each shock with probability α, independently of other shocks, so that its probability of surviving \(k\) shocks is \(α^k\). What is the probability that the system is surviving at time t?

Problem 3

Determine numerical values to three decimal places for \(P\left( X = k \right)\), \(k = 0, 1, 2\), when

(a) \(X\) has a binomial distribution with parameters \(n = 20\) and \(p = 0.06\).

(b) \(X\) has a binomial distribution with parameters \(n = 40\) and \(p = 0.03\).

(c) \(X\) has a Poisson distribution with parameter \(λ = 1.2\).

Problem 4

Suppose that 100 tags, numbered \(1, 2, . . . , 100\), are placed into an urn, and 10 tags are drawn successively, with replacement. Let A be the event that no tag is drawn twice. Show that

Use the approximation \(1 − x ≈ e^{−x}\) for \(x ≈ 0\) to get

The law of rare events says that if

where \(X_i \sim \Bernoulli(p_i)\), then let \(Y \overset{d}{=} \Poisson(\mu)\) where \(\mu = \sum_{i=1}^n p_i\), then

Find

1. The appropriate values for the parameters \(p_i\), \(n, \, k, \mu\) that correctly represents our problem.
2. Identify $Y$ and the approximate the probability of the event $A$ you were asked to compute above using $Y$.
3. Identify the worst-case error in the law of rare events for this particular problem.

HW 11 due Apr 17

Problem 1

Consider the Markov chain on \(\{0,1\}\) whose transition probability matrix is

1. Verify that \((\pi_0,\pi_1) = (\beta/(\alpha + \beta), \alpha/(\alpha + \beta))\) is a stationary distribution.

2. Show that the first return distribution to state \(0\) is given by \(f_{00}^{(1)} = 1-\alpha\) and \(f_{00}^{(n)} = \alpha \beta (1 - \beta)^{n-2}\) for \(n=2,3,\ldots\).

3. Calculate the mean return time \(m_0 = \sum_{n=1}^{\infty} n f_{00}^{(n)}\) and verify that \(\pi_0 = 1/m_0\).

Problem 2

Let \(\{\alpha_i \colon i=1,2,\ldots \}\) be a probability distribution, and consider the Markov chain whose transition probability matrix is

What condition on the probability distribution \(\{\alpha_i \colon i=1,2,\ldots \}\) is necessary and sufficient in order that a limiting distribution exist, and what is this limiting distribution? Assume \(\alpha_1 > 0\) and \(\alpha_2 > 0\) so that chain is aperiodic.

Problem 3

Given the transition matrix

1. Draw a picture of the Markov chain, and determine the communication classes.

2. Determine the period(s) of the communication class(es).

3. Determine the limits as \(n \to \infty\) of \(P_{i0}^{(n)}\) for \(i=0,1,\ldots,4\).

Hint: Read section 4.5 of the textbook

Problem 4

A Markov chain on states 0,1,\ldots has transition probabilities

Find the stationary distribution.

Problem 5

1. Consider the reflected random walk on \(\mathbb{Z}^+\) (if you’re at state \(0\), then you go to state \(1\) with probability one), with probability \(p\) of going forward, and probability \(q\) of going backwards. If \(p > q\), determine the probability that you return to the origin.

2. Now with \(p < q\), determine the stationary distribution. What is the mean return time to \(k\) given that you start at state \(k\)? If start in state \(i \neq k\), let \(T_{ik}\) be the first hitting time of \(k\). Determine \(E[T_{ik}]\).

HW 10 due Apr 10

Problem 1

Consider the Markov Chain given by the transition matrix

on the state space $0,1,2,3$. Recall the first return pmf $f_{ii}^{(n)}$ defined in the textbook and in our notes. Using the equation

determine \(f^{(n)}_{00}\) for \(n = 1,2,3,4,5\).

Problem 2

Let \(X\) be a random variable taking values in the nonnegative integers. Show that

Problem 3

An individual either drives their car or walks in going from their home to their office in the morning, and from their office to their home in the afternoon. They use the following strategy: If it is raining in the morning, then they drive the car, provided it is at home to be taken. Similarly, if it is raining in the afternoon and their car is at the office, then they drive the car home. They walk on any morning or afternoon that it is not raining or if the car is not where they are. Assume that, independent of the past, it rains during successive mornings and afternoons with constant probability p. In the long run, on what fraction of days does our person walk in the rain? What if they own two cars?

Problem 4

1. Show that if $i$ and $j$ communicate, then $d(i) = d(j)$.

2. Show that a finite, irreducible, aperiodic Markov chain is recurrent.

Problem 5

Problem of the points.

Blaise Pascal and Pierre Fermat discussed the problem of the points in 1654. Suppose you are your friend have contributed a sum S to a pot of money, and are playing until the first person wins $n$ rounds of a game. Assume that it is a game of chance and skill, but you and your friend are equally likely to win each round. Suppose the game stops after $r$ rounds due to an earthquake. Then, you must divide the pot $S$ fairly based on the fact that you have won $a < n$ rounds, and your friend has won $r - a < n$. Give a formula for the amount of money you should get in terms of $r, n$ and $a$.

HW 9 due Apr 03

Problem 1

Suppose $\{a_i\}$ is a sequence such that $a_i \to a$ as $i \to \infty$. Show that the Cesaro mean converges to $a$ as well:

I expect an \(\epsilon-\delta\) proof

Problem 2

1. Determine the long run / limiting distribution for the Markov chain whose transition probability matrix is

   \[P = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac14 & \frac14 & \frac14 & \frac14 \\ 0 & 0 & \frac12 & \frac12 \end{pmatrix}\]
   Show solution.

