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Chapter 7: Problem 30

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

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 \(1 ?\)

A machine in use is replaced by a new machine either when it fails or when it reaches the age of \(T\) years. If the lifetimes of successive machines are independent with a common distribution \(F\) having density \(f\), show that (a) the long-run rate at which machines are replaced equals $$ \left[\int_{0}^{T} x f(x) d x+T(1-F(T))\right]^{-1} $$ (b) the long-run rate at which machines in use fail equals $$ \frac{F(T)}{\int_{0}^{T} x f(x) d x+T[1-F(T)]} $$

Let \(\left\\{N_{1}(t), t \geqslant 0\right\\}\) and \(\left\\{N_{2}(t), t \geqslant 0\right\\}\) be independent renewal processes. Let \(N(t)=N_{1}(t)+N_{2}(t)\) (a) Are the interarrival times of \(\\{N(t), t \geqslant 0\\}\) independent? (b) Are they identically distributed? (c) Is \(\\{N(t), t \geqslant 0\\}\) a renewal process?

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 $$ P\\{Y(t)>x \mid A(t)=s\\} $$

For a renewal reward process consider $$ W_{n}=\frac{R_{1}+R_{2}+\cdots+R_{n}}{X_{1}+X_{2}+\cdots+X_{n}} $$ where \(W_{n}\) represents the average reward earned during the first \(n\) cycles. Show that \(W_{n} \rightarrow E[R] / E[X]\) as \(n \rightarrow \infty\).
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