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To prove Equation ( \(7.24)\), define the following notation: \(X_{i}^{j} \equiv\) time spent in state \(i\) on the \(j\) th visit to this state; \(N_{i}(m) \equiv\) number of visits to state \(i\) in the first \(m\) transitions In terms of this notation, write expressions for (a) the amount of time during the first \(m\) transitions that the process is in state \(i ;\) (b) the proportion of time during the first \(m\) transitions that the process is in state \(i\) Argue that, with probability 1 , (c) \(\sum_{j=1}^{N_{i}(m)} \frac{X_{i}^{j}}{N_{i}(m)} \rightarrow \mu_{i}\) as \(m\) (d) \(\mathrm{N}_{i}(m) / m \rightarrow \pi_{i} \quad\) as \(m \rightarrow \infty\). (e) Combine parts (a), (b), (c), and (d) to prove Equation (7.24).

Short Answer

Expert verified
In summary, we have found expressions for the time spent in state i and the proportion of time spent in state i during the first m transitions. We have also shown that the limits of the average time spent in state i converge to μ_i and the ratio of the number of visits to state i converges to π_i, both with probability 1. By combining these results, we have proved the equation (7.24): \[ \lim_{m\to\infty} \frac{\sum_{j=1}^{N_{i}(m)} X_{i}^{j}}{m} = \mu_{i} \cdot \pi_{i} \]

Step by step solution

01

(a) Time spent in state i during m transitions

To find an expression for the amount of time spent in state i during the first m transitions, we need to add the time spent in state i for each of the N_i(m) visits. The total time spent in state i during the first m transitions can be given by the sum: \[ \sum_{j=1}^{N_{i}(m)} X_{i}^{j} \]
02

(b) Proportion of time spent in state i during m transitions

To find the proportion of time spent in state i during the first m transitions, we need to divide the time spent in state i by the total time (m). The proportion of time spent in state i during the first m transitions can be given by: \[ \frac{\sum_{j=1}^{N_{i}(m)} X_{i}^{j}}{m} \]
03

(c) Limit of the average time in state i

We need to show that, with probability 1, the limit of the average time spent in state i converges to μ_i as m goes to infinity. This can be given by: \[ \lim_{m\to\infty} \sum_{j=1}^{N_{i}(m)} \frac{X_{i}^{j}}{N_{i}(m)} = \mu_{i} \] This holds because, as m increases and the process spends more time in state i, the average time spent in state i converges to the expected time spent in state i, which is μ_i.
04

(d) Limit of the ratio of visits to state i

We need to show that, with probability 1, the limit of the ratio of the number of visits to state i converges to π_i as m goes to infinity. This can be given by: \[ \lim_{m\to\infty} \frac{N_{i}(m)}{m} = \pi_{i} \] This limit shows that, as the total number of transitions m increases (goes to infinity), the proportion of visits to state i converges to its stationary distribution, which is π_i.
05

(e) Proving the equation (7.24)

Combining parts (a), (b), (c), and (d), we can write the desired equation (7.24) as: \[ \lim_{m\to\infty} \frac{\sum_{j=1}^{N_{i}(m)} X_{i}^{j}}{m} = \lim_{m\to\infty} \mu_{i} \cdot \frac{N_{i}(m)}{m} = \mu_{i} \cdot \pi_{i} \] This equation shows that as m goes to infinity, the proportion of time spent in state i during the first m transitions converges to the product of the expected time spent in state i (μ_i) and its stationary distribution (π_i), ultimately proving the equation (7.24).

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Most popular questions from this chapter

In a semi-Markov process, let \(t_{i j}\) denote the conditional expected time that the process spends in state \(i\) given that the next state is \(j\). (a) Present an equation relating \(\mu_{i}\) to the \(t_{i j}\). (b) Show that the proportion of time the process is in \(i\) and will next enter \(j\) is equal to \(P_{i} P_{i j} t_{i j} / \mu_{i}\)

A system consists of two independent machines that each function for an exponential time with rate \(\lambda .\) There is a single repairperson. If the repairperson is idle when a machine fails, then repair immediately begins on that machine; if the repairperson is busy when a machine fails, then that machine must wait until the other machine has been repaired. All repair times are independent with distribution function \(G\) and, once repaired, a machine is as good as new. What proportion of time is the repairperson idle?

Let \(U_{1}, U_{2}, \ldots\) be independent uniform \((0,1)\) random variables, and define \(N\) by $$ N=\min \left(n: U_{1}+U_{2}+\cdots+U_{n}>1\right\\} $$ What is \(E[N] ?\)

Let \(h(x)=P\left(\sum_{i=1}^{T} X_{i}>x\right\\}\) where \(X_{1}, X_{2}, \ldots\) are independent random variables having distribution function \(F_{e}\) and \(T\) is independent of the \(X_{i}\) and has probability mass function \(P[T=n\\}=\rho^{n}(1-\rho), n \geqslant 0 .\) Show that \(h(x)\) satisfies Equation (7.53). Hint: Start by conditioning on whether \(T=0\) or \(T>0\).

Consider a semi-Markov process in which the amount of time that the process spends in each state before making a transition into a different state is exponentially distributed. What kind of process is this?

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