   Solution: \(\begin{equation} \begin{split} \begin{pmatrix} \pi_{0}\\ \pi_{1}\\ \pi_{2}\\ \pi_{3} \end{pmatrix} &=(\pi_{0}\,\pi_{1}\, \pi_{2}\, \pi_{3}) \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \end{split} \end{equation}\)

   \(\Rightarrow \begin{pmatrix} \pi_{0}\\ \pi_{1}\\ \pi_{2}\\ \pi_{3} \end{pmatrix} =\begin{pmatrix} \frac{1}{10}\\ \frac{1}{10}\\ \frac{2}{5}\\ \frac{2}{5} \end{pmatrix}\)

2. (4.1.11) Suppose a production process changes states according to a Markov Process whose transition probability matrix is given by

   \[P = \begin{pmatrix} 0.3 & 0.5 & 0 & 0.2 \\ 0.5 & 0.2 & 0.2 & 0.1 \\ 0.2 & 0.3 & 0.4 & 0.1 \\ 0.1 & 0.2 & 0.4 & 0.3 \end{pmatrix}\]
   1. Determine the limiting probabilities \(\pi_0\) and \(\pi_3\) given that \(\pi_1 = 119/379\) and \(\pi_2 = 81/379\)
   2. Suppose that states \(0\) and \(1\) are “in-control” while states \(2\) and \(3\) are deemed out of control. In the long run, what fraction of time is the process out of control?
   3. In the long run, what fraction of transitions are from an in-control state to an out of control state.
   Show solution.

   Solution:

   From transition probability matrix

   \[\begin{cases} \pi_{2}=0.2\pi_{1}+0.4\pi_{2}+0.4\pi_{3}\\ \pi_{0}+\pi_{1}+\pi_{2}+\pi_{3}=1 \end{cases}\]

   $\Rightarrow \pi_{0}=\frac{117}{379}, \pi_{3}=\frac{62}{379}$

   $\pi_{2}+\pi_{3}=\frac{143}{379}$ is the fraction of time being out of control.

   \[\begin{equation} \begin{split} E&[\frac{1}{n}\sum_{k=1}^{n}1_{\{X_{k}=0, X_{k+1}=2\}}+1_{\{X_{k}=0, X_{k+1}=3\}} + 1_{\{X_{k}=1, X_{k+1}=2\}}+1_{\{X_{k}=1, X_{k+1}=3\}}]\\ =&\frac{1}{n}\sum_{k=1}^{n}\big(P\{X_{k}=0, X_{k+1}=2\}+P\{X_{k}=0, X_{k+1}=3\}+P\{X_{k}=1, X_{k+1}=2\}\\ &+P\{X_{k}=1, X_{k+1}=3\}\big) \end{split} \end{equation}\]

   Let $a_{k}=P\{X_{k}=0, X_{k+1}=2\}+P\{X_{k}=0, X_{k+1}=3\}+P\{X_{k}=1, X_{k+1}=2\}+P\{X_{k}=1, X_{k+1}=3\}$

   Then, using conditional probability, $a_{k}=p_{02}P(X_{k}=0)+p_{03}P(X_{k}=0)+p_{12}P(X_{k}=1)+p_{13}P(X_{k}=1)$

   \[\begin{equation} \underset{k\rightarrow\infty}{\lim}a_{k}=(p_{02}+p_{03})\pi_{0}+(p_{12}+p_{13})\pi_{1}=\frac{591}{3790} \end{equation}\]

   By problem 1.1, we have the answer be $\frac{591}{3790}$.

Problem 3

Consider the simple random walk with probabilities $p$ and $q = 1 - p$ of going forwards or backwards on the state space $1,2,3,4$. The states $1$ and $4$ are absorbing.

1. What are the communication classes?
2. What is the periodicity of each communication class?
3. Let \(T\) be a transient class in this system. Show that \(P^{(n)}_{ij} \to 0\) for any \(ij\) in the communication class.

Problem 4

(Problem 4.3.3 from K&P)

A Markov chain on states ${0, 1, 2, 3, 4, 5}$ has transition probability matrix

(a)

(b)

Find all communicating classes; which classes are transient and which are recurrent?

Problem 5

Show that a finite chain that is aperiodic and irreducible is regular and recurrent. I expect a proper proof from the definitions.

HW 8 due Mar 27

Problem 1

Suppose I start a chain on $N$ states with $P(X_0 = 1) = \pi_1,P(X_0 = 2) = \pi_2,\ldots$ where $\pi = (\pi_1,\ldots,\pi_N)$ is the stationary probability satisfying $\pi = \pi P$, where $P$ is a regular transition probability. What is $P(X_k = 1),P(X_k = 2),\ldots$ for fixed $k > 1$. Hint: first do the case $k=1$.

Problem 2

A Markov chain has transition probability

Determine its limiting distribution. What proportion of time, in the long run, does it spend in each state?

Problem 3

Consider the gambler’s ruin problem on states $0,\ldots,N$, with probability $p$ of going from state $i$ to $i+1$ and probability $q$ of going from state $i$ to $i-1$ for $i = 1,\ldots,N-1$. Fix a state $k$, where $0 < k < N$ and let $W_{ik}$ be the mean total visits to state $k$ starting from $i$ (before being absorbed into state $0$ or $N$. Formally, the definition is

where $1_A$ the usual indicator rv of the event $A$. Show that

for $i = 1,\ldots,N-1$ and $W_{0k} = W_{Nk} = 0$. $\delta_{ik}$ is $1$ if $k=i$ and $0$ otherwise.

You may try to solve the recurrence for fun!

Problem 4

Determine the following limits in terms of the transition probability matrix $P$ and the limiting distribution $\pi$ of a finite-state regular (irreducible, aperiodic) Markov chain \(\{X_n\}\).

1. \[\lim_{n \to \infty} \Prob( X_{n+1} = j | X_0 = i)\]
2. \[\lim_{n \to \infty} \Prob( X_{n} = k, X_{n+1} = j | X_0 = i)\]
3. \[\lim_{n \to \infty} \Prob( X_{n-1} = k, X_n = j | X_0 = i)\]

Problem 5

Consider a Markov chain with probability

Assume $0 < p_0 < 1$, $\sum_{i=0}^N p_i = 1$ and find the limiting distribution.

HW 7 due Mar 20

Problem 1

A Markov chain has the transition probability matrix

with states labeled \(\{0,1,2\}\). It starts in state \(X_0 = 0\). Eventually, the process will end up in state 2. What is the probability that the time \(T = \min\{n ≥ 0; X_n = 2\}\) is an odd number?

