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Chapter 7: Problem 27

For a renewal process, let \(A(t)\) be the age at time \(t\). Prove that if \(\mu<\infty\), then with probability 1 $$ \frac{A(t)}{t} \rightarrow 0 \quad \text { as } t \rightarrow \infty $$

Short Answer

Expert verified
To prove that if \(\mu < \infty\), then \(\frac{A(t)}{t} \rightarrow 0\) as \(t \rightarrow \infty\) with probability 1, we condition the expected age function on the event that the first interarrival time \(S_1\) is greater than \(t\). After computing the conditional expectations, we find an upper bound for \(E[A(t)]\). Given that \(\mu = E[X]<\infty\), the Strong Law of Large Numbers (SLLN) ensures that the fraction of time the system has been in renewal goes to \(\frac{1}{\mu}\) as \(t\) goes to infinity. Thus, we have \(\lim_{t \to \infty} \frac{A(t)}{t} = 0\) with probability 1, completing the proof.

Step by step solution

01

Define the renewal process and relevant terms

A renewal process is a point process where events occur at random instants of time, and the time between successive events (interarrival times) are independent and identically distributed (i.i.d.) random variables. Let \(A(t)\) represent the age of the system at time \(t\), which is the time elapsed since the last renewal. Additionally, let \(X\) be the random variable representing interarrival times. We are given that the mean time between renewals, \(\mu=E[X]<\infty\).
02

Find the expected value of the age function, E[A(t)]

The expected value of \(A(t)\) is given by: \[ E[A(t)] = \int_0^t (t-s) f_X(t-s) \, ds, \] where \(f_X(t-s)\) is the probability density function of the interarrival time \(X\).
03

Condition on the event {\(t

We will condition on the event that the first interarrival time, \(S_1 > t\): \[ E[A(t)] = E[A(t)|t<S_1]P(t<S_1) + E[A(t)|t\geq S_1]P(t\geq S_1). \]
04

Compute the conditional expectations

For the conditional expectations, we have: \[ E[A(t)|t<S_1] = t, \quad P(t<S_1) = 1 - F_X(t), \] and \[ E[A(t)|t\geq S_1] \leq E[X], \quad P(t\geq S_1) = F_X(t), \] where \(F_X(t)\) is the cumulative distribution function of interarrival time \(X\).
05

Calculate an upper bound for E[A(t)]

Using the previously computed components, we can now calculate an upper bound for \(E[A(t)]\): \[ E[A(t)] \leq t (1-F_X(t)) + E[X]F_X(t). \]
06

Use the Strong Law of Large Numbers (SLLN)

Under the given condition that \(\mu = E[X]<\infty\), the fraction of time the system has been in renewal goes to 1 as \(t\) goes to infinity, i.e., \[ \lim_{t \to \infty} \frac{N(t)}{t}=\frac{1}{\mu}, \] where \(N(t)\) is the number of renewals up to time \(t\). This is a result of the Strong Law of Large Numbers (SLLN).
07

Conclude the proof

Since the fraction of time the system has been in renewal goes to \(\frac{1}{\mu}\) with a probability of 1, we have: \[ \frac{A(t)}{t} = 1 - \frac{N(t)}{t}, \] and taking the limit as \(t \to \infty\), \[ \lim_{t \to \infty} \frac{A(t)}{t} = 0 \quad\text{with probability 1}. \] Thus, we have proved the desired result that if \(\mu < \infty\), then \(\frac{A(t)}{t} \rightarrow 0\) as \(t \rightarrow \infty\) with probability 1.

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Most popular questions from this chapter

For an interarrival distribution \(F\) having mean \(\mu\), we define the equilibrium distribution of \(F\), denoted \(F_{e}\), by $$ F_{\mathrm{e}}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y $$ (a) Show that if \(F\) is an exponential distribution, then \(F=F_{t^{*}}\) (b) If for some constant \(c\), $$ F(x)=\left\\{\begin{array}{ll} 0, & x

To prove Equation (7.24), define the following notation: \(X_{i}^{\prime}=\) time spent in state \(i\) on the \(j\) th visit to this state; \(N_{i}(m)=\) 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 \(\underline{L}\) Argue that, with probability 1 , (c) \(\sum_{j=1}^{N_{1}(m)} \frac{X_{i}^{\prime}}{N_{i}(m)} \rightarrow \mu_{l} \quad\) as \(m \rightarrow \infty\) (d) \(N_{1}(m) / m \rightarrow n_{i}\) as \(m \rightarrow \infty\). (e) Combine parts (a), (b), (c), and (d) to prove Equation (7.24).

If \(A(t)\) and \(Y(t)\) are respectively the age and the excess at time \(t\) of a renewal process having an interarrival distribution \(F\), calculate $$ P[Y(t)>x|A(t)=s| $$

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 \(\mid N=n]\) 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 $$ \left.P\left(X_{i}=1\right]=p=1-P \mid X_{i}=0\right], \quad i \geq 1 $$ 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 cquation, let us define the indicator variables \(I_{i}, i \geq 1\) by $$ I_{i}=\left\\{\begin{array}{ll} 1, & \text { if } i \leq N \\ 0, & \text { if } i>N \end{array}\right. $$ (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: \(I_{1}\) 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)?

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2) ?\) (b) exponentially distributed with mean \(1 ?\)
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