Chapter 5

Renewal processes

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Problem 5.1

The purpose of this exercise is to show that for an arbitrary renewal process, N(t)N(t), the number of renewals in (0,t](0, t] is a (non-defective) rv.

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(a)

Let X1,X2,…X_{1}, X_{2}, \ldots be a sequence of IID inter-renewal rv s. Let Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} be the corresponding renewal epochs for each n≥1n \geq 1. Assume that each XiX_{i} has a finite expectation X‾>0\overline{X}>0 and, for any given t>0t>0, use the WLLN to show that lim⁡n→∞P(Sn≤t)=0\lim_{n \rightarrow \infty } \mathbb {P}\left(S_{n} \leq t\right) = 0.

(b)

Use (a) to show that lim⁡n→∞P(N(t)≥n)=0\lim_{n \rightarrow \infty } \mathbb {P}\left(N(t) \geq n\right) = 0 for each t>0t>0 and explain why this means that N(t)N(t) is a rv, i.e., is not defective.

(c)

Now suppose that the XiX_{i} do not have a finite mean. Consider truncating each XiX_{i} to Xˇi\check{X}_{i}, where for any given b>0b>0, Xˇi=min⁡(Xi,b)\check{X}_{i} = \min (X_{i}, b). Let Nˇ(t)\check{N}(t) be the renewal counting process for the inter-renewal intervals Xˇi\check{X}_{i}. Show that Nˇ(t)\check{N}(t) is non-defective for each t>0t>0. Show that N(t)≤Nˇ(t)N(t) \leq \check{N}(t) and thus that N(t)N(t) is non-defective.

Problem 5.2

This exercise shows that, for an arbitrary renewal process, N(t)N(t), the number of renewals in (0,t](0, t], is non-defective and has finite expectation.

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(a)

Let the inter-renewal intervals have the CDF FX(x)\mathsf{F}_{X}(x), with, as usual, FX(0)=0\mathsf{F}_{X}(0) = 0. Using whatever combination of mathematics and common sense is comfortable for you, show that for any ϵ\epsilon, 0<ϵ<10<\epsilon <1, there is a δ>0\delta >0 such that FX(δ)≤1−ϵ\mathsf{F}_{X}(\delta ) \leq 1-\epsilon. In other words, you are to show that a positive rv must lie in some range of positive values bounded away from 0 with positive probability.

(b)

Show that P(Sn≤δ)≤(1−ϵ)n\mathbb {P}\left(S_{n} \leq \delta \right) \leq (1-\epsilon )^{n}.

(c)

Show that E[N(δ)]≤1/ϵ\mathbb {E}\left[N(\delta )\right] \leq 1/\epsilon.

(d)

For the ϵ,δ\epsilon , \delta of (a), show that for every integer kk, E[N(kδ)]≤k/ϵ\mathbb {E}\left[N(k \delta )\right] \leq k/\epsilon and thus that E[N(t)]≤(t+δ)/ϵδ\mathbb {E}\left[N(t)\right] \leq (t+\delta )/\epsilon \delta for any t>0t>0.

(e)

Use the result here to show that N(t)N(t) is non-defective.

Problem 5.3

Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be the inter-renewal intervals of a renewal process generalized to allow for inter-renewal intervals of size 0 and let P(Xi=0)=α\mathbb {P}\left(X_{i} = 0\right) = \alpha, 0<α<10<\alpha <1. Let {Yi;i≥1}\left\{ Y_{i}; i \geq 1\right\} be the sequence of non-zero interarrival intervals. For example, if X1=x1>0X_{1} = x_{1}>0, X2=0X_{2} = 0, X3=x3>0,…X_{3} = x_{3}>0, \ldots, then Y1=x1Y_{1} = x_{1}, Y2=x3,…Y_{2} = x_{3}, \ldots.

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(a)

Find the CDF of each YiY_{i} in terms of that of the XiX_{i}.

(b)

Find the PMF of the number of arrivals of the generalized renewal process at each epoch at which arrivals occur.

(c)

Explain how to view the generalized renewal process as an ordinary renewal process with inter-renewal intervals {Yi;i≥1}\left\{ Y_{i}; i \geq 1\right\} and bulk arrivals at each renewal epoch.

(d)

When a generalized renewal process is viewed as an ordinary renewal process with bulk arrivals, what is the distribution of the bulk arrivals? (The point of this part is to illustrate that bulk arrivals on an ordinary renewal process are considerably more general than generalized renewal processes.)

Problem 5.4

Is it true for a renewal process that:

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(a)

N(t)<nN(t)<n if and only if Sn>tS_{n}>t?

(b)

N(t)≤nN(t) \leq n if and only if Sn≥tS_{n} \geq t?

(c)

N(t)>nN(t)>n if and only if Sn<tS_{n}<t?

Problem 5.5

(This shows that convergence WP1 implies convergence in probability.) Let {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} be a sequence of rv s that converges to 0 WP1. For any positive integers mm and kk, let

A(m,k)={ω:∣Yn(ω)∣≤1/kfor all n≥m}. A(m, k) = \left\{ \omega : \left|Y_{n}(\omega )\right| \leq 1/k \quad \text{for all } n \geq m\right\} .
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(a)

Show that if lim⁡n→∞Yn(ω)=0\lim_{n \rightarrow \infty } Y_{n}(\omega ) = 0 for some given ω\omega, then (for any given kk) ω∈A(m,k)\omega \in A(m, k) for some positive integer mm.

(b)

Show that for all k≥1k \geq 1

P(⋃m=1∞A(m,k))=1. \mathbb {P}\left(\bigcup _{m=1}^{\infty } A(m, k)\right) = 1.
(c)

Show that, for all m≥1m \geq 1, A(m,k)⊆A(m+1,k)A(m, k) \subseteq A(m+1, k). Use this (plus (1.9)) to show that

lim⁡m→∞P(A(m,k))=1. \lim _{m \rightarrow \infty } \mathbb {P}\left(A(m, k)\right) = 1.
(d)

Show that if ω∈A(m,k)\omega \in A(m, k), then ∣Ym(ω)∣≤1/k\left|Y_{m}(\omega )\right| \leq 1/k. Use this (plus (c)) to show that

lim⁡m→∞P(∣Ym∣>1/k)=0. \lim _{m \rightarrow \infty } \mathbb {P}\left(\left|Y_{m}\right|>1/k\right) = 0.

