Chapter 7: Problem 46
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?
Chapter 7: Problem 46
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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Get started for freeConsider a renewal process having the gamma \((n, \lambda)\) interarrival
distribution, and let \(Y(t)\) denote the time from \(t\) until the next renewal.
Use the theory of semi-Markov processes to show that
$$
\lim _{t \rightarrow \infty} P(Y(t)
If the mean-value function of the renewal process \(\\{N(t), t \geqslant 0\\}\) is given by \(m(t)=\) \(t / 2, t \geqslant 0\), what is \(P[N(5)=0\\} ?\)
Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables. The nonnegative integer valued random variable \(N\) is said to be a stopping time for the sequence if the event \(\\{N=n\\}\) is independent of \(X_{n+1}, X_{n+2}, \ldots .\) The idea being that the \(X_{i}\) are observed one at a time-first \(X_{1}\), then \(X_{2}\), and so on-and \(N\) represents the number observed when we stop. Hence, the event \(\\{N=n\\}\) corresponds to stopping after having observed \(X_{1}, \ldots, X_{n}\) and thus must be independent of the values of random variables yet to come, namely, \(X_{n+1}, X_{n+2}, \ldots\) (a) Let \(X_{1}, X_{2}, \ldots\) be independent with $$ P\left[X_{i}=1\right\\}=p=1-P\left(X_{i}=0\right\\}, \quad i \geqslant 1 $$ Define $$ \begin{aligned} &N_{1}=\min \left[n: X_{1}+\cdots+X_{n}=5\right\\} \\ &N_{2}=\left\\{\begin{array}{ll} 3, & \text { if } X_{1}=0 \\ 5, & \text { if } X_{1}=1 \end{array}\right. \\ &N_{3}=\left\\{\begin{array}{ll} 3, & \text { if } X_{4}=0 \\ 2, & \text { if } X_{4}=1 \end{array}\right. \end{aligned} $$ Which of the \(N_{i}\) are stopping times for the sequence \(X_{1}, \ldots ?\) An important result, known as Wald's equation states that if \(X_{1}, X_{2}, \ldots\) are independent and identically distributed and have a finite mean \(E(X)\), and if \(N\) is a stopping time for this sequence having a finite mean, then $$ E\left[\sum_{i=1}^{N} X_{i}\right]=E[N] E[X] $$ To prove Wald's equation, let us define the indicator variables \(I_{i}, i \geqslant 1\) by $$ I_{i}=\left\\{\begin{array}{ll} 1, & \text { if } i \leqslant N \\ 0, & \text { if } i>N \end{array}\right. $$ (b) Show that $$ \sum_{i=1}^{N} X_{i}=\sum_{i=1}^{\infty} X_{i} I_{i} $$ From part (b) we see that $$ \begin{aligned} E\left[\sum_{i=1}^{N} X_{i}\right] &=E\left[\sum_{i=1}^{\infty} X_{i} I_{i}\right] \\ &=\sum_{i=1}^{\infty} E\left[X_{i} I_{i}\right] \end{aligned} $$ where the last equality assumes that the expectation can be brought inside the summation (as indeed can be rigorously proven in this case). (c) Argue that \(X_{i}\) and \(I_{i}\) are independent. Hint: \(I_{i}\) equals 0 or 1 depending on whether or not we have yet stopped after observing which random variables? (d) From part (c) we have $$ E\left[\sum_{i=1}^{N} X_{i}\right]=\sum_{i=1}^{\infty} E[X] E\left[I_{i}\right] $$ Complete the proof of Wald's equation. (e) What does Wald's equation tell us about the stopping times in part (a)?
Satellites are launched according to a Poisson process with rate \(\lambda .\) Each satellite will, independently, orbit the earth for a random time having distribution \(F\). Let \(X(t)\) denote the number of satellites orbiting at time \(t\). (a) Determine \(P\\{X(t)=k\\}\). Hint: Relate this to the \(M / G / \infty\) queue. (b) If at least one satellite is orbiting, then messages can be transmitted and we say that the system is functional. If the first satellite is orbited at time \(t=0\), determine the expected time that the system remains functional. Hint: \(\quad\) Make use of part (a) when \(k=0\).
For the renewal process whose interarrival times are uniformly distributed over \((0,1)\), determine the expected time from \(t=1\) until the next renewal.
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