Problem 2

In a Markov chain, let \(S = \{1,\ldots,N\}\) and suppose \(\emptyset \neq A \subset S\) is the set of absorbing states. Consider \(T\) the absorption time \(T = \inf \{ k \geq 0 \colon X_k \in A \}\). Assume

We can prove this later, but we will have to make assumptions about the reachability of the absorbing states. Otherwise, we could be in a situation where from some state \(k\), it is just not possible to get absorbed and thus \(v_k = \infty\).

We are going to prove the equation we derived in class more rigorously. For all \(i \not\in A\)

where \(p_{ij} = P(X_{k+1} = j \vert X_k = i)\). In matrix form, this is the equation \(\vec{v} = \vec{1} + Q \vec{v}\).

1. Fix \(i \not\in A\). Show

   \[\begin{equation} v_i = \sum_{j=1}^\infty j P(T = j | X_0 = i) = \sum_{j=1}^\infty \sum_{k \in S} j P(T = j | X_1 = k) p_{ik} \label{eq:hw7 p2 1} \end{equation}\]

2. Show that for \(j = 1\), \(P(T = 1 \vert X_1 = k) = 1\) if \(k \in A\) and \(0\) otherwise.

3. Show that for $j > 1$, if $k \not\in A$

   \[\begin{align} P(T = j | X_1 = k) & = P(X_2 \not\in A, \ldots, X_{j-1} \not\in A, X_j \in A | X_1 = k) \nonumber \\ & = P(T = j-1 | X_0 = k) \label{eq:hw7 p2 2} \end{align}\]

   This lies at the heart of the proof, and the second equality should be shown using the Markov property and time homogeneity.

4. Substitute the result of \(\ref{eq:hw7 p2 2}\) into \(\ref{eq:hw7 p2 1}\) and show

   \[v_i = \sum_{k \not\in A} \sum_{j=1}^\infty (j+1) P(T = j | X_0 = k) p_{ik} + \sum_{k \in A} p_{ik},\]

   and use this to complete the proof.

Problem 3

An urn contains five tags, of which three are red and two are green. A tag is randomly selected from the urn and replaced with a tag of the opposite color. This continues until only tags of a single color remain in the urn. Let \(X_n\) denote the number of red tags in the urn after the nth draw, with \(X_0 = 3\). What is the probability that the game ends with the urn containing only red tags?

Problem 4

1. Consider the two state Markov chain with \(p_{01} = a\) and \(p_{10} = b\), where \(a,b \in (0,1)\). Let \(P\) be its transition matrix. Show (by induction) that

   \[P^n = \frac{1}{a+b} \begin{pmatrix} b & a \\ b & a \\ \end{pmatrix} + \frac{(1-a-b)^n}{a + b} \begin{pmatrix} a & -a \\ -b & b \\ \end{pmatrix}\]

2. Determine \(P^n\) for \(n=2,3,4,5\) whose transition probability matrix is

   \[P = \begin{pmatrix} 0.4 & 0.6 \\ 0.7 & 0.3 \\ \end{pmatrix}\]

   Also determine \(\lim_{n \to \infty} P^n\).

HW 6 due Mar 6

Problem 1 (Karlin and Pinsky, 4th ed, 3.1.12)

An MC has transition probability matrix

Find

1. $\Prob(X_2 = 1, X_1 = 1 | X_0 = 0)$
2. If the initial probability vector is $(\Prob(X_0 = 0), \Prob(X_0 = 1), \Prob(X_0 = 2)) = (\frac12,\frac13,\frac16)$, find $\Prob(X_2 = 1, X_1 = 1, X_0 = 0)$
3. Find the vector $(\Prob(X_2 = 0), \Prob(X_2 = 1), \Prob(X_2 = 2))$ using matrix multiplication (and the same initial probability vector from the previous part).

Problem 2

(3.1.1 from K and P) A simplified model for the spread of a disease (Covid-19) goes this way: The total population size is $N=5$, of which some are diseased (have the disease) and some are healthy. The selection is such that an encounter between any one pair of individuals is just as likely between any other pair. At any time instance, let us assume that only one pair of individuals (randomly, uniformly selected) may meet. If one of these individuals is diseased and the other is not, then the disease is transmitted from the diseased to the healthy with probability $\alpha = 0.1$. Otherwise, no disease transmission takes place. Let $X_n$ denote the number of diseased persons in the population at the end of the nth period. Specify the transition probability of this Markov chain.

Problem 3

(3.4.4 from K and P) A coin is tossed repeatedly until two successive heads appear. Find the mean number of tosses required.

1. Define a Markov chain (let \(X_n\) be the MC that represents the current number of successive heads that have appeared) and find its transition matrix.
2. Condition on the first-step and compute the mean number of tosses required.

Problem 4

1. A red die is rolled a single time. A green die is rolled repeatedly. The game stops at the first roll where the sum of the two dice is either 4 or 7. What is the probability that the game stops with a sum of 4?

2. Suppose the craps dice are shaved on faces 1-6 such that the pmf of a single die is

   \[\begin{align*} p(2) & = p(3) = p(4) = p(5) = 1/6 - \e\\ p(1) & = p(6) = 1/6 + 2\e \end{align*}\]

   where \(\e = 0.02\). Find the win probability using the method described in class (see also KP, section 2.2).

HW 5 Due Wed Feb 27

Problem 1

1. Let \(X\) have a Poisson distribution with parameter \(λ > 0\). Suppose $λ$ itself is random, following an exponential density with parameter θ.

   1. What is the marginal distribution of X?
   2. Determine the conditional density for λ given X = k.

2. Recall Polya’s Urn. An Urn contains a red balls and b green balls to start with. A ball is drawn at random and it is returned to the urn along with another ball of the same color. Let \(X_n\) be the fraction of red balls.

   1. Show that \(X_n\) is a martingale.
   2. Simulate \(X_n\) and draw graphs of its various realizations for different values of \(a\) and \(b\).
   3. Show, numerically, that \(X_n\) converges to some random number \(X_\infty\). Draw pictures of the distribution of \(X_\infty\) for different initial data \(a\) and \(b\).