Since k≥1k \geq 1 is arbitrary, this shows that {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} converges in probability.

Problem 5.6

A town starts a mosquito control program and the rv ZnZ_{n} is the number of mosquitoes at the end of the nnth year (n=0,1,2,…n = 0, 1, 2, \ldots). Let XnX_{n} be the growth rate of the mosquito population in year nn; i.e., Zn=XnZn−1Z_{n} = X_{n} Z_{n-1}; n≥1n \geq 1. Assume that {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} is a sequence of IID rv s with the PMF P(X=2)=1/2\mathbb {P}\left(X = 2\right) = 1/2; P(X=1/2)=1/4\mathbb {P}\left(X = 1/2\right) = 1/4; P(X=1/4)=1/4\mathbb {P}\left(X = 1/4\right) = 1/4. Suppose that Z0Z_{0}, the initial number of mosquitoes, is some known constant and assume for simplicity and consistency that ZnZ_{n} can take on non-integer values.

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(a)

Find E[Zn]\mathbb {E}\left[Z_{n}\right] as a function of nn and find lim⁡n→∞E[Zn]\lim_{n \rightarrow \infty } \mathbb {E}\left[Z_{n}\right].

(b)

Let Wn=log⁡2XnW_{n} = \log_{2} X_{n}. Find E[Wn]\mathbb {E}\left[W_{n}\right] and E[log⁡2(Zn/Z0)]\mathbb {E}\left[\log_{2}(Z_{n}/Z_{0})\right] as a function of nn.

(c)

There is a constant α\alpha such that lim⁡n→∞(1/n)[log⁡2(Zn/Z0)]=α\lim_{n \rightarrow \infty }(1/n)[\log_{2}(Z_{n}/Z_{0})] = \alpha WP1. Find α\alpha and explain how this follows from the SLLN.

(d)

Using (c), show that lim⁡n→∞Zn=β\lim_{n \rightarrow \infty } Z_{n} = \beta WP1 for some β\beta and evaluate β\beta.

(e)

Explain carefully how the result in (a) and the result in (d) are compatible. What you should learn from this problem is that the expected value of the log of a product of IID rv s might be more significant than the expected value of the product itself.

Problem 5.7

In this exercise, you will find an explicit expression for {ω:lim⁡nYn(ω)=0}\left\{ \omega : \lim_{n} Y_{n}(\omega ) = 0\right\}. You need not be mathematically precise.

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(a)

Let {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} be a sequence of rv s. Using the definition of convergence for a sequence of numbers, justify the following set equivalences:

{ω:lim⁡nYn(ω)=0}=⋂k=1∞{ω:there exists an m such that ∣Yn(ω)∣≤1/k for all n≥m}=⋂k=1∞⋃m=1∞{ω:Yn(ω)≤1/k for all n≥m}=⋂k=1∞⋃m=1∞⋂n=m∞{ω:Yn(ω)≤1/k}. \begin{aligned} \left\{ \omega : \lim _{n} Y_{n}(\omega ) = 0\right\} & = \bigcap _{k=1}^{\infty }\left\{ \omega : \text{there exists an } m \text{ such that } \left|Y_{n}(\omega )\right| \leq 1/k \text{ for all } n \geq m\right\} \\ & = \bigcap _{k=1}^{\infty } \bigcup _{m=1}^{\infty }\left\{ \omega : Y_{n}(\omega ) \leq 1/k \text{ for all } n \geq m\right\} \\ & = \bigcap _{k=1}^{\infty } \bigcup _{m=1}^{\infty } \bigcap _{n=m}^{\infty }\left\{ \omega : Y_{n}(\omega ) \leq 1/k\right\} . \end{aligned}
(b)

Explain how this shows that {ω:lim⁡nYn(ω)=0}\left\{ \omega : \lim_{n} Y_{n}(\omega ) = 0\right\} must be an event.

(c)

Use De Morgan's laws to show that the complement of the above equivalence is

{ω:lim⁡nYn(ω)=0}c=⋃k=1∞⋂m=1∞⋃n=m∞{ω:Yn(ω)>1/k}. \left\{ \omega : \lim _{n} Y_{n}(\omega ) = 0\right\} ^{c} = \bigcup _{k=1}^{\infty } \bigcap _{m=1}^{\infty } \bigcup _{n=m}^{\infty }\left\{ \omega : Y_{n}(\omega )>1/k\right\} .
(d)

Show that for {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} to converge WP1, it is necessary and sufficient to satisfy

P(⋂m=1∞⋃n=m∞{Yn>1/k})=0for all k≥1. \mathbb {P}\left(\bigcap _{m=1}^{\infty } \bigcup _{n=m}^{\infty }\left\{ Y_{n}>1/k\right\} \right) = 0 \qquad \text{for all } k \geq 1.
(e)

Show that for {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} to converge WP1, it is necessary and sufficient to satisfy

lim⁡m→∞P(⋃n=m∞{Yn>1/k})=0for all k≥1. \lim _{m \rightarrow \infty } \mathbb {P}\left(\bigcup _{n=m}^{\infty }\left\{ Y_{n}>1/k\right\} \right) = 0 \qquad \text{for all } k \geq 1.
Problem 5.8

Consider the event ⋂m≥1⋃n≥mAn\bigcap_{m \geq 1} \bigcup_{n \geq m} A_{n} where A1,A2,…A_{1}, A_{2}, \ldots are arbitrary events.

