Question

Consider a system that can be in either state 1 or 2 or \(3 .\) Each time the system enters state \(i\) it remains there for a random amount of time having mean \(\mu_{i}\) and then makes a transition into state \(j\) with probability \(P_{i j} .\) Suppose $$ P_{12}=1, \quad P_{21}=P_{23}=\frac{1}{2}, \quad P_{31}=1 $$ (a) What proportion of transitions takes the system into state \(1 ?\) (b) If \(\mu_{1}=1, \mu_{2}=2, \mu_{3}=3\), then what proportion of time does the system spend in each state?

Short answer

(a) The proportion of transitions that take the system into state 1 is \(\frac{5}{9}\). (b) The proportion of time the system spends in each state is as follows: in state 1, it spends \(\frac{4}{9}\) of the time; in state 2, \(\frac{4}{9}\) of the time; and in state 3, \(\frac{1}{9}\) of the time.

Step-by-step solution

Find Long-Term State Probabilities

We will first find the long-term probabilities of being in each state. To do this, we will use the given transition probabilities to create the following system of linear equations: \( \pi_1 = \pi_1 P_{11} + \pi_2 P_{21} + \pi_3 P_{31} \) \( \pi_2 = \pi_1 P_{12} + \pi_2 P_{22} + \pi_3 P_{32} \) \( \pi_3 = \pi_1 P_{13} + \pi_2 P_{23} + \pi_3 P_{33} \) along with the condition that the probabilities must sum to 1: \( \pi_1 + \pi_2 + \pi_3 = 1 \)

Substitute the Given Transition Probabilities

Next, we need to substitute the given transition probabilities into the equations we obtained in step 1. In this case, \( P_{12} = 1 \) \( P_{21} = P_{23} = \frac{1}{2} \) \( P_{31} = 1 \) It follows that \(P_{11} = P_{22} = P_{33} = 0\), since all outgoing probabilities from a state must add up to 1. Then, the linear equations become: \( \pi_1 = \pi_2 \cdot \frac{1}{2} + \pi_3 \) \( \pi_2 = \pi_1 \) \( \pi_3 = \pi_2 \cdot \frac{1}{2} \)

Solve the Linear Equations

Now, we will solve the linear equations along with the condition that the probabilities must sum to 1: 1. From Equation 2, we know \( \pi_2 = \pi_1 \). 2. Substitute Equation 1 into Equation 3: \( \pi_3 = (\pi_2 \cdot \frac{1}{2}) \cdot \frac{1}{2} = \pi_1 \cdot \frac{1}{4} \). 3. Use the condition that the probabilities sum to 1: \( \pi_1 + \pi_1 + \pi_1 \cdot \frac{1}{4} = 1 \). Solving this last equation gives us \( \pi_1 = \frac{4}{9} \). Then, we have \( \pi_2 = \pi_1 = \frac{4}{9} \) and \( \pi_3 = \pi_1 \cdot \frac{1}{4} = \frac{1}{9} \).

Answer Part (a)

The proportion of transitions that take the system into state 1 is equal to the sum of the probabilities of transitions from any other state to state 1: Proportion of transitions to state 1 = \( \pi_2 P_{21} + \pi_3 P_{31} \) Substituting the values, we get: Proportion of transitions to state 1 = \( \frac{4}{9} \cdot \frac{1}{2} + \frac{1}{9} \cdot 1 = \frac{5}{9} \).

Answer Part (b)

Finally, to answer part (b), we will use the given mean times spent in each state. We know that: \( \mu_1 = 1 \) \( \mu_2 = 2 \) \( \mu_3 = 3 \) The proportion of time spent in each state is equal to the long-term probability of that state multiplied by the mean time spent in that state, divided by the total time spent in all states: Proportion of time in state i = \( \frac{\pi_i \cdot \mu_i}{(\pi_1 \cdot \mu_1) + (\pi_2 \cdot \mu_2) + (\pi_3 \cdot \mu_3)} \) Calculating for each state: 1. State 1: \( \frac{(\frac{4}{9}) \cdot 1}{(\frac{4}{9} \cdot 1) + (\frac{4}{9} \cdot 2) + (\frac{1}{9} \cdot 3)} = \frac{4}{9} \) 2. State 2: \( \frac{(\frac{4}{9}) \cdot 2}{(\frac{4}{9} \cdot 1) + (\frac{4}{9} \cdot 2) + (\frac{1}{9} \cdot 3)} = \frac{8}{9} \cdot \frac{1}{2} = \frac{4}{9} \) 3. State 3: \( \frac{(\frac{1}{9}) \cdot 3}{(\frac{4}{9} \cdot 1) + (\frac{4}{9} \cdot 2) + (\frac{1}{9} \cdot 3)} = \frac{3}{9} \cdot \frac{1}{3} = \frac{1}{9} \) Hence, the system spends \(\frac{4}{9}\) of the time in state 1, \(\frac{4}{9}\) in state 2, and \(\frac{1}{9}\) in state 3.

Key concepts

Transition Probabilities

In stochastic processes, transition probabilities are essentially the odds of a system changing its state from one to another. Given our system having states 1, 2, and 3, we assign probabilities, such as P₁₂ for the system transitioning from state 1 to state 2. This concept is crucial because it determines the future state of the system based on its current state.

Understanding how to calculate these probabilities is essential. As shown in the step-by-step solution, we use the given transition probabilities to form a system of linear equations. The solution to these equations reveals the long-term behavior of our system and can predict the likelihood of the system being in a particular state at a specific time. For search engine optimization, remember that phrases like 'calculate transition probabilities' and 'what are transition probabilities' are frequently searched by students needing assistance.

State Probabilities

State probabilities, denoted by π_i, represent the long-term proportion of time a stochastic process is expected to be in state i. To determine these, we need to understand the steady-state behavior of the system. In our solution, we first set up equations based on the balance of probabilities entering and leaving each state.

By solving these equations, we find the proportions or steady-state probabilities, indicating how much time, on average, the system spends in each state. This is essential for predicting the system's performance over time and for making informed decisions based on these probabilities. Students searching online might use terms like 'finding state probabilities' or 'steady-state behavior in stochastic processes' to grasp this concept better.

Markov Chains

Markov chains are a type of stochastic process that possess the 'memorylessness' property, meaning the next state only depends on the current state and not on the sequence of events that preceded it. Our system is a classic example of a Markov chain, where the transition to a new state depends only on the current state and not on how the system got there.

Understanding Markov chains is fundamental in fields like economics, game theory, and genetics because they simplify complex stochastic processes. For someone struggling with this concept, searching for 'basics of Markov chains' or 'Markov chain examples' can yield results that explain the concept using everyday situations.

Mean Sojourn Time

The mean sojourn time refers to the average amount of time a system spends in a particular state before moving on to another state. The Greek letter μ denotes the mean sojourn time, and as denoted in our exercise, μ₁, μ₂, and μ₃ represent the mean sojourn times for states 1, 2, and 3, respectively.

By combining the state probabilities and the mean sojourn times, you can calculate the proportion of time the system spends in each state over the long run. This can provide insights into the efficiency and performance of different states within a system. A useful tip for those seeking solutions online could be to search 'how to calculate mean sojourn time' or 'importance of mean sojourn time in stochastic processes'.