/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Consider the following process. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability \(1 / 2\) ). During the rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us. (a) Given that she reports a head on the \(n\) th toss, what is the probability that a head is thrown on the \((n+1)\) st toss? (b) Consider this process as having two states, heads and tails. By computing the other three transition probabilities analogous to the one in part (a), write down a "transition matrix" for this process. (c) Now assume that the process is in state "heads" on both the \((n-1)\) st and the \(n\) th toss. Find the probability that a head comes up on the \((n+1)\) st toss. (d) Is this process a Markov chain?

Short Answer

Expert verified
(a) \(\frac{5}{6}\); (b) Transition Matrix: \[ \begin{bmatrix} \frac{5}{6} & \frac{1}{2} \\ \frac{1}{6} & \frac{1}{2} \end{bmatrix} \]; (c) Probability: \(\frac{5}{6}\); (d) Yes, it is a Markov Chain.

Step by step solution

01

Understand the Coin Choice

We have two coins: a fair coin (FH) with one head and one tail, and a biased coin (HH) with two heads. Our friend chooses one of these coins at random, with a probability of \(\frac{1}{2}\) for each coin.
02

Calculate Probability for Part (a)

If a head is shown on the \(n\)-th toss, it could be because either the fair coin showed heads or the biased coin was picked. The probability of getting a head when the fair coin shows heads is \(\frac{1}{2}\), and when she picks the biased coin, it's 1. The probability that the chosen coin was the biased one given a head was tossed can be computed using Bayes' theorem. The probability of the next toss being a head, given the previous toss was a head, is \(P(H_{n+1} | H_n) = \frac{1}{3} \times \frac{1}{2} + \frac{2}{3} \times 1 = \frac{5}{6}\).
03

Transition Probabilities for Part (b)

Consider the states as "Heads" and "Tails". From "Heads", the transition to "Heads" (HH) is \(\frac{5}{6}\) (calculated in Step 2) and transition to "Tails" (HT) is \(\frac{1}{6}\). If the state is "Tails", you can transition to "Tails" (TT) \(\frac{1}{2}\) and to "Heads" (TH) \(\frac{1}{2}\) using only the fair coin. So, the transition matrix \(T\) is: \[ T = \begin{bmatrix} \frac{5}{6} & \frac{1}{2} \ \frac{1}{6} & \frac{1}{2} \end{bmatrix} \]
04

Probability of Heads on Two Consecutive Tosses for Part (c)

Given state "Heads" for the \((n-1)\)-th and \(n\)-th tosses, the probability the \((n+1)\)-th toss is a head is just the HH transition probability, \(\frac{5}{6}\). This does not depend on the \((n-1)\)-th if \(n\)-th is "Heads."
05

Determine if it's a Markov Chain for Part (d)

A Markov Chain is defined by the property that the future state depends only on the current state, not any previous states. The transition probabilities in this process depend only on the current state, so this process can be considered a Markov Chain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transition Matrix
When discussing Markov chains, a transition matrix is a key concept that succinctly represents the probabilities of moving from one state to another in a process. In the exercise, there are two states: "Heads" and "Tails."
To form a transition matrix, we need to calculate transition probabilities such as the probability of remaining in the "Heads" state or moving to the "Tails" state after tossing the coins.
For example, if the current state is "Heads", the probability of tossing a "Head" again is calculated as the sum of choosing a fair coin showing heads or a biased coin. Similarly, transitions from "Tails" are dependent on using only the fair coin, as the biased coin can't show tails.
  • The transition from "Heads" to "Heads" is \( \frac{5}{6} \).
  • The transition from "Heads" to "Tails" is \( \frac{1}{6} \).
  • The transition from "Tails" to "Heads" is \( \frac{1}{2} \).
  • The transition from "Tails" to "Tails" is \( \frac{1}{2} \).
This can be assembled into a matrix notation, which reveals the process dynamics: \[ T = \begin{bmatrix} \frac{5}{6} & \frac{1}{2} \ \frac{1}{6} & \frac{1}{2} \end{bmatrix} \] Each entry in the matrix corresponds to the probability of moving from one state to another, encapsulating the entire transition regime of the process.
Probability
Probability plays a fundamental role in understanding the behavior of Markov chains. In this particular exercise, probability is used to determine the chance of observing a specific outcome based on prior events.
To solve these problems, knowing how to transition between states by applying probability theory is crucial. For example, the probability of going from "Heads" on the nth toss to another "Heads" on the \( (n+1) \)-th toss can be derived using known transitional states and their probabilities.
In essence, we deal with conditional probabilities that inform us on likely outcomes conditional upon previous events. By comprehending these probabilities, we can predict future states of the process with better accuracy.
Bayes' Theorem
Bayes' Theorem provides a way to compute conditional probabilities and is an integral part of the problem-solving framework. It allows us to reverse the conditions of probability statements and is vital when analyzing outcomes in processes like tossing coins.
In the exercise, after a "Head" result, Bayes' Theorem helps us compute the likelihood that a biased coin was chosen. It adjusts probabilities based on reported outcomes and the nature of the specific coin being used.
  • This theorem lets us reassess initial conditions (such as which coin might have been chosen) once new evidence (like seeing a "Head") is available.
  • It provides deeper insights into the likelihoods involved, for example, after observing a head, one might reconsider the chances of it having come from the biased coin.
Applying Bayes' Theorem in these contexts equips us with a powerful way to approach complex probabilistic problems.
States in Processes
In the context of Markov chains, understanding states is crucial. A state represents a specific condition or position of the system at a given time. In the exercise provided, states can be labeled as "Heads" or "Tails" based on coin toss outcomes.
These states help in tracking the progression of the random process. They allow for the quantification and prediction of the system's behavior over time.
  • A process must account for each state's occurrences and movements to accurately predict future states.
  • States in a Markov chain must meet the condition of memorylessness, where transition to next state relies only on the current state.
Identifying these states is essential for defining how the process transitions over time, and for setting up the corresponding transition matrix, which defines the shifting likelihoods between these states.