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(a)

Show that

lim⁡m→∞P(⋃n≥mAn)=0⟺P(⋂m≥1⋃n≥mAn)=0. \lim _{m \rightarrow \infty } \mathbb {P}\left(\bigcup _{n \geq m} A_{n}\right) = 0 \qquad \Longleftrightarrow \qquad \mathbb {P}\left(\bigcap _{m \geq 1} \bigcup _{n \geq m} A_{n}\right) = 0.
(b)

Show that if ∑m=1∞P(Am)<∞\sum_{m=1}^{\infty } \mathbb {P}\left(A_{m}\right)<\infty, then lim⁡m→∞P(⋃n≥mAn)=0\lim_{m \rightarrow \infty } \mathbb {P}\left(\bigcup_{n \geq m} A_{n}\right) = 0.

(c)

Show that if ∑m=1∞P(Am)<∞\sum_{m=1}^{\infty } \mathbb {P}\left(A_{m}\right)<\infty, then P(⋂m≥1⋃n≥mAn)=0\mathbb {P}\left(\bigcap_{m \geq 1} \bigcup_{n \geq m} A_{n}\right) = 0. This well-known result is called the Borel–Cantelli lemma.

The set {⋂m⋃n≥mAn}\left\{ \bigcap_{m} \bigcup_{n \geq m} A_{n}\right\} is often referred to as the set of ω\omega that are contained in infinitely many of the AnA_{n}. Without trying to be precise about what this latter statement means, explain why it is a good way to think about {⋂m⋃n≥mAn}\left\{ \bigcap_{m} \bigcup_{n \geq m} A_{n}\right\}.

Problem 5.9

Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be the inter-renewal intervals of a renewal process and assume that E[Xi]=∞\mathbb {E}\left[X_{i}\right] = \infty. Let b>0b>0 be an arbitrary number and Xˇi\check{X}_{i} be a truncated random variable defined by Xˇi=Xi\check{X}_{i} = X_{i} if Xi≤bX_{i} \leq b and Xˇi=b\check{X}_{i} = b otherwise.

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(a)

Show that for any constant M>0M>0, there is a bb sufficiently large so that E[Xˇi]≥M\mathbb {E}\left[\check{X}_{i}\right] \geq M.

(b)

Let {Nˇ(t);t≥0}\left\{ \check{N}(t); t \geq 0\right\} be the renewal counting process with inter-renewal intervals {Xˇi;i≥1}\left\{ \check{X}_{i}; i \geq 1\right\} and show that for all t>0t>0, Nˇ(t)≥N(t)\check{N}(t) \geq N(t).

(c)

Show that for all sample functions N(t,ω)N(t, \omega ), except a set of probability 0, N(t,ω)/t≤2/MN(t, \omega )/t \leq 2/M for all sufficiently large tt. Note: Since MM is arbitrary, this means that lim⁡t→∞N(t)/t=0\lim_{t \rightarrow \infty } N(t)/t = 0 with probability 1.

Problem 5.10

The first part of the proof of the SLLN showed that if E[X4]<∞\mathbb {E}\left[X^{4}\right]<\infty, then E[X2]<∞\mathbb {E}\left[X^{2}\right]<\infty. Here we generalize this to show that for 1≤m<n1 \leq m<n, if E[∣Xn∣<∞]\mathbb {E}\left[\left|X^{n}\right|<\infty \right], then E[∣Xm∣<∞]\mathbb {E}\left[\left|X^{m}\right|<\infty \right].

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(a)

For any x>0x>0, show that xm≤1+xnx^{m} \leq 1+x^{n}.

(b)

For any rv XX, show that if E[∣Xn∣]<∞\mathbb {E}\left[\left|X^{n}\right|\right]<\infty, then E[∣Xm∣]≤1+E[∣Xn∣]\mathbb {E}\left[\left|X^{m}\right|\right] \leq 1+\mathbb {E}\left[\left|X^{n}\right|\right].

Problem 5.11

Let Y(t)=SN(t)+1−tY(t) = S_{N(t)+1}-t be the residual life at time tt of a renewal process. First consider a renewal process in which the interarrival time has density fX(x)=e−x\mathsf{f}_{X}(x) = e^{-x}; x≥0x \geq 0, and next consider a renewal process with density

fX(x)=3(x+1)4;x≥0. \mathsf{f}_{X}(x) = \frac{3}{(x+1)^{4}}; \qquad x \geq 0.

For each of the above densities, use renewal-reward theory to find:

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(a)

the time average of Y(t)Y(t);

(b)

the time average of Y2(t)Y^{2}(t) (i.e., lim⁡T→∞(1/T)∫0TY2(t) dt\lim_{T \rightarrow \infty }(1/T) \int_{0}^{T} Y^{2}(t) \, dt).

For the exponential density, verify your answers by finding E[Y(t)]\mathbb {E}\left[Y(t)\right] and E[Y2(t)]\mathbb {E}\left[Y^{2}(t)\right] directly.

Problem 5.12

Consider a variation of an M/G/1 queueing system in which there is no facility to save waiting customers. Assume customers arrive according to a Poisson process of rate λ\lambda. If the server is busy, the customer departs and is lost forever; if the server is not busy, the customer enters service with a service time CDF denoted by FY(y)\mathsf{F}_{Y}(y).

Successive service times (for those customers that are served) are IID and independent of arrival times. Assume that customer number 0 arrives and enters service at time t=0t = 0.

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(a)

Show that the sequence of times S1,S2,…S_{1}, S_{2}, \ldots at which successive successful customers enter service are the renewal times of a renewal process. Show that each inter-renewal interval Xi=Si−Si−1X_{i} = S_{i}-S_{i-1} (where S0=0S_{0} = 0) is the sum of two independent rv s, Yi+UiY_{i}+U_{i} where YiY_{i} is the iith service time; find the probability density of UiU_{i}.

