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Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

Short Answer

Expert verified
The long-run proportion of time, π_i, represents the fraction of time spent in state i in the long run. It can be thought of as the steady-state probability of being in state i. Both the proportion of transitions into state i and the proportion of transitions from state i will be equal to π_i in the long run. The term π_i P_ij represents the fraction of transitions from state i to state j in the long run. The term ∑_i (π_i P_ij) is the long-run proportion of transitions that end up in state j. Since the long-run proportions of transitions into state j must be equal to the long-run proportions of transitions that end up in state j, the balance condition equation is: \[ \pi_{j}=\sum_{i} \pi_{i} P_{i j} \]

Step by step solution

01

(a) Proportion of transitions

The long-run proportion of time π_i represents the fraction of time the Markov chain spends in state i in the long run. It can be thought of as the steady-state probability of being in state i. Since the Markov chain keeps transitioning between states, this proportion π_i essentially represents the probability of being in state i at any given time in the long run. Now, since the chain is spending a proportion π_i of its time in state i, it naturally implies that both the proportion of transitions into state i and the proportion of transitions from state i will be same as π_i in the long run. This occurs in order to maintain that equilibrium state (steady-state) in which the proportion of time in state i remains constant over time.
02

(b) Proportion of transitions and the property

The term π_i P_ij represents the proportion of transitions that have two specific properties: the chain being in state i and transitioning from state i to state j. In other words, this term represents the fraction of transitions in the Markov chain where the chain transits from state i to state j in the long run.
03

(c) Proportion of transitions and the property

The term ∑_i (π_i P_ij) sums up the product of π_i and P_ij over all states i. This represents the proportion of transitions in the Markov chain where the chain ends up in state j, regardless of which state it started from. In other words, ∑_i (π_i P_ij) is the long-run proportion of transitions that end up in state j.
04

(d) Deriving the equation

Since we know that the long-run proportion of transitions into state j (π_j) must be equal to the long-run proportion of transitions that end up in state j (∑_i (π_i P_ij)), we can write the equation as: \[ \pi_{j}=\sum_{i} \pi_{i} P_{i j} \] This equation represents the balance condition in a Markov chain, which helps in finding the steady-state distribution of the states.

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Key Concepts

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

Steady-State Probability
Understanding the concept of steady-state probability provides valuable insights for analyzing Markov chains. This term refers to the probability that a Markov chain will be found in a particular state if observed over an extended period. Imagine you're monitoring the patterns of weather changes; steady-state probability could represent the likelihood that it will be sunny in the long-run analysis of weather data. It's the balance point where the state's incoming and outgoing transitions even out over time.

The power of steady-state probability lies in its predictive ability. It equips us with a stable point of reference, regardless of our starting point in the chain. The steadiness implies that, after a large number of transitions, the likelihood of being in any given state stabilizes and does not change with additional transitions. This deeply signifies the Markov chain's reliance on its current state, rather than its history.
Transition Proportions

Equilibrium of Movement

Transition proportions capture the essence of Markov chain dynamics. These proportions detail how often a system moves from one state to another over time. For clarification, consider the example of a board game where players advance based on dice rolls: the probability of moving from one square to another would be analogous to a transition proportion in a Markov chain.

More formally, transition proportions are represented by the transition matrix, where each entry denotes the probability of moving from one state to another in one step. It's worth emphasizing that because Markov chains are memoryless, these probabilities are consistent regardless of the prior states occupied by the system. Transition proportions play a crucial role in understanding how states interact with one another and are foundational for determining steady-state probabilities.
Long-Run Proportion

Forecasting the Future

The long-run proportion takes the concept of steady-state a step further. It addresses the enduring presence of a state within a Markov chain, representing the percentage of time a system is expected to reside in a given state after many steps or transitions. This long-term perspective can be likened to observing a habitual coffee drinker's pattern over many years to predict the likelihood of them visiting a certain coffee shop.

Essentially, long-run proportions quantify the notion of equilibrium over infinite time, and by doing so, they facilitate predictions about the future behavior of the system. This concept is significant for fields that rely on such predictions, such as economics or population studies, where long-term trends and patterns are a focus of the analysis.
Balance Condition

The Scales of Equilibrium

The balance condition is like a mathematical fulcrum that keeps the Markov chain in equilibrium. It is a fundamental concept that ensures the long-run proportion of time spent in each state is constant. The balance condition mirrors the principle of conservation in physics, where the incoming 'flow' of probability into a state must equal the outgoing 'flow'.

To encapsulate this idea, the balance equation \[ \pi_{j}=\sum_{i} \pi_{i} P_{i j} \] is used to maintain the equilibrium across the states. This pivotal equation reveals that the long-run proportion of being in state 'j' \( \pi_{j} \) must equal the sum of all the proportions of being in each state 'i' and transitioning to 'j'. The balance condition serves as a cornerstone for determining the steady-state probabilities, and thus, it is a critical tool for analyzing Markov chains in various applications.

