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Chapter 9: Problem 7

Let \(Y_{n}\) be the number of heads shown in \(n\) tosses of a coin. Let \(Z_{n}=Y_{n}\) modulo 10 . Show that \(\left(Z_{n}\right.\); \(n \geq 0)\) is a Markov chain; find its transition probabilities and stationary distribution.

Short Answer

Expert verified
\((Z_n, n \geq 0)\) is a Markov chain with uniform transition probabilities and stationary distribution \(\pi_i = 0.1\) for each state \(i\).

Step by step solution

01

Define the Variables and Problem Context

First, let's interpret the problem statement. We are given a sequence \(Y_n\), which represents the number of heads in \(n\) coin tosses. \(Z_n = Y_n \mod 10\) then represents the remainder when \(Y_n\) is divided by 10. We need to show \((Z_n, n \geq 0)\) is a Markov chain, find its transition probabilities, and identify the stationary distribution.
02

Determine the State Space

The state space \(S\) of \(Z_n\) is the set of all possible remainders when dividing by 10, which is \(\{0, 1, 2, \ldots, 9\}\). Since after each coin toss, the number of heads can change by either 0 or 1 and impacts the current remainder, the states of \(Z_n\) will be among these 10 values.
03

Establish the Markov Property

The sequence \((Z_n)\) has the Markov property if the future state depends only on the present state, not on the past states. Since the remainder of heads after the next toss adds either 0 or 1 to the current count \(Y_n\), the probability of transitioning to the next state depends only on \(Z_n\), confirming the Markov property.
04

Find the Transition Probabilities

For any state \(i \in \{0, 1, \, \ldots, 9\}\), the transition probability \(P_{ij}\) is given by:1. \(P_{i,i} = \frac{1}{2}\), indicating a head keeps you in the current state modulo 10 with 50% chance.2. \(P_{i,i+1} = \frac{1}{2}\), indicating a head changes the state to \(i+1\) modulo 10 with 50% chance.3. For state 9, \(P_{9,0} = \frac{1}{2}\), since moving from 9 to the next state in modulo 10 results in 0.
05

Solve for the Stationary Distribution

The stationary distribution \(\pi\) satisfies \(\pi_j = \sum_{i}\pi_i P_{ij}\), for all states \(j\). Given symmetry and uniformity of transition probabilities, the stationary distribution is uniform: \(\pi_i = 0.1\) for each state \(i\in \{0, 1, \, \ldots, 9\}\). This satisfies normalization \(\sum_{i=0}^{9}\pi_i = 1\).
06

Conclusions

The process \((Z_n, n \geq 0)\) forms a Markov chain with state space \(\{0, 1, \ldots, 9\}\), equal probability transitions \(P_{i,i}=\frac{1}{2}\), and \(P_{i,i+1}=\frac{1}{2}\) modulo 10. The stationary distribution is uniform across all states: each state has probability 0.1.

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Key Concepts

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

Transition Probabilities
In a Markov chain, transition probabilities describe how likely it is to move from one state to another in a single time step. For the sequence \( (Z_n) \), each state represents a remainder when the number of heads in coin tosses is divided by 10. Transition probabilities account for the possibilities of getting heads or tails on the next coin toss.

  • When the state is \( i \), there is a 50% chance (probability \( \frac{1}{2} \)) to stay in the same remainder, i.e., catching a tail.
  • There is also a 50% chance to move to \( i+1 \) if heads appear, except when \( i = 9 \), where moving forward wraps around to 0.
This simple structure ensures that each transition just depends on a fair coin toss, reflecting the symmetry and randomness of this stochastic process.
Stationary Distribution
The stationary distribution provides a stable long-term probability for each state in a Markov chain. It is the distribution that remains unchanged as the system evolves over time. For the Markov chain \((Z_n)\), the stationary distribution can be derived by considering the uniform transition probabilities that do not favor any state over another.