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Most popular questions from this chapter

Consider the game of tennis when deuce is reached. If a player wins the next point, he has advantage. On the following point, he either wins the game or the game returns to deuce. Assume that for any point, player A has probability .6 of winning the point and player B has probability .4 of winning the point. (a) Set this up as a Markov chain with state 1: A wins; 2: B wins; 3: advantage \(\mathrm{A} ; 4:\) deuce; 5 : advantage \(\mathrm{B}\). (b) Find the absorption probabilities. (c) At deuce, find the expected duration of the game and the probability that \(\mathrm{B}\) will win.

Find the matrices \(\mathbf{P}^{2}, \mathbf{P}^{3}, \mathbf{P}^{4},\) and \(\mathbf{P}^{n}\) for the Markov chain determined by the transition matrix \(\mathbf{P}=\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) .\) Do the same for the transition matrix \(\mathbf{P}=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right) .\) Interpret what happens in each of these processes.

Here is an elegant method due to Guibas and Odlyzko \(^{10}\) to obtain the expected time to reach a pattern, say \(\mathrm{HTH},\) for the first time. Let \(f(n)\) be the number of sequences of length \(n\) which do not have the pattern HTH. Let \(f_{p}(n)\) be the number of sequences that have the pattern for the first time after \(n\) tosses. To each element of \(f(n),\) add the pattern HTH. Then divide the resulting sequences into three subsets: the set where HTH occurs for the first time at time \(n+1\) (for this, the original sequence must have ended with \(\mathrm{HT}\) ); the set where HTH occurs for the first time at time \(n+2\) (cannot happen for this pattern); and the set where the sequence HTH occurs for the first time at time \(n+3\) (the original sequence ended with anything except \(\mathrm{HT}\) ). Doing this, we have $$ f(n)=f_{p}(n+1)+f_{p}(n+3) $$ Thus, $$ \frac{f(n)}{2^{n}}=\frac{2 f_{p}(n+1)}{2^{n+1}}+\frac{2^{3} f_{p}(n+3)}{2^{n+3}} $$ If \(T\) is the time that the pattern occurs for the first time, this equality states that $$ P(T>n)=2 P(T=n+1)+8 P(T=n+3) $$ Show that if you sum this equality over all \(n\) you obtain $$ \sum_{n=0}^{\infty} P(T>n)=2+8=10 $$ Show that for any integer-valued random variable $$ E(T)=\sum_{n=0}^{\infty} P(T>n) $$ and conclude that \(E(T)=10 .\) Note that this method of proof makes very clear that \(E(T)\) is, in general, equal to the expected amount the casino pays out and avoids the martingale system theorem used by Li.

Consider an absorbing Markov chain with state space \(S .\) Let \(f\) be a function defined on \(S\) with the property that $$ f(i)=\sum_{j \in S} p_{i j} f(j) $$ or in vector form $$ \mathbf{f}=\mathbf{P f} $$ Then \(f\) is called a harmonic function for \(\mathbf{P}\). If you imagine a game in which your fortune is \(f(i)\) when you are in state \(i,\) then the harmonic condition means that the game is fair in the sense that your expected fortune after one step is the same as it was before the step. (a) Show that for \(f\) harmonic $$ \mathbf{f}=\mathbf{P}^{n} \mathbf{f} $$ for all \(n\). (b) Show, using (a), that for \(f\) harmonic $$ \mathbf{f}=\mathbf{P}^{\infty} \mathbf{f} $$ where $$ \mathbf{P}^{\infty}=\lim _{n \rightarrow \infty} \mathbf{P}^{n}=\left(\begin{array}{c|c} \mathbf{0} & \mathbf{B} \\ \hline \mathbf{0} & \mathbf{I} \end{array}\right) $$ (c) Using (b), prove that when you start in a transient state \(i\) your expected final fortune $$ \sum_{k} b_{i k} f(k) $$ is equal to your starting fortune \(f(i)\). In other words, a fair game on a finite state space remains fair to the end. (Fair games in general are called martingales. Fair games on infinite state spaces need not remain fair with an unlimited number of plays allowed. For example, consider the game of Heads or Tails (see Example 1.4). Let Peter start with 1 penny and play until he has 2. Then Peter will be sure to end up 1 penny ahead.)

Let \(\mathbf{P}\) be a regular transition matrix and let \(\mathbf{w}\) be the unique non-zero fixed vector of \(\mathbf{P}\). Show that no entry of \(\mathbf{w}\) is 0 .

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