(b)

Assume that a reward (actually a cost in this case) of one unit is incurred for each customer turned away. Sketch the expected reward function as a function of time for the sample function of inter-renewal intervals and service intervals shown below; the expectation is to be taken over those (unshown) arrivals of customers that must be turned away.  {#fig-1 width="70%"}

(c)

Let ∫0tR(τ) dτ\int_{0}^{t} R(\tau ) \, d\tau denote the accumulated reward (i.e., cost) from 0 to tt and find the limit as t→∞t \rightarrow \infty of (1/t)∫0tR(τ) dτ(1/t) \int_{0}^{t} R(\tau ) \, d\tau. Explain (without any attempt to be rigorous or formal) why this limit exists with probability 1.

(d)

In the limit of large tt, find the expected reward from time tt until the next renewal. Hint: Sketch this expected reward as a function of tt for a given sample of inter-renewal intervals and service intervals; then find the time average.

(e)

Now assume that the arrivals are deterministic, with the first arrival at time 0 and the nnth arrival at time n−1n-1. Does the sequence of times S1,S2,…S_{1}, S_{2}, \ldots at which subsequent customers start service still constitute the renewal times of a renewal process? Draw a sketch of arrivals, departures, and service time intervals. Again find lim⁡t→∞(∫0tR(τ) dτ)/t\lim_{t \rightarrow \infty }\left(\int_{0}^{t} R(\tau ) \, d\tau \right)/t.

Problem 5.13

Let Z(t)=t−SN(t)Z(t) = t-S_{N(t)} be the age of a renewal process and Y(t)=SN(t)+1−tY(t) = S_{N(t)+1}-t be the residual life. Let FX(x)\mathsf{F}_{X}(x) be the CDF of the inter-renewal interval and find the following as a function of FX(x)\mathsf{F}_{X}(x):

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(a)

P(Y(t)>x∣Z(t)=s)\mathbb {P}\left(Y(t)>x \mid Z(t) = s\right);

(b)

P(Y(t)>x∣Z(t+x/2)=s)\mathbb {P}\left(Y(t)>x \mid Z(t+x/2) = s\right);

(c)

P(Y(t)>x∣Z(t+x)>s)\mathbb {P}\left(Y(t)>x \mid Z(t+x)>s\right) for a Poisson process.

Problem 5.14

Let FZ(z)\mathsf{F}_{Z}(z) be the fraction of time (over the limiting interval (0,∞)(0, \infty )) that the age of a renewal process is at most zz. Show that FZ(z)\mathsf{F}_{Z}(z) satisfies

FZ(z)=1X‾∫x=0zP(X>x) dxWP1. \mathsf{F}_{Z}(z) = \frac{1}{\overline{X}} \int _{x=0}^{z} \mathbb {P}\left(X>x\right) \, dx \qquad \text{WP1.}
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Problem 5.15
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(a)

Let JJ be a stopping rule and I{J≥n}\mathbb {I}_{\left\{ J \geq n\right\} } be the indicator rv of the event {J≥n}\left\{ J \geq n\right\}. Show that J=∑n≥1I{J≥n}J = \sum_{n \geq 1} \mathbb {I}_{\left\{ J \geq n\right\} }.

(b)

Show that I{J≥1}≥I{J≥2}≥…\mathbb {I}_{\left\{ J \geq 1\right\} } \geq \mathbb {I}_{\left\{ J \geq 2\right\} } \geq \ldots, i.e., show that for each n≥1n \geq 1, I{J≥n}(ω)≥I{J≥n+1}(ω)\mathbb {I}_{\left\{ J \geq n\right\} }(\omega ) \geq \mathbb {I}_{\left\{ J \geq n+1\right\} }(\omega ) for each ω∈Ω\omega \in \Omega (except perhaps for a set of probability 0).

Problem 5.16
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(a)

Use Wald's equality to compute the expected number of trials of a Bernoulli process up to and including the kkth success.

(b)

Use elementary means to find the expected number of trials up to and including the first success. Use this to find the expected number of trials up to and including the kkth success. Compare with (a).

Problem 5.17

A gambler with an initial finite capital of d>0d>0 dollars starts to play a dollar slot machine. At each play, either his dollar is lost or is returned with some additional number of dollars. Let XiX_{i} be his change of capital on the iith play. Assume that {Xi;i=1,2,…}\left\{ X_{i}; i = 1, 2, \ldots \right\} is a set of IID rv s taking on integer values {−1,0,1,…}\left\{ -1, 0, 1, \ldots \right\}. Assume that E[Xi]<0\mathbb {E}\left[X_{i}\right]<0. The gambler plays until losing all his money (i.e., the initial dd dollars plus subsequent winnings).

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(a)

Let JJ be the number of plays until the gambler loses all his money. Is the weak law of large numbers sufficient to argue that lim⁡n→∞P(J>n)=0\lim_{n \rightarrow \infty } \mathbb {P}\left(J>n\right) = 0 (i.e., that JJ is a random variable) or is the strong law necessary?

(b)

Find E[J]\mathbb {E}\left[J\right].

Problem 5.18

Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be IID binary random variables with PX(0)=PX(1)=1/2\mathsf{P}_{X}(0) = \mathsf{P}_{X}(1) = 1/2. Let JJ be a positive integer-valued random variable defined on the above sample space of binary sequences and let SJ=∑i=1JXiS_{J} = \sum_{i=1}^{J} X_{i}. Find the simplest example you can in which JJ is not a stopping trial for {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} and where E[X]E[J]≠E[SJ]\mathbb {E}\left[X\right] \mathbb {E}\left[J\right] \neq \mathbb {E}\left[S_{J}\right].

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Problem 5.19

Let J=min⁡{n∣Sn≤b or Sn≥a}J = \min \left\{ n \mid S_{n} \leq b \text{ or } S_{n} \geq a\right\}, where aa is a positive integer, bb is a negative integer, and Sn=X1+X2+⋯+XnS_{n} = X_{1}+X_{2}+\cdots +X_{n}. Assume that {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} is a set of zero-mean IID rv s that can take on only the set of values {−1,0,+1}\left\{ -1, 0, +1\right\}, each with positive probability.