3. Let \(\{\xi_i\}\) be iid \(\Bernoulli(p)\) rvs for \(0 < p < 1\). Let \(X_0 = 1\) and \(X_n = p^{-n} \prod_{i=1}^n \xi_i\). Show that $X_n$ defines a martingale, and determine the limit of $X_n$ as $X_n \to \infty$.

Problem 2

1. Suppose $X$, $Y$ and $Z$ are discrete random variables with a joint pmf \(p(u,v,w)\). Show that

   \[\E[ \E[X|Y,Z] | Z ] = \E[X | Z]\]

   Hint: Start with the definition \(\E[X|Y=y,Z=z] = \sum_x x\, p_{X|Y,Z}(x|y,z)\) where

   \[p_{X|Y,Z}(x|y,z) = \frac{p_{X,Y,Z}(x,y,z)}{p_{Y,Z}(y,z)}\]

2. If \(X_n\) is a martingale, compute \(\E[X_n]\) in terms of \(\E[X_1]\). Hint: Start with \(n=2\)

Problem 3

1. A Markov chain \(X_0, X_1, \ldots\) on states \(0,1,2\) has the transition probability matrix

   \[P = \begin{pmatrix} 0.1 & 0.2 & 0.7 \\ 0.9 & 0.1 & 0 \\ 0.1 & 0.8 & 0.1 \end{pmatrix}\]

   and initial distribution \(p_0 = \Pr\{X_0 = 0\} = 0.3, \quad p_1 = \Pr\{X_0 = 1\} = 0.4, \quad p_2 = \Pr\{X_0 = 2\} = 0.3.\)

   Determine \(\Pr\{X_0 = 0, X_1 = 1, X_2 = 2\}\).

2. A Markov chain $X_0, X_1, X_2, \ldots$ has the transition probability matrix

Determine the conditional probabilities \(\Pr\{X_2 = 1, X_3 = 1 \mid X_1 = 0\} \quad \text{and} \quad \Pr\{X_1 = 1, X_2 = 1 \mid X_0 = 0\}.\)

HW 4 Due Fri Feb 20

Problem 1

1. Suppose $N$ is a geometric random variable taking values in $0,1,\ldots$ with parameter $p$. (By inspecting the range of $N$, we see that $N$ is in fact, a shifted geometric random variable that represents the number of failures before the first success.) Let

   \[X = \sum_{i=1}^N \xi_i\]

   where ${\xi_i}$ are iid random variables independent of $N$, and $X = 0$ if $N = 0$. Find the cdf of $X$. Hint be mindful of the $N=0$ case. Assume that $\xi_i$ are iid $\Exponential(1)$. Simplify your answers as much as possible.

   Show solution.

   Solution:

   Let $f$ be the density function of $\xi_{i}$. Then

   \[f(z)= \begin{cases} e^{-z} & z\geq 0\\ 0 & z<0 \end{cases}\]

   We know that a sum of independent exponentials produces a Gamma distribution (see page 30, section 1.4.4 in Karlin and Pinsky 4th edition). Then, the density of

   \[X_k = \sum_{i=1}^k \xi_i\]

   is

   \[f_k(z)= \begin{cases} \frac{1}{(k-1)!} z^{k-1} e^{-z} & z\geq 0\\ 0 & z<0 \end{cases}\]

   Let us find the cdf of $X$.

   \[\begin{align*} P(X \leq t) & = \sum_{k=0}^{\infty} P(X \leq t| N = k) \Prob(N = k)\\ & = \sum_{k=0}^{\infty} P\left( \sum_{i=1}^k \xi_i \leq t| N = k \right) p(1-p)^{k}\\ & = P\left( 0 \leq t | N = 0 \right) p + \sum_{k=1}^{\infty} P\left( X_k \leq t | N = k \right) p(1-p)^{k} \end{align*}\]

   Now we have to divide the cases up into $t \geq 0$ and $t < 0$ since when $N = 0$, we have $X = 0$.

   \[P\left( X \leq t | N = 0 \right) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \\ \end{cases}\]

   Thus, the only interesting case is when $t \geq 0$; we have

   \[\begin{align*} F_X(t) & = p + \sum_{k=1}^{\infty} \int_0^{t} \frac{1}{(k-1)!} z^{k-1} e^{-z} dz \quad p(1-p)^{k} \\ & = p + p (1-p) \int_0^{t} \sum_{k=1}^{\infty} \frac{1}{(k-1)!} (z(1-p))^{k-1} e^{-z} dz \\ & = p + p (1-p) \int_0^{t} e^{z(1-p)} e^{-z} dz \\ & = p + p (1-p) \int_0^{t} e^{-p z)} dz \\ & = p + (1-p)(1 - e^{-pt }) \\ & = 1 - (1-p) e^{-pt } \end{align*}\]

   Putting this together, we get

   \(F_X(t) = \begin{cases} 0 & t < 0 \\ 1 - (1-p) e^{-pt } & t \geq 0 \\ \end{cases}\)

2. Suppose that three contestants on a quiz show are each given the same question and that each answers it correctly, independently of the others with probability $p$. But the difficulty of the question is itself a random variable. So suppose that $p$ is $\Uniform(0,1)$. What is the probability that exactly two of the contestants answer correctly?

   Show solution.

   Solution:

   Let $X$ be the number of contestants answering question correctly.

   \(\begin{equation} \begin{split} P(X=2) &=\int_{0}^{1}P(X=2\mid p=u)f_p(u) du\\ &=\int_{0}^{1}\binom{3}{2}u^{2}(1-u)du\\ &=\frac{1}{4} \end{split} \end{equation}\)

3. Let $X_0,X_1,\ldots$ be iid nonnegative random variables having a pdf. Let $N$ be the first index $k$ for which $X_k > X_0$. $N = 1$ if $X_1 > X_0$, $N=2$ if $X_1 \leq X_0, X_2 > X_0$ and so on. Find the pmf for $N$ and the mean $\E[N]$. (Interpretation: $X_0$, $X_1$ are bids on the car you are trying to sell. $N$ is the index of the first bid that is better than the initial bid.)

