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Chapter 4: Problem 42

Let \(A\) be a set of states, and let \(A^{c}\) be the remaining states. (a) What is the interpretation of $$ \sum_{i \in A} \sum_{j \in A^{r}} \pi_{i} P_{i j} ? $$ (b) What is the interpretation of $$ \sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} ? $$ (c) Explain the identity $$ \sum_{i \in A} \sum_{j \in A^{c}} \pi_{i} P_{i j}=\sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} $$

Short Answer

Expert verified
In summary, the expressions in parts (a) and (b) represent the total probability of transitioning from any state in set \(A\) to any state in set \(A^{c}\), and vice versa, respectively. The identity in part (c) holds because, in an equilibrium Markov chain, the flow of probabilities between states in sets \(A\) and \(A^{c}\) must be balanced to maintain a constant probability distribution.

Step by step solution

01

Part (a) Interpretation

The given expression is: \[ \sum_{i \in A} \sum_{j \in A^{c}} \pi_{i} P_{i j} \] In this sum, the outer summation goes through all states \(i\) in set \(A\) and the inner summation goes through all states \(j\) in the complement set \(A^{c}\). The expression inside the summation, \(\pi_{i} P_{i j}\), represents the probability of being in state \(i\) and transitioning to state \(j\). The interpretation of the given expression is the total probability of transitioning from any state in set \(A\) to any state in set \(A^{c}\).
02

Part (b) Interpretation

The given expression is: \[ \sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} \] This sum is similar to the sum in part (a), but the roles of \(A\) and \(A^{c}\) are reversed. The outer summation goes through all states \(i\) in the complement set \(A^{c}\) and the inner summation goes through all states \(j\) in the set \(A\). The interpretation of the given expression is the total probability of transitioning from any state in set \(A^{c}\) to any state in set \(A\).
03

Part (c) Explanation of the Identity

We need to explain why the following identity holds: \[ \sum_{i \in A} \sum_{j \in A^{c}} \pi_{i} P_{i j}=\sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} \] Consider a Markov chain in equilibrium, which means the probability distribution of states remains constant over time. In this situation, the probability of transitioning from any state in \(A\) to any state in \(A^{c}\) must be equal to the probability of transitioning from any state in \(A^{c}\) to any state in \(A\). This equality maintains the balance between sets \(A\) and \(A^{c}\) and ensures that the overall probability distribution remains constant. Therefore, the identity holds because it represents the balanced flow of probabilities between the states in sets \(A\) and \(A^{c}\) in an equilibrium Markov chain.

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Key Concepts

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

State Transition Probabilities
Markov chains rely heavily on understanding state transition probabilities. These are the likelihoods of moving from one state to another within a system. In mathematical terms, if a system is in a state \( i \), the transition probability \( P_{ij} \) indicates the probability of moving to state \( j \) in one time step.
For example, if we have three different states in a system: A, B, and C, the probability of moving from A to B during each transition could be denoted as \( P_{AB} \). Transitioning to any other state from A would have its own probability like \( P_{AC} \).
  • These probabilities are captured in a transition matrix, where each entry \( P_{ij} \) expresses the chance of moving from state \( i \) to state \( j \).
  • The sum of transition probabilities from any given state to all possible next states should equal 1.
Equilibrium State
An equilibrium state in a Markov chain is a stable condition where the system's state probabilities do not change over time. This occurs when the probability distribution reaches a point where further transitions do not alter the overall likelihood distribution. Essentially, the system stops "wandering" among states.
This concept can be visualized as water in a bathtub. Initially, the water might be sloshing around, moving from one side to the other. However, after a while, it stabilizes, reaching equilibrium. The water doesn't stop moving entirely鈥攋ust as the state of a system doesn't cease to have state transitions鈥攂ut the overall distribution remains constant.
  • To achieve equilibrium, the inflow and outflow probabilities for each state pair must be balanced.
  • This condition ensures that over time, the system behaves predictably and consistently.
Probability Distribution
Probability distributions in Markov chains detail how probable various states are at any point in time, with each state having an associated probability of occurrence. This distribution is a vector that reflects the likelihood of the process being in any particular state.
To visualize this, imagine rolling a die. Each face of the die is a different state. Initially, if you haven't started rolling the die, the probability distribution might be uniform: each face has an equal chance of landing up, i.e., 1/6. As the die is rolled multiple times, the observed states (outcomes) might form a different distribution, such as higher frequencies for certain numbers.
  • The sum of all state probabilities in the distribution must equal 1.
  • An initial probability distribution can evolve by applying the state transition probabilities iteratively.
Equilibrium Markov Chain
An equilibrium Markov chain is one where the probability distribution of the states does not change over time, despite the inherent transitions between states. At equilibrium, the flow of probability into any set of states equals the flow out, maintaining a steady state.
This concept is foundational for understanding many real-world systems, from traffic flow patterns to the spread of diseases.
To establish if a Markov chain is at equilibrium, examine the probability flows in and out of each state. If the total inflow equals the total outflow for each state, the chain is at equilibrium.
  • An equilibrium Markov chain ensures that the chance of being in any given state remains stable over time.
  • This balance means systems modeled by Markov chains can have predictable long-term behavior.
  • Real-world applications of this concept include inventory management, weather forecasting, and financial modeling.