  • The equation \( \pi_j = \sum_{i} \pi_i P_{ij} \) must hold true for all states \( j \).
  • In this case, all states \( i = 0, 1, \, \ldots, 9 \) are equally likely in the long run because the coin is fair.
Thus, each state in the stationary distribution has equal probability \( \pi_i = 0.1 \). This probability sums to 1 across all ten states, confirming the uniformity and feasibility of the distribution.
State Space
The state space of a Markov chain is a fundamental concept defining all possible states the process can occupy. For the \( (Z_n) \) sequence, the state space is directly related to the possible remainders when dividing the number of heads by 10.

  • The state space is \( \{0, 1, 2, \ldots, 9\} \), capturing each remainder possible when numbers are taken modulo 10.
  • Each state represents a different remainder, which results from the varying number of heads obtained over continued coin tosses.
This well-defined set of ten states allows us to apply the principles of Markov chains such as transition probabilities and stationary distribution effectively.
Markov Property
The Markov property is a key characteristic of processes where future states depend only on the current state, not past states. This property assures that the process is memoryless, making calculations straightforward and assumptions cleaner.

  • For the sequence \((Z_n)\), the future state is determined solely by the current remainder and the outcome of the next coin toss.
  • The Markov property holds because each transition probability depends only on \( Z_n \), not on the sequence of previous coin tosses.
This foundational feature ensures the chain's simplicity, as predicting the next state requires only the present information, aligning closely with real-world scenarios modeled by Markov chains.

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Most popular questions from this chapter

Murphy's Law Let \(\left(X_{n} ; n \geq 1\right)\) be an irreducible aperiodic persistent chain. Let \(s-\) \(\left(s_{1}, \ldots, s_{m}\right)\) be any finite sequence of states of the chain such that \(p_{s_{1} s_{2}} p_{s_{2} s_{3}} \ldots p_{s_{m-1} s_{m}}>0\). Show that with probability 1 the sequence \(s\) occurs in finite time. Explain the implications.

At each time \(n=0,1,2, \ldots\) a number \(Y_{n}\) of particles is injected into a chamber, where \(\left(Y_{n} ; n \geq 0\right)\) are independent Poisson random variables with parameter \(\lambda\). The lifetimes of particles are independent and geometric with parameter \(p\). Let \(X_{n}\) be the number of particles in the chamber at time \(n\). Show that \(X_{n}\) is a Markov chain; find its transition probabilities and the stationary distribution.

Lumping Let \(X\) have state space \(S\) and suppose that \(S=\cup_{k} A_{k}\), where \(A_{i} \cap A_{j}=\phi\) for \(i \neq j\). Let \(\left(Y_{n} ; n \geq 0\right)\) be a process that takes the value \(y_{k}\) whenever the chain \(X\) lies in \(A_{k}\). Show that \(Y\) is also a Markov chain if \(p_{i_{1} j}=p_{i_{2} j}\) for any \(i_{1}\) and \(i_{2}\) in the same set \(A_{k}\).

"Motto" is a coin-tossing game at the start of which each player chooses a sequence of three letters, each of which is either \(H\) or \(T\) (his "motto"). \(\Lambda\) fair coin is then tossed repeatedly, and the results recorded as a sequence of \(H\) s and \(T \mathrm{~s}\) ( \(H\) for "heads," \(T\) for "tails"). The winner is the first player whose motto occurs as three consecutive letters in this sequence. Four players \(A, B, C, D\) choose as their mottoes, respectively, \(H H T, T H H, T T H\), and \(H T T\). Show that if only \(A\) and \(B\) take part in a game then \(B\) has probability \(\frac{3}{4}\) of winning. With what probability does \(C\) win if he plays a game with \(B\) as the only opponent? If all four players take part simultaneously, what are the respective probabilities of each player winning? (You may assume that if a fair coin is tossed repeatedly then with probability 1 any motto will occur eventually.)

Let \(\left(f_{k} ; k \geq 0\right)\) be a probability mass function. Suppose the Markov chain \(X\) has transition probabilities $$ p_{j k}= \begin{cases}f_{j-k+1} & \text { for } k>0, j-k+1 \geq 0 \\\ \sum_{i=j+1}^{\infty} f_{i} & \text { for } k=0 \\ 0 & \text { otherwise. }\end{cases} $$ Show that \(X\) is recurrent and nonnull if \(\sum_{k=1}^{\infty} k f_{k}>1\).
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