   Show solution.

   Solution:

   Let $f$ and $F$ be pdf and cdf of $X_{i}$ resp.

   \[\begin{equation} \begin{split} P(N & = 1) = P(X_{i}\leq X_{0}) \\ &=\int_{0}^{\infty}dx_{i}\int_{x_{i}}^{\infty}dx_{0}f(x_{i})f(x_{0})\\ &=\int_{0}^{\infty}dx_{i}f(x_{i})(1-F(x_{i}))\\ &=1-\frac{1}{2}F(x_{i})^{2}\biggr\rvert_{0}^{\infty}\\ &=\frac{1}{2} \end{split} \end{equation}\]

   Before we tackle the general $k \geq 2$ case, we define the tail of $X_0$ as $T(t) = 1 - F(t)$, and hence $dT = T’(t)dt = -f(t)dt$.

   \[\begin{equation} \begin{split} P(N & = k) = P(X_{1}\leq X_{0}, \cdots, X_{k-1} \leq X_0, X_k > X_0) \\ &=\int_{0}^{\infty} P(X_{1}\leq t , \cdots, X_{k-1} \leq t , X_k > t|X_0 = t) f(t) dt\\ &=\int_{0}^{\infty} F^{k-1}(t) T(t) f(t) dt\\ & = \int_0^{\infty} F^{k-1} (1 - F) f dt \\ & = \int_0^{\infty} F^{k-1} f - \int_0^{\infty} F^{k} f dt \\ & := I_k - I_{k-1} \\ \end{split} \end{equation}\]

   where we have used the last line to define the integral $I_k$. Now we evaluate the integral $I_k$ for $k \geq 2$ using integration by parts (when in doubt…).

   \[\begin{align*} I_k & = \int_0^{\infty} F^{k-1} f dt \\ & = - \int_0^{\infty} F^{k-1} T'(t) dt \\ & = - F^{k-1} T(t) \big|_{0}^{\infty} + \int_0^{\infty} (k-1)F^{k-2} T dt \\ & = \int_0^{\infty} (k-1)F^{k-2} (1-F) dt \\ & = (k-1)\int_0^{\infty} F^{k-2} dt - (k-1)\int_0^{\infty} F^{k-1} dt \\ & = (k-1) I_{k-1} - (k-1)I_{k} dt \end{align*}\]

   and therefore moving the $(k-1)T_k$ to the LHS of the above equation

   \[\begin{align*} I_k & = \frac{(k-1)}{k} I_{k-1} \\ & = \frac{(k-1)}{k} \frac{k-2}{k-1} I_{k-2} \\ & = \frac{(k-1)}{k} \frac{k-2}{k-1} \cdots I_{1} \\ & = \frac{1}{k} \\ \end{align*}\]

   where we have used $I_1 = 1$. Thus

   \[\begin{align*} \Prob(N = k) & = \frac{1}{k} - \frac{1}{(k+1)}\\ & = \frac{1}{k(k+1)} \quad k \geq 2 \end{align*}\]

   Then,

   \(\begin{equation} \begin{split} E[N] &=\sum_{k=1}^{\infty}kP(N=k)\\ &= \frac12 + \sum_{k=2}^{\infty} \frac{1}{k+1} &= \infty \end{split} \end{equation}\)

Problem 2

1. Suppose $X_1,X_2,X_3$ are discrete random variables. Show that

   \[\E[ X_1 + X_2 + X_3 | X_2, X_3 ] = X_2 + X_3 + \E[X_1 | X_2,X_3]\]
   Show solution.

   Solution:

   \[\begin{equation} \begin{split} E[X_{1}+X_{2}+X_{3}\mid X_{2}, X_{3}] &=E[X_{1}\mid X_{2},X_{3}]+E[X_{2}\mid X_{2},X_{3}]+E[X_{3}\mid X_{2},X_{3}]\\ &=X_{2}+X_{3}+E[X_{1}\mid X_{2},X_{3}] \end{split} \end{equation}\]

   We could also do this the following way for discrete $X_1,X_2$ and $X_3$. The conditional expectation is a FUNCTION of the values of $X_2,X_3$. So

   \[\begin{align*} \E[ X_1 + X_2 + X_3 | X_2 = a_2,X_3 = a_3] & = \sum_{a_1} (a_1 + a_2 + a_3) p_{X_1|X_2,X_3}(a_1 | a_2,a_3)\\ & = \sum_{a_1} (a_1 + a_2 + a_3) \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)}\\ & = \sum_{a_1} a_1 \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)} \\ & \qquad + \sum_{a_1} (a_2 + a_3) \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)}\\ & = \E[ X_1 | X_2 = a_2,X_3 = a_3] + a_2 + a_3 \end{align*}\]

   using the law of total probability in the last equation.

Problem 3

Conditional dependence and independence.

1. Draw \(P\) uniformly randomly from \([0,1]\) and then draw two independent samples \(X_1\) and \(X_2\) with \(\Bernoulli(P)\) distribution. Compute the (unconditional) correlation coefficient of \(X_1\) and \(X_2\).

2. Let \(N\) have a Poisson distribution with parameter \(\lambda > 0\). Suppose that, conditioned on \(N\), we draw a random variable \(X \sim \Binomial(N,p)\). Let \(Y = N - X\). Show that \(X\) and \(Y\) have Poisson distributions with parameters \(\lambda p\) and \((1 - p) \lambda\) and that \(X\) and \(Y\) are independent.

3. Let \(X\) and \(Y\) be jointly distributed random variables whose joint probability mass function is given in the following table:

Show that the covariance between \(X\) and \(Y\) is zero even though \(X\) and \(Y\) are not independent.

HW 3 Due Fri Feb 13

Problem 1: Conditional probability redux.

Let $A_0, A_1, \ldots, A_n$ be events of non-zero probability.

Show that \(\Prob\left(\bigcap_{i=0}^n A_i \right) = \Prob(A_0) \Prob(A_1 | A_0) \cdots \Prob(A_n | A_{n-1} \ldots A_0 )\)

Problem 2

1. Suppose $U$ has uniform distribution in $[0,1]$. Derive the density function for the random variables

   1. $Y = \log(1-U)$
   2. $W_n = U^n$ for $n \geq 1$

   Refer to section 1.2.6 in your textbook.