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(a)

Is JJ a stopping rule? Why or why not?

(b)

What are the possible values of SJS_{J}?

(c)

Find an expression for E[SJ]\mathbb {E}\left[S_{J}\right] in terms of pp, aa, and bb, where p=P(SJ≥a)p = \mathbb {P}\left(S_{J} \geq a\right).

(d)

Find an expression for E[SJ]\mathbb {E}\left[S_{J}\right] from Wald's equality. Use this to solve for pp.

Problem 5.20

Show that the interchange of expectation and sum in

E[SJ]=E[∑n=1∞XnI{J≥n}]=∑n=1∞E[XnI{J≥n}](5.34) \mathbb {E}\left[S_{J}\right] = \mathbb {E}\left[\sum _{n=1}^{\infty } X_{n} \mathbb {I}_{\left\{ J \geq n\right\} }\right] = \sum _{n=1}^{\infty } \mathbb {E}\left[X_{n} \mathbb {I}_{\left\{ J \geq n\right\} }\right] \tag {5.34}

(from the proof of Wald's equality, Theorem 5.5.3) is valid if E[J]<∞\mathbb {E}\left[J\right]<\infty.

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Problem 5.21

Consider a G/G/1 queue with inter-arrival sequence {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} and service time sequence {Vi;i≥0}\left\{ V_{i}; i \geq 0\right\}. Assume that E[X]<∞\mathbb {E}\left[X\right]<\infty and that E[V]<E[X]\mathbb {E}\left[V\right]<\mathbb {E}\left[X\right].

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(a)

Let Zn=∑i=1n(Xi−Vi−1)Z_{n} = \sum_{i=1}^{n}(X_{i}-V_{i-1}) and let InI_{n} be the sum of the intervals between 0 and Sn=∑i=1nXiS_{n} = \sum_{i=1}^{n} X_{i} during which the server is empty. Show that In≥ZnI_{n} \geq Z_{n} and explain the difference between them. Show that if arrival nn enters an empty system, then In=ZnI_{n} = Z_{n}.

(b)

Assume that the inter-arrival distribution is bounded in the sense that for some b<∞b<\infty, P(X>b)=0\mathbb {P}\left(X>b\right) = 0. Show that the duration of each idle period for the G/G/1 queue has a duration less than or equal to bb.

(c)

Show that the number of idle intervals for the server up to time SnS_{n} is at least Zn/bZ_{n}/b. Show that the number of renewals up to time SnS_{n} (Nr(Sn)N^{r}(S_{n})) at which an arrival sees an empty system is at least (Zn/b)(Z_{n}/b).

(d)

Show that the renewal counting process satisfies lim⁡t→∞Nr(t)/t=a\lim_{t \rightarrow \infty } N^{r}(t)/t = a WP1 for some a>0a>0. Show that the mean number of arrivals per arrival to an empty system is 1/a<∞1/a<\infty.

(e)

Assume everything above except to remove the assumption that P(X>b)=0\mathbb {P}\left(X>b\right) = 0 for some bb. Use the truncation method on XX to establish (d) for this new case.

Problem 5.22

A company has three promising approaches to the design of a new wireless application. Unknown to the company, the first approach requires two weeks, the second, four weeks, and the third eight weeks, but only the first will be successful. The company hires one programmer after another for the design, stopping when a design is successful. Let XiX_{i} be the time taken by the iith programmer, assuming that each programmer independently chooses an approach with equal probability (i.e., pXi(x)=1/3\mathsf{p}_{X_{i}}(x) = 1/3 for x=2,4x = 2, 4, and 8). Let JJ be the number of the first programmer who is successful and let T=∑i=1JXiT = \sum_{i=1}^{J} X_{i} be the time of the first success.

?
(a)

Use Wald's equality to find E[T]\mathbb {E}\left[T\right].

(b)

Compute E[∑i=1JXi∣J=n]\mathbb {E}\left[\sum_{i=1}^{J} X_{i} \mid J = n\right] and show that it is not equal to E[∑i=1nXi]\mathbb {E}\left[\sum_{i=1}^{n} X_{i}\right].

(c)

Use (b) for a second derivation of E[T]\mathbb {E}\left[T\right].

(d)

Find E[T]\mathbb {E}\left[T\right] if the company tells successive programmers which approaches have been tried unsuccessfully earlier.

Problem 5.23
?
(a)

Consider a renewal process for which the inter-renewal intervals have the PMF pX(1)=pX(2)=1/2\mathsf{p}_{X}(1) = \mathsf{p}_{X}(2) = 1/2. Let m(t)=E[N(t)]m(t) = \mathbb {E}\left[N(t)\right] and use elementary combinatorics to show that m(1)=1/2m(1) = 1/2, m(2)=5/4m(2) = 5/4, and m(3)=15/8m(3) = 15/8.

(b)

Use elementary means to show that E[SN(1)]=1/2\mathbb {E}\left[S_{N(1)}\right] = 1/2 and E[SN(1)+1]=9/4\mathbb {E}\left[S_{N(1)+1}\right] = 9/4. Verify (5.37) in this case (i.e., for t=1t = 1) and show why N(1)N(1) is not a stopping trial. Note also that the expected duration, E[SN(1)+1−SN(1)]\mathbb {E}\left[S_{N(1)+1}-S_{N(1)}\right] is not equal to X‾\overline{X}.

(c)

Generalize (a) so that P(X=1)=1−p\mathbb {P}\left(X = 1\right) = 1-p and P(X=2)=p\mathbb {P}\left(X = 2\right) = p. Let Wn=N(n)−N(n−1)W_{n} = N(n)-N(n-1), i.e., WnW_{n} is 1 if an arrival occurs at time nn and is 0 if no arrival occurs at nn. Show that W‾n\overline{W}_{n} satisfies the difference equation W‾n=1−pW‾n−1\overline{W}_{n} = 1-p \overline{W}_{n-1} for n≥1n \geq 1 where by convention W‾0=1\overline{W}_{0} = 1. Use this to show that

W‾n=1−(−p)n+11+p.(A.91) \overline{W}_{n} = \frac{1-(-p)^{n+1}}{1+p}. \tag {A.91}

From this, solve for m(n)m(n) for n≥1n \geq 1 and verify the result in (a).