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Most popular questions from this chapter

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.

A Markov chain \(\left\\{X_{n}, n \geqslant 0\right\\}\) with states \(0,1,2\), has the transition probability matrix $$ \left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ 0 & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{2} & 0 & \frac{1}{2} \end{array}\right] $$ If \(P\left\\{X_{0}=0\right\\}=P\left\\{X_{0}=1\right\\}=\frac{1}{4}\), find \(E\left[X_{3}\right]\).

In a Markov decision problem, another criterion often used, different than the expected average return per unit time, is that of the expected discounted return. In this criterion we choose a number \(\alpha, 0<\alpha<1\), and try to choose a policy so as to maximize \(E\left[\sum_{i=0}^{\infty} \alpha^{i} R\left(X_{i}, a_{i}\right)\right]\) (that is, rewards at time \(n\) are discounted at rate \(\left.\alpha^{n}\right)\) Suppose that the initial state is chosen according to the probabilities \(b_{i} .\) That is, $$ P\left\\{X_{0}=i\right\\}=b_{i}, \quad i=1, \ldots, n $$ For a given policy \(\beta\) let \(y_{j a}\) denote the expected discounted time that the process is in state \(j\) and action \(a\) is chosen. That is, $$ y_{j a}=E_{\beta}\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left[X_{n}=j, a_{n}=a\right\\}}\right] $$ where for any event \(A\) the indicator variable \(I_{A}\) is defined by $$ I_{A}=\left\\{\begin{array}{ll} 1, & \text { if } A \text { occurs } \\ 0, & \text { otherwise } \end{array}\right. $$ (a) Show that $$ \sum_{a} y_{j a}=E\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left\\{X_{n}=j\right\\}}\right] $$ or, in other words, \(\sum_{a} y_{j a}\) is the expected discounted time in state \(j\) under \(\beta\). (b) Show that $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Hint: For the second equation, use the identity $$ I_{\left\\{X_{n+1}=j\right\\}}=\sum_{i} \sum_{a} I_{\left\\{X_{n-i}, a_{n-a}\right\rangle} I_{\left\\{X_{n+1}=j\right\\}} $$ Take expectations of the preceding to obtain $$ E\left[I_{\left.X_{n+1}=j\right\\}}\right]=\sum_{i} \sum_{a} E\left[I_{\left\\{X_{n-i}, a_{n-a}\right\\}}\right] P_{i j}(a) $$ (c) Let \(\left\\{y_{j a}\right\\}\) be a set of numbers satisfying $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Argue that \(y_{j a}\) can be interpreted as the expected discounted time that the process is in state \(j\) and action \(a\) is chosen when the initial state is chosen according to the probabilities \(b_{j}\) and the policy \(\beta\), given by $$ \beta_{i}(a)=\frac{y_{i a}}{\sum_{a} y_{i a}} $$ is employed. Hint: Derive a set of equations for the expected discounted times when policy \(\beta\) is used and show that they are equivalent to Equation \((4.38) .\) (d) Argue that an optimal policy with respect to the expected discounted return criterion can be obtained by first solving the linear program $$ \begin{array}{ll} \operatorname{maximize} & \sum_{j} \sum_{a} y_{j a} R(j, a), \\ \text { such that } & \sum_{j} \sum_{a} y_{j a}=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a}=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \\ y_{j a} \geqslant 0, \quad \text { all } j, a \end{array} $$ and then defining the policy \(\beta^{*}\) by $$ \beta_{i}^{*}(a)=\frac{y_{i a}^{*}}{\sum_{a} y_{i a}^{*}} $$ where the \(y_{j a}^{*}\) are the solutions of the linear program.

A DNA nucleotide has any of four values. A standard model for a mutational change of the nucleotide at a specific location is a Markov chain model that supposes that in going from period to period the nucleotide does not change with probability \(1-3 \alpha\), and if it does change then it is equally likely to change to any of the other three values, for some \(0<\alpha<\frac{1}{3}\). (a) Show that \(P_{1,1}^{n}=\frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). (b) What is the long-run proportion of time the chain is in each state?

In the gambler's ruin problem of Section 4.5.1, suppose the gambler's fortune is presently \(i\), and suppose that we know that the gambler's fortune will eventually reach \(N\) (before it goes to 0 ). Given this information, show that the probability he wins the next gamble is $$ \begin{array}{ll} \frac{p\left[1-(q / p)^{i+1}\right]}{1-(q / p)^{i}}, & \text { if } p \neq \frac{1}{2} \\ \frac{i+1}{2 i}, & \text { if } p=\frac{1}{2} \end{array} $$

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