   Show solution.

   Solution: 1. As $U\sim \Uniform[0,1]$, range of $Y$ is [-$\infty$,0]. Given $t\leq 0$,

   \[\begin{equation} \begin{split} F_{Y}(t) &=P(Y\leq t)=P(\log(1-U)\leq t)\\ &=P(1-U\leq e^{t})=P(1-e^{t}\leq U)\\ &=e^{t} \end{split} \end{equation}\]

   So $f_{Y}(t)=e^{t}$ for $t\in[-\infty,0]$ and zero elsewhere.

   2. Range of $W_{n}$ is $[0,1]$. Given $t\in[0,1]$, \begin{equation} \begin{split} P(W_{n}\leq t) &=P(U^{n}\leq t)
   &=P(U\leq t^{1/n})
   &=t^{\frac{1}{n}} \end{split} \end{equation} So $f_{W_{n}}(t)=\frac{1}{n}t^{\frac{1}{n}-1}$ for $t\in[0,1]$ and zero elsewhere.

2. If $X \sim \Exponential(2)$ and $Y \sim \Exponential(3)$ are independent, find $\Prob(X < Y)$.

   Show solution.

   Solution: \(\begin{equation} \begin{split} P(X<Y) &=\int_{0}^{\infty}\int_{x}^{\infty}2e^{-2x}3e^{-3y}dydx &=\int_{0}^{\infty}2e^{-2x}\int_{x}^{\infty}3e^{-3y}dydx &=\frac{2}{5} \end{split} \end{equation}\)

Problem 3

1. A quarter is tossed repeatedly until a head appears. Let $N$ be the trial number when this head appears. Then, a dollar is tossed $N$ times. Let $X$ count the number of times the dollar comes up tails. Determine $P(X = 0)$, $P(X = 1)$ and $\E[X]$.

   Show solution.

   Solution:

   \[\begin{equation} \begin{split} P(X=0) &=\sum_{n=1}^{\infty}P(N=n, \text{no tails}) \\ &=\sum_{n=1}^{\infty}\big(\frac{1}{2}\big)^{n}\binom{n}{0}\big(\frac{1}{2}\big)^{n} \\ &= \frac{1}{4} \sum_{n=0}^{\infty}\big(\frac{1}{4}\big)^{n} \\ &= \frac{1}{4} \frac{1}{1-1/4} \end{split} \end{equation}\] \[\begin{equation} \begin{split} P(X=1) &=\sum_{n=1}^{\infty}P(N=n, \text{one tail})\\ &=\sum_{n=1}^{\infty}\big(\frac{1}{2}\big)^{n}\binom{n}{1}\big(\frac{1}{2}\big)^{n}\\ &=\sum_{n=0}^{\infty}\frac{n}{4^{n}} \end{split} \end{equation}\]

   Let $S=P(X=1)$, then

   \[\begin{equation} S-\frac{1}{4}S=\sum_{n=0}^{\infty}\big(\frac{1}{4}\big)^{n}=\frac{1}{3} \end{equation}\]

   So $S=P(X=1)=\frac{4}{9}$. The way to see this in general is by defining

   \[\begin{align} S(p) & = \sum_{n=0}^{\infty} p^n = \frac{1}{1-p} \\ S'(p) & = \sum_{n=0}^{\infty} n p^{n-1} = \frac{1}{(1-p)^2}\\ p S'(p) & = \sum_{n=0}^{\infty} n p^n = \frac{p}{(1-p)^2} \end{align}\]

   and then substituting $p=1/4$.

   \[\begin{equation} \begin{split} E[X] &=\sum_{k=0}^{\infty}kP(X=k)=\sum_{k=0}^{\infty}\sum_{n=1}^{\infty}P(X=k\mid N=n)P(N=n)\\ &=\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}k\binom{n}{k}\big(\frac{1}{2}\big)^{n}P(N=n)\\ &=\sum_{n=1}^{\infty}P(N=n)\sum_{k=0}^{\infty}k\binom{n}{k}\big(\frac{1}{2}\big)^{n}\\ &=\sum_{n=1}^{\infty}\frac{n}{2}P(X=n)=\frac{1}{2}\sum_{n=1}^{\infty}nP(N=n)\\ &= \frac12 E[N] = 1 \end{split} \end{equation}\]

   since $N$ is geometric taking values in ${1,\ldots,}$ and wikipedia says $E[N] = 1/p$.

   In the above we used that if $Z \sim \Binomial(n,p)$ then

   \[\E[Z] = \sum_{k=0}^n k \binom{n}{k} p^{n-k} (1-p)^k = n p\]

   with $p = 1/2$.

2. Let $N$ have Poisson distribution with parameters $\lambda = 1$. Conditioned on $N = n$, let $X$ have a uniform distribution over the integers $0,1,\ldots,n+1$. What is the marginal distribution for $X$?

   Show solution.

   Solution

   \[\begin{equation} \begin{split} P(X=t) &=\sum_{n=0}^{\infty}P(X=t\mid N=n)P(N=n)\\ &=e^{-1}\sum_{n=0}^{\infty}\frac{1}{n+2}\frac{1}{n!}=e^{-1} \sum_{n=t-1}^{\infty}\frac{n+1}{(n+2)!}\\ &=e^{-1}\sum_{n=t+1}^{\infty}\frac{n-1}{n}=e^{-1}\sum_{n=t+1}^{\infty}\frac{n-1}{n!}\\ &=e^{-1}\sum_{n=t+1}^{\infty}\frac{1}{(n-1)!}-e^{-1}\sum_{n=t+1}^{\infty}\frac{1}{n!}\\ &=\frac{e^{-1}}{t!} \end{split} \end{equation}\]

Problem 4

Do men have more sisters than women have? In a certain society, all married couples use the following strategy to determine the number of children that they will have: If the first child is a girl, they have no more children. If the first child is a boy, they have a second child. If the second child is a girl, they have no more children. If the second child is a boy, they have exactly one additional child. (We ignore twins, assume sexes are equally likely, and the sex of distinct children are independent random variables, etc.)