Problem 5.24

Let {N(t);t>0}\left\{ N(t); t>0\right\} be a renewal counting process generalized to allow for inter-renewal intervals {Xi}\left\{ X_{i}\right\} of duration 0. Let each XiX_{i} have the PMF P(Xi=0)=1−ϵ\mathbb {P}\left(X_{i} = 0\right) = 1-\epsilon; P(Xi=1/ϵ)=ϵ\mathbb {P}\left(X_{i} = 1/\epsilon \right) = \epsilon.

?
(a)

Sketch a typical sample function of {N(t);t>0}\left\{ N(t); t>0\right\}. Note that N(0)N(0) can be non-zero (i.e., N(0)N(0) is the number of zero interarrival times that occur before the first non-zero interarrival time).

(b)

Evaluate E[N(t)]\mathbb {E}\left[N(t)\right] as a function of tt.

(c)

Sketch E[N(t)]/t\mathbb {E}\left[N(t)\right]/t as a function of tt.

(d)

Evaluate E[SN(t)+1]\mathbb {E}\left[S_{N(t)+1}\right] as a function of tt (do this directly, and then use Wald's equality as a check on your work).

(e)

Sketch the lower bound E[N(t)]/t≥1/E[X]−1/t\mathbb {E}\left[N(t)\right]/t \geq 1/\mathbb {E}\left[X\right]-1/t on the same graph with (c).

(f)

Sketch E[SN(t)+1−t]\mathbb {E}\left[S_{N(t)+1}-t\right] as a function of tt and find the time average of this quantity.

(g)

Evaluate E[SN(t)]\mathbb {E}\left[S_{N(t)}\right] as a function of tt; verify that E[SN(t)]≠E[X]E[N(t)]\mathbb {E}\left[S_{N(t)}\right] \neq \mathbb {E}\left[X\right] \mathbb {E}\left[N(t)\right].

Problem 5.25

Let {N(t);t>0}\left\{ N(t); t>0\right\} be a renewal counting process and let m(t)=E[N(t)]m(t) = \mathbb {E}\left[N(t)\right] be the expected number of arrivals up to and including time tt. Let {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} be the sequence of inter-renewal intervals and assume that FX(0)=0\mathsf{F}_{X}(0) = 0.

?
(a)

For all x>0x>0 and t>xt>x show that E[N(t)∣X1=x]=E[N(t−x)]+1\mathbb {E}\left[N(t) \mid X_{1} = x\right] = \mathbb {E}\left[N(t-x)\right]+1.

(b)

Use (a) to show that m(t)=FX(t)+∫0tm(t−x) dFX(x)m(t) = \mathsf{F}_{X}(t)+\int_{0}^{t} m(t-x) \, d\mathsf{F}_{X}(x) for t>0t>0. This equation is the renewal equation, derived differently in (5.55).

(c)

Suppose that XX is an exponential rv of parameter λ\lambda. Evaluate Lm(s)L_{m}(s) from (5.57); verify that the inverse Laplace transform is λt\lambda t; t≥0t \geq 0.

Problem 5.26
?
(a)

Let the inter-renewal interval of a renewal process have a second-order Erlang density, fX(x)=λ2xexp⁡(−λx)\mathsf{f}_{X}(x) = \lambda^{2} x \exp (-\lambda x). Evaluate the Laplace transform of m(t)=E[N(t)]m(t) = \mathbb {E}\left[N(t)\right].

(b)

Use this to evaluate m(t)m(t) for t≥0t \geq 0. Verify that your answer agrees with (5.59).

(c)

Evaluate the slope of m(t)m(t) at t=0t = 0 and explain why that slope is not surprising.

(d)

View the renewals here as being the even numbered arrivals in a Poisson process of rate λ\lambda. Sketch m(t)m(t) for the process here and show one half the expected number of arrivals for the Poisson process on the same sketch. Explain the difference between the two.

Problem 5.27
?
(a)

Let N(t)N(t) be the number of arrivals in the interval (0,t](0, t] for a Poisson process of rate λ\lambda. Show that the probability that N(t)N(t) is even is [1+exp⁡(−2λt)]/2[1+\exp (-2 \lambda t)]/2.

(b)

Let N~(t)\widetilde{N}(t) be the number of even numbered arrivals in (0,t](0, t]. Show that N~(t)=N(t)/2−Iodd(t)/2\widetilde{N}(t) = N(t)/2-\mathbb {I}_{\mathrm{odd}}(t)/2 where Iodd(t)\mathbb {I}_{\mathrm{odd}}(t) is a random variable that is 1 if N(t)N(t) is odd and 0 otherwise.

(c)

Use (a) and (b) to find E[N~(t)]\mathbb {E}\left[\widetilde{N}(t)\right]. Note that this is m(t)m(t) for a renewal process with second-order Erlang inter-renewal intervals.

Problem 5.28
?
(a)

Consider a function r(z)≥0r(z) \geq 0 defined as follows for 0≤z<∞0 \leq z<\infty: For each integer n≥1n \geq 1 and each integer k,1≤k<nk, 1 \leq k<n, r(z)=1r(z) = 1 for n+k/n≤z≤n+k/n+2−nn+k/n \leq z \leq n+k/n+2^{-n}. For all other zz, r(z)=0r(z) = 0. Sketch this function and show that r(z)r(z) is not directly Riemann integrable.

(b)

Evaluate the Riemann integral ∫0∞r(z) dz\int_{0}^{\infty } r(z) \, dz.

(c)

Suppose r(z)r(z) is decreasing, i.e., that r(z)≥r(y)r(z) \geq r(y) for all y>z>0y>z>0. Show that if r(z)r(z) is Riemann integrable, it is also directly Riemann integrable.

