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Consider a Markov chain in steady state. Say that a \(k\) length run of zeroes ends at time \(m\) if $$ X_{m-k-1} \neq 0, \quad X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0, X_{m} \neq 0 $$ Show that the probability of this event is \(\pi_{0}\left(P_{0,0}\right)^{k-1}\left(1-P_{0,0}\right)^{2}\), where \(\pi_{0}\) is the limiting probability of state 0 .

Short Answer

Expert verified
In conclusion, the probability of a k-length run of zeroes ending at time m for a Markov chain in steady state is given by \(\pi_{0}(P_{0,0})^{k-1}(1-P_{0,0})^{2}\).

Step by step solution

01

Define the conditions for the event to happen

A k-length run of zeroes ends at time m if the following conditions are met: 1. \(X_{m-k-1} \neq 0\), meaning the state at time m - k - 1 is not 0. 2. \(X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0\), meaning the states in the range [m - k, m - 1] are all zeroes. 3. \(X_m \neq 0\), meaning the state at time m is not 0. Notice that because of the Markov property, these events are independent.
02

Compute the probability of the first condition

Since the Markov chain is in steady state, we have the limiting probability of state 0 as \(\pi_0\). Therefore, the probability of \(X_{m-k-1}\) being not equal to 0 is given by: \[P(X_{m-k-1} \neq 0) = 1 - \pi_{0}\]
03

Compute the probability of the second condition

For the consecutive k-1 zeros in the range [m - k, m - 1], we have \((k - 1)\) independent transitions from state 0 to state 0. The probability of each individual transition is given by \(P_{0,0}\), so the probability of the whole sequence happening is the product of the probabilities for each of these transitions, which is: \[P(X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0) = P_{0,0}^{k-1}\]
04

Compute the probability of the third condition

Finally, the probability of \(X_m\) being not equal to 0 is given by: \[P(X_m \neq 0) = 1 - P_{0,0}\]
05

Combine the probabilities

Now that we have the probabilities for the three conditions, we need to combine them to find the probability of the entire event. Since the events are independent, we can simply multiply their probabilities together: \[P(Event) = P(X_{m-k-1} \neq 0) \cdot P(X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0) \cdot P(X_m \neq 0)\] \[P(Event) = (1 - \pi_{0}) \cdot (P_{0,0})^{k-1} \cdot (1 - P_{0,0})\]
06

Adjust the result to include the limiting probability of state 0

The result above only accounts for a specific stretch of k zeroes within the Markov chain. To account for the entire chain, we need to consider the limiting probability of state 0, denoted as \(\pi_0\). Therefore, the final probability of the event will be: \[P(Event) = \pi_{0}\left(P_{0,0}\right)^{k-1}\left(1-P_{0,0}\right)^{2}\] In conclusion, the probability of a k-length run of zeroes ending at time m for a Markov chain in steady state is given by \(\pi_{0}(P_{0,0})^{k-1}(1-P_{0,0})^{2}\).