1. What is the probability distribution for the number of children in a family?
2. What is the probability distribution for the number of girl children in a family?
3. A male child is chosen at random from all of the male children in the population. What is the probability distribution for the number of sisters of this child?
4. What is the probability distribution for the number of brothers of the previously chosen male child?

HW 2 Due Fri Feb 6

1. Let $N$ cards carry the distinct numbers $x_1,\ldots,x_N$. If two cards are drawn at random without replacement, show that the correlation coefficient $\rho$ between the numbers appearing on the two cards in $-1/(N-1)$.

   Show solution.

   Solution

   Let $X,Y$ be the number on the first and second cards respectively.

   \[\begin{equation} \begin{split} P(X=x_{n}) &=\frac{1}{N}\\ P(Y=x_{n}) &=P(X=x_{n}, Y=x_{n})+P(X\neq x_{n}, Y=x_{n})=P(X\neq x_{n}, Y=x_{n})\\ & =P(Y=x_{n}\mid X\neq x_{n})\times P(X\neq x_{n}) =\frac{1}{N-1}\times \frac{N-1}{N}\\ &=\frac{1}{N} \end{split} \end{equation}\]

   So we know $E[X]=E[Y]$ and $E[X^{2}]=E[Y^{2}]$. Let

   \[\begin{equation} T=\sum_{n=1}^{N}x_{n} \quad \text{ and } \quad S=\sum_{n=1}^{N}x_{n}^{2} \end{equation}\]

   Then

   \[\begin{equation} \begin{split} E[X]&=E[Y]=\sum_{n=1}^{N}x_{n}P(X=x_{n})=\frac{T}{N}\\ E[X^{2}]&=E[Y^{2}]=\sum_{n=1}^{N}x_{n}^{2}P(X=x_{n})=\frac{S}{N} \end{split} \end{equation}\]

   We know $\rho$ is the correlation coefficient of $X,Y$, the formula of $\rho$ is

   \[\begin{equation} \rho=\frac{\Cov(X,Y)}{\sigma_{X}\sigma_{Y}} =\frac{\E[XY]-\E[X]\E[Y]}{\sigma_{X}\sigma_{Y}} \end{equation}\]

   Also we know

   \[\begin{equation} \sigma_{X}=(\E[X^{2}]-(\E[X])^{2})^{\frac{1}{2}}=(\frac{S}{N}-\frac{T^{2}}{N^{2}})^{\frac{1}{2}}=\sigma_{Y} \end{equation}\]

   There is only $\E[XY]$ left to compute.

   \[\begin{equation} \begin{split} \E[XY] &=\sum_{n=1}^{N}x_{n}\sum_{m\neq n}x_{m}P(X=x_{n}, Y=x_{m})\\ &=\sum_{n=1}^{N}x_{n}(T-x_{n})\frac{1}{N-1}\frac{1}{N}\\ &=\frac{T^{2}}{N(N-1)}-\frac{S}{N(N-1)} \end{split} \end{equation}\]

   Bring those formulas back to $\rho$, we have

   \(\begin{equation} \rho=-\frac{1}{N-1} \end{equation}\)

2. Let $X$ and $Y$ be independent random variables having distribution $F_x$ and $F_y$ respectively.

   • Let $Z = \max(X,Y)$. Express $F_Z(t)$ in terms of $F_X(s)$ and $F_Y(u)$.
   • Let $W = \min(X,Y)$. Express $F_W(t)$ in terms of $F_X(s)$ and $F_Y(u)$.
   Show solution.

   Solution

   Since $X,Y$ are independent,

   \[\begin{equation} \begin{split} F_{Z}(t) &=P(Z\leq t)\\ &=P(\max(X,Y)\leq t)\\ &=P(X\leq t, Y\leq t)\\ &=P(X\leq t)P(Y\leq t)\\ &=F_{X}(t)F_{Y}(t)\\ \end{split} \end{equation}\]

   since if $\max(X,Y) \leq t$ then both $X \leq t$ and $Y \leq t$.

   \(\begin{equation} \begin{split} F_{W}(t)&=P(W\leq t)\\ &=1-P(W>t)\\ &=1-P(\min(X,Y)>t)\\ &=1-P(X>t, Y>t)\\ &=1-P(X>t)P(Y>t)\\ &=1-(1-P(X\leq t))(1-P(Y\leq t))\\ &=1-(1-F_{X}(t))(1-F_{Y}(t))\\ &=F_{X}(t)+F_{Y}(t)-F_{X}(t)F_{Y}(t) \end{split} \end{equation}\)

3. Let $U$ be Poisson distributed with parameter $\lambda$ and let $V = 1/(1+U)$. Find the expected value of $V$.

   Show solution.

   Solution

   \[U\sim\Poisson(\lambda), \qquad P(U=k)=\frac{\lambda^{k}e^{-\lambda}}{k!}\]

   \(\begin{equation} \begin{split} \E[V] &=\sum_{k=0}^{\infty}\frac{1}{1+k} P(U=k) \\ &=\sum_{k=0}^{\infty}\frac{1}{1+k}\frac{\lambda^{k}e^{-\lambda}}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{1}{1+k}\frac{\lambda^{k}}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k}}{(k+1)!}\\ &=\frac{e^{-\lambda}}{\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k+1}}{(k+1)!}\\ &=\frac{e^{-\lambda}}{\lambda}\sum_{n=1}^{\infty}\frac{\lambda^{k}}{k!}\\ &=\frac{e^{-\lambda}}{\lambda}(e^{\lambda}-1)\\ &=\frac{1-e^{-\lambda}}{\lambda} \end{split} \end{equation}\)

4. Let $U,V,W$ be independent random variables with equal variances $\sigma^2$. Let $X = U + V$ and let $Y = V - W$. Find the covariance of $X$ and $Y$.

   Show solution.