(d)

Suppose f(z)≥0f(z) \geq 0, defined for z≥0z \geq 0, is decreasing and Riemann integrable and that f(z)≥r(z)f(z) \geq r(z) for all z≥0z \geq 0. Show that r(z)r(z) is Directly Riemann integrable.

(e)

Let XX be a nonnegative rv for which E[X2]<∞\mathbb {E}\left[X^{2}\right]<\infty. Show that xFXc(x)x \mathsf{F}_{X}^{c}(x) is directly Riemann integrable.

Problem 5.29

Let Z(t),Y(t),X~(t)Z(t), Y(t), \widetilde{X}(t) denote the age, residual life, and duration at time tt for a renewal counting process {N(t);t>0}\left\{ N(t); t>0\right\} in which the interarrival time has a density given by f(x)f(x). Find the following probability densities; assume steady state:

?
(a)

fY(t)(y∣Z(t+s/2)=s)\mathsf{f}_{Y(t)}(y \mid Z(t+s/2) = s) for given s>0s>0;

(b)

fY(t),Z(t)(y,z)\mathsf{f}_{Y(t), Z(t)}(y, z);

(c)

fY(t)∣X~(t)(y∣x)\mathsf{f}_{Y(t) \mid \widetilde{X}(t)}(y \mid x);

(d)

fZ(t)∣Y(t−s/2)(z∣s)\mathsf{f}_{Z(t) \mid Y(t-s/2)}(z \mid s) for given s>0s>0;

(e)

fY(t)(y∣Z(t+s/2)≥s)\mathsf{f}_{Y(t)}(y \mid Z(t+s/2) \geq s) for given s>0s>0.

Problem 5.30
?
(a)

Find lim⁡t→∞{E[N(t)]−t/X‾}\lim_{t \rightarrow \infty }\left\{ \mathbb {E}\left[N(t)\right]-t/\overline{X}\right\} for a renewal counting process {N(t);t>0}\left\{ N(t); t>0\right\} with inter-renewal times {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\}.

(b)

Evaluate your result for the case in which XX is an exponential rv (you already know what the result should be in this case).

(c)

Evaluate your result for a case in which E[X]<∞\mathbb {E}\left[X\right]<\infty and E[X2]=∞\mathbb {E}\left[X^{2}\right] = \infty. Explain (very briefly) why this does not contradict the elementary renewal theorem.

Problem 5.31

Customers arrive at a bus stop according to a Poisson process of rate λ\lambda. Independently, buses arrive according to a renewal process with the inter-renewal interval CDF FX(x)\mathsf{F}_{X}(x). At the epoch of a bus arrival, all waiting passengers enter the bus and the bus leaves immediately. Let R(t)R(t) be the number of customers waiting at time tt.

?
(a)

Draw a sketch of a sample function of R(t)R(t).

(b)

Given that the first bus arrives at time X1=xX_{1} = x, find the expected number of customers picked up; then find E[∫0xR(t) dt]\mathbb {E}\left[\int_{0}^{x} R(t) \, dt\right], again given the first bus arrival at X1=xX_{1} = x.

(c)

Find lim⁡t→∞1t∫0tR(τ) dτ\lim_{t \rightarrow \infty } \frac{1}{t} \int_{0}^{t} R(\tau ) \, d\tau (with probability 1). Assuming that FX\mathsf{F}_{X} is a non-arithmetic distribution, find lim⁡t→∞E[R(t)]\lim_{t \rightarrow \infty } \mathbb {E}\left[R(t)\right]. Interpret what these quantities mean.

(d)

Use (c) to find the time-average expected wait per customer.

(e)

Find the fraction of time that there are no customers at the bus stop. (Hint: this part is independent of (a), (b), and (c); check your answer for E[X]≪1/λ\mathbb {E}\left[X\right] \ll 1/\lambda.)

Problem 5.32

Consider the same setup as in Exercise 5.31 except that now customers arrive according to a non-arithmetic renewal process independent of the bus arrival process. Let 1/λ1/\lambda be the expected inter-renewal interval for the customer renewal process. Assume that both renewal processes are in steady state (i.e., either we look only at t≫0t \gg 0, or we assume that they are equilibrium processes). Given that the nnth customer arrives at time tt, find the expected wait for customer nn. Find the expected wait for customer nn without conditioning on the arrival time.

?
Problem 5.33

Let {N1(t);t>0}\left\{ N_{1}(t); t>0\right\} be a Poisson counting process of rate λ\lambda. Assume that the arrivals from this process are switched on and off by arrivals from a non-arithmetic renewal counting process {N2(t);t>0}\left\{ N_{2}(t); t>0\right\} (see figure below). The two processes are independent.  {#fig-1 width="80%"} Let {NA(t);t≥0}\left\{ N_{A}(t); t \geq 0\right\} be the switched process; that is NA(t)N_{A}(t) includes arrivals from {N1(t);t>0}\left\{ N_{1}(t); t>0\right\} while N2(t)N_{2}(t) is even and excludes arrivals from {N1(t);t>0}\left\{ N_{1}(t); t>0\right\} while N2(t)N_{2}(t) is odd.

?
(a)

Is NA(t)N_{A}(t) a renewal counting process? Explain your answer and if you are not sure, look at several examples for N2(t)N_{2}(t).

(b)

Find lim⁡t→∞1tNA(t)\lim_{t \rightarrow \infty } \frac{1}{t} N_{A}(t) and explain why the limit exists with probability 1. Hint: Use symmetry -- i.e., look at N1(t)−NA(t)N_{1}(t)-N_{A}(t). To show why the limit exists, use the renewal-reward theorem. What is the appropriate renewal process to use here?