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Key Concepts

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

Steady State
In the world of Markov chains, a steady state represents a situation where the probabilities of being in each state have settled into a stable pattern over time. Think of it as reaching a calm, unchanging phase. This happens when the distribution of states no longer changes as time progresses. At this point, the probabilities are called steady state probabilities. Each state in the Markov chain has its own steady state probability, noting how likely it is for the process to be in that state in the long run.
In our exercise, we're dealing with what happens when the Markov chain is already in this steady state. This allows us to use consistent probabilities, like the limiting probability for state 0, denoted as \(\pi_0\). This \(\pi_0\) tells us the likelihood of being in state 0 at any point in time once the steady state has been reached.
Limiting Probability
Limiting probability is a key concept when working with Markov chains. It represents the probability that the chain is in a particular state after a long time. In simple terms, it's the likelihood that the process ends up in a particular state as time approaches infinity.
For each state in a Markov chain, there is a corresponding limiting probability. This probability is constant once the chain has reached its steady state. For state 0, this probability is called \(\pi_0\). Knowing this probability is crucial because it allows us to predict the long-term behavior of the system. In our exercise, \(\pi_0\) helps us determine how often we would encounter a run of zeroes in the long term.
Transition Probability Matrix
The transition probability matrix is like a roadmap for a Markov chain. It contains all the information you need to know about how the chain moves from one state to another. This matrix details the probabilities of transitioning from any given state to any other state, including itself.
The transition probabilities are denoted by \(P_{i,j}\), where \(i\) is the initial state and \(j\) is the state you move to. For instance, \(P_{0,0}\) is the probability of staying in state 0. In our exercise, \(P_{0,0}\) is crucial for figuring out the likelihood of staying in state 0 for several steps—specifically for a complete run of zeroes. Understanding the transition probability matrix enables us to glimpse the dynamism of the Markov chain in action.
Independent Events
In probability, events are termed independent if the occurrence of one event does not affect the occurrence of another. This is a powerful principle in statistics and Markov chains. When dealing with a Markov chain in a steady state, certain events can be treated as independent because the present state is all you need to predict the future.
  • In our exercise, the independence of certain transitions is key. The probability that certain events happen consecutively in a Markov chain relies on this principle.
  • The independence helps us to multiply probabilities directly to evaluate compound events—like a stretch of zeroes that ends at a particular time.
Understanding independent events simplifies complex probability calculations and allows for smooth assessments of various scenarios within a Markov chain.

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Most popular questions from this chapter

Consider a population of individuals each of whom possesses two genes that can be either type \(A\) or type \(a\). Suppose that in outward appearance type \(A\) is dominant and type \(a\) is recessive. (That is, an individual will have only the outward characteristics of the recessive gene if its pair is aa.) Suppose that the population has stabilized, and the percentages of individuals having respective gene pairs \(A A, a a\), and \(A a\) are \(p, q\), and \(r .\) Call an individual dominant or recessive depending on the outward characteristics it exhibits. Let \(S_{11}\) denote the probability that an offspring of two dominant parents will be recessive; and let \(S_{10}\) denote the probability that the offspring of one dominant and one recessive parent will be recessive. Compute \(S_{11}\) and \(S_{10}\) to show that \(S_{11}=S_{10}^{2} .\) (The quantities \(S_{10}\) and \(S_{11}\) are known in the genetics literature as Snyder's ratios.)

Suppose in the gambler's ruin problem that the probability of winning a bet depends on the gambler's present fortune. Specifically, suppose that \(\alpha_{i}\) is the probability that the gambler wins a bet when his or her fortune is \(i\). Given that the gambler's initial fortune is \(i\), let \(P(i)\) denote the probability that the gambler's fortune reaches \(N\) before 0 . (a) Derive a formula that relates \(P(i)\) to \(P(i-1)\) and \(P(i+1)\). (b) Using the same approach as in the gambler's ruin problem, solve the equation of part (a) for \(P(i)\) (c) Suppose that \(i\) balls are initially in urn 1 and \(N-i\) are in urn 2, and suppose that at each stage one of the \(N\) balls is randomly chosen, taken from whichever urn it is in, and placed in the other urn. Find the probability that the first urn becomes empty before the second.

For a time reversible Markov chain, argue that the rate at which transitions from \(i\) to \(j\) to \(k\) occur must equal the rate at which transitions from \(k\) to \(j\) to \(i\) occur.

It follows from the argument made in Exercise 38 that state \(i\) is null recurrent if it is recurrent and \(\pi_{i}=0 .\) Consider the one-dimensional symmetric random walk of Example 4.18. (a) Argue that \(\pi_{i}=\pi_{0}\) for all \(i\). (b) Argue that all states are null recurrent.

A professor continually gives exams to her students. She can give three possible types of exams, and her class is graded as either having done well or badly. Let \(p_{i}\) denote the probability that the class does well on a type \(i\) exam, and suppose that \(p_{1}=0.3, p_{2}=0.6\), and \(p_{3}=0.9 .\) If the class does well on an exam, then the next exam is equally likely to be any of the three types. If the class does badly, then the next exam is always type \(1 .\) What proportion of exams are type \(i, i=1,2,3 ?\)

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