   Solution

   As $U,V,W$ are independent,

   \[\begin{equation} \begin{split} \Cov(X,Y) &=\E[XY]-\E[X]\E[Y]\\ &=\E[(U+V)(V-W)]-\E[U+V]\E[V-W]\\ &=\E[UV+V^{2}-UW-VW]-(\E[U]\E[V]+(\E[V])^{2}-\E[U]\E[W]-\E[V]\E[W])\\ &=\E[V^{2}]-(\E[V])^{2}\\ &=\Var(V)\\ &=\sigma^{2} \end{split} \end{equation}\]

   Another way to do this is to write

   \(\begin{align} \Cov(X,Y) & = \Cov(U,V) - \Cov(U,W) + \Cov(V,V) \\ & \qquad + \Cov(U,V) - \Cov(U,W) + \Cov(V,V) + \Cov(V,W) \\ & = 0 - 0 + \Var(V) + 0 \\ & = \sigma^2 \end{align}\)

HW 1 Due Fri Jan 30

Homework will be a little long this week, since it is mostly review.

1. Which topics about stochastic processes most interest you? (E.g. Brownian motion, stock market modeling, statistical physics, etc.)

2. Given that you end up working as hard as you think you can work this semester, what grade do think you can achieve in this class? How important is the letter grade to you?

3. Would you prefer a challenging class at a good pace so you can learn a lot, or do you prefer an easy-paced class so you can learn the material well?

4. Of the three general principles for modeling probabilities stated in K&P chapter 1, in your opinion, which one applies best to the modeling of share prices in the stock market?

5. Plot the distribution function

   \[F(x) = \begin{cases} 0 & x \leq 0\\ x^3 & 0 < x < 1\\ 1 & x \geq 1 \end{cases}\]
   • Determine the corresponding density function $f(x)$ in the three regions.
   • What is the mean of the distribution?
   • If $X$ is a random variable with distribution $F$, then evaluate $\mathbb{P}(1/4 \leq X \leq 3/4)$.
   Show solution.

   Solution: The density function is

   \[f(x)= \begin{cases} 3x^{2} & 0<x<1 \\ 0 & \text{otherwise} \end{cases}\] \[\begin{equation*} E(X)=\int_{0}^{1}xf(x)dx=\frac{3}{4} \end{equation*}\]

   \(\begin{equation*} P(\frac{1}{4}\leq X \leq \frac{3}{4})=F(\frac{3}{4})-F(\frac{1}{4})=\frac{13}{32} \end{equation*}\)

6. Determine the distribution function, mean and variance corresponding to the triangular density

   \(f(x) = \begin{cases} x & 0 \leq x \leq 1,\\ 2 - x & 1 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}\)

   Show solution.

   Solution

   \[F(x)= \begin{cases} 0 & x<0 \\ \frac{1}{2}x^{2} & 0\leq x \leq1 \\ -\frac{1}{2}x^{2}+2x-1 & 1<x\leq 2 \\ 1 & x>2 \end{cases}\] \[\begin{equation*} E[X]=1 \end{equation*}\]

   \(\begin{equation*} \text{Var}(X)=\frac{1}{6} \end{equation*}\)

7. Let $1_{A}$ be the indicator random variable associated with an event $A$, defined to be one if $A$ occurs, and zero otherwise. Show

   • $1_{A^c} = 1 - 1_{A}$
   • $1_{A \cap B} = 1_{A} 1_{B} = \min( 1_{A}, 1_{B} )$
   • $1_{A \cup B} = \max( 1_{A}, 1_{B} )$.
   Show solution.

   Solution:

   If $\omega \in A^{c}$, then

   \[\begin{equation*} 1_{A^{c}}(\omega)=1=1-1_{A}(\omega) \end{equation*}\]

   If $\omega \in A$, then

   \[\begin{equation*} 1_{A}(\omega)=1=1-1_{A^{c}}(\omega) \end{equation*}\]

   So $1_{A^{c}}=1-1_{A}$

   If $\omega\in A\cap B$, then

   \[\begin{equation*} 1_{A\cap B}(\omega)=1=1_{A}1_{B}(\omega) \end{equation*}\]

   If $\omega \notin A\cap B$, then $\omega$ is either in $X-(A\cup B)$, $A-B$ or $B-A$, we have

   \[\begin{equation*} 1_{A\cap B}(\omega)=0=1_{A}1_{B}(\omega) \end{equation*}\]

   So

   \[1\_{A\cap B}=1_{A}1_{B}\]

   Note that $\min(1_{A},1_{B})$ has only two possible values ${1,0}$.

   \[\begin{equation*} \min(1_{A},1_{B})(\omega)=1 \Longleftrightarrow 1_{A}(\omega)=1=1_{B}(\omega) \Longleftrightarrow \omega\in A\cap B \end{equation*}\]

   So

   \[\begin{equation*} 1_{A\cap B}=1_{A}1_{B}=\min(1_{A},1_{B}) \end{equation*}\]

   Note that $\max(1_{A},1_{B})$ has only two possible values ${1,0}$.

   \(\begin{equation*} \omega\in A\cup B\Longleftrightarrow 1_{A}(\omega)=1\text{ or }1_{B}(\omega)=1 \Longleftrightarrow \max(1_{A},1_{B})(\omega)=1 \end{equation*}\)

8. A pair of dice is tossed. If the two outcomes are equal, the dice are tossed again, and the process is repeated. If the dice is unequal, their sum is recorded. Determine the probability mass function for the sum.

   Show solution.

   Solution:

   There are only 30 possible outcomes. Among 30 outcomes, there are 2 outcomes of having 3, 4, 10 and 11 as their sum. 4 outcomes of having 5,6,8 and 9 as sum. And 6 outcomes of having 7 as sum.
   Define $X$= sum of two dice. Then sample space is ${3,4,5,6,7,8,9,10,11}$ And we have the following probability

   \(\begin{equation*} \begin{split} P(X=3) &=P(X=4)=P(X=10)=P(X=11)=\frac{1}{15}\\ P(X=5) &=P(X=6)=P(X=8)=P(X=9)=\frac{2}{15}\\ P(X=7) &=\frac{3}{15} \end{split} \end{equation*}\)