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Most popular questions from this chapter

Each day, one of \(n\) possible elements is requested, the \(i\) th one with probability \(P_{i}, i \geqslant 1, \sum_{1}^{n} P_{i}=1\). These elements are at all times arranged in an ordered list which is revised as follows: The element selected is moved to the front of the list with the relative positions of all the other elements remaining unchanged. Define the state at any time to be the list ordering at that time and note that there are \(n !\) possible states. (a) Argue that the preceding is a Markov chain. (b) For any state \(i_{1}, \ldots, i_{n}\) (which is a permutation of \(1,2, \ldots, n\) ), let \(\pi\left(i_{1}, \ldots, i_{n}\right)\) denote the limiting probability. In order for the state to be \(i_{1}, \ldots, i_{n}\), it is necessary for the last request to be for \(i_{1}\), the last non- \(i_{1}\) request for \(i_{2}\), the last non- \(i_{1}\) or \(i_{2}\) request for \(i_{3}\), and so on. Hence, it appears intuitive that $$ \pi\left(i_{1}, \ldots, i_{n}\right)=P_{i_{1}} \frac{P_{i_{2}}}{1-P_{i_{1}}} \frac{P_{i_{3}}}{1-P_{i_{1}}-P_{i_{2}}} \cdots \frac{P_{i_{n-1}}}{1-P_{i_{1}}-\cdots-P_{i_{n-2}}} $$ Verify when \(n=3\) that the preceding are indeed the limiting probabilities.

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(i, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1 .\) Find the limiting probabilities of this Markov chain.

A group of \(n\) processors is arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long-run probabilities.

Let \(P^{(1)}\) and \(P^{(2)}\) denote transition probability matrices for ergodic Markov chains having the same state space. Let \(\pi^{1}\) and \(\pi^{2}\) denote the stationary (limiting) probability vectors for the two chains. Consider a process defined as follows: (i) \(X_{0}=1\). A coin is then flipped and if it comes up heads, then the remaining states \(X_{1}, \ldots\) are obtained from the transition probability matrix \(P^{(1)}\) and if tails from the matrix \(P^{(2)}\). Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If \(p=P\\{\) coin comes up heads \(\\}\), what is \(\lim _{n \rightarrow \infty} P\left(X_{n}=i\right)\) ? (ii) \(X_{0}=1\). At each stage the coin is flipped and if it comes up heads, then the next state is chosen according to \(P^{(1)}\) and if tails comes up, then it is chosen according to \(P^{(2)}\). In this case do the successive states constitute a Markov chain? If so, determine the transition probabilities. Show by a counterexample that the limiting probabilities are not the same as in part (i).

In Example 4.3, Gary is in a cheerful mood today. Find the expected number of days until he has been glum for three consecutive days.
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