(c)

Now suppose that {N1(t);t>0}\left\{ N_{1}(t); t>0\right\} is a non-arithmetic renewal counting process but not a Poisson process and let the expected inter-renewal interval be 1/λ1/\lambda. For any given δ\delta, find lim⁡t→∞E[NA(t+δ)−NA(t)]\lim_{t \rightarrow \infty } \mathbb {E}\left[N_{A}(t+\delta )-N_{A}(t)\right] and explain your reasoning. Why does your argument in (b) fail to demonstrate a time average for this case?

Problem 5.34

An M/G/1 queue has arrivals at rate λ\lambda and a service time distribution given by FY(y)\mathsf{F}_{Y}(y). Assume that λ<1/E[Y]\lambda <1/\mathbb {E}\left[Y\right]. Epochs at which the system becomes empty define a renewal process. Let FZ(z)\mathsf{F}_{Z}(z) be the CDF of the inter-renewal intervals and let E[Z]\mathbb {E}\left[Z\right] be the mean inter-renewal interval.

?
(a)

Find the fraction of time that the system is empty as a function of λ\lambda and E[Z]\mathbb {E}\left[Z\right]. State carefully what you mean by such a fraction.

(b)

Apply Little's theorem, not to the system as a whole, but to the number of customers in the server (i.e., 0 or 1). Use this to find the fraction of time that the server is busy.

(c)

Combine your results in (a) and (b) to find E[Z]\mathbb {E}\left[Z\right] in terms of λ\lambda and E[Y]\mathbb {E}\left[Y\right]; give the fraction of time that the system is idle in terms of λ\lambda and E[Y]\mathbb {E}\left[Y\right].

(d)

Find the expected duration of a busy period.

Problem 5.35

Consider a sequence X1,X2,…X_{1}, X_{2}, \ldots of IID binary rv s with P(Xn=1)=p1\mathbb {P}\left(X_{n} = 1\right) = p_{1} and P(Xn=0)=p0=1−p1\mathbb {P}\left(X_{n} = 0\right) = p_{0} = 1-p_{1}. A renewal is said to occur at time n≥2n \geq 2 if Xn−1=0X_{n-1} = 0 and Xn=1X_{n} = 1.

?
(a)

Show that {N(n);n>0}\left\{ N(n); n>0\right\} is a renewal counting process where N(n)N(n) is the number of renewals up to and including time nn.

(b)

What is the probability that a renewal occurs at time nn, n≥2n \geq 2?

(c)

Find the expected inter-renewal interval; use Blackwell's theorem.

(d)

Now change the definition of renewal; a renewal now occurs at time nn if Xn−1=1X_{n-1} = 1 and Xn=1X_{n} = 1. Show that {N∗(n);n≥0}\left\{ N^{*}(n); n \geq 0\right\} is a delayed renewal counting process where Nn∗N_{n}^{*} is the number of renewals up to and including nn for this new definition of renewal.

(e)

Find E[Yi]\mathbb {E}\left[Y_{i}\right] for i≥2i \geq 2 for the case in (d).

(f)

Find E[Y1]\mathbb {E}\left[Y_{1}\right] for the case in (d). Hint: Show that E[Y1∣X1=1]=1+E[Y2]\mathbb {E}\left[Y_{1} \mid X_{1} = 1\right] = 1+\mathbb {E}\left[Y_{2}\right] and E[Y1∣X1=0]=1+E[Y1]\mathbb {E}\left[Y_{1} \mid X_{1} = 0\right] = 1+\mathbb {E}\left[Y_{1}\right].

(g)

Looking at your results above for the strings (0,1) and (1,1), show that for an arbitrary string a=(a1,…,ak)\mathbf{a} = (a_{1}, \ldots , a_{k}), the arrival process of successive occurrences of the string is a renewal process if no proper suffix of a\mathbf{a} is a prefix of a\mathbf{a}. Otherwise it is a delayed renewal process.

(h)

Suppose a string a=(a1,…ak)\mathbf{a} = (a_{1}, \ldots a_{k}) of length kk has no proper suffixes equal to a prefix. Show that the time to the first renewal satisfies

E[Y1]=1∏ℓ=1kpaℓ.(A.103) \mathbb {E}\left[Y_{1}\right] = \frac{1}{\prod _{\ell =1}^{k} p_{a_{\ell }}}. \tag {A.103}
(i)

Suppose the string a=(a1,…,ak)\mathbf{a} = (a_{1}, \ldots , a_{k}) has at least one proper suffix equal to a prefix, and suppose ii is the length of the longest such suffix. Show that the expected time until the first occurrence of a\mathbf{a} is given by

E[Y1]=1∏ℓ=1kpaℓ+E[Ui],(A.104) \mathbb {E}\left[Y_{1}\right] = \frac{1}{\prod _{\ell =1}^{k} p_{a_{\ell }}}+\mathbb {E}\left[U_{i}\right], \tag {A.104}

where E[Ui]\mathbb {E}\left[U_{i}\right] is the expected time until the first occurrence of the string (a1,…,ai)(a_{1}, \ldots , a_{i}).

(j)

Show that the expected time until the first occurrence of a=(a1,…,ak)\mathbf{a} = (a_{1}, \ldots , a_{k}) is given by

E[Y1]=∑i=1kIi∏ℓ=1ipaℓ,(A.105) \mathbb {E}\left[Y_{1}\right] = \sum _{i=1}^{k} \frac{\mathbb {I}_{i}}{\prod _{\ell =1}^{i} p_{a_{\ell }}}, \tag {A.105}

where, for 1≤i≤k1 \leq i \leq k, Ii\mathbb {I}_{i} is 1 if the prefix of a\mathbf{a} of length ii is equal to the suffix of length ii.

(k)

Use (i) to find, first, the expected time until the first occurrence of (1,1,1,1,1,1,0)(1,1,1,1,1,1,0) and, second, that of (1,1,1,1,1,1)(1,1,1,1,1,1). Use (4.31) (the recursive first-occurrence-time relation E[time to 111…1]=1+p1E[time to 111…1]+p0E[time to 111…10]\mathbb {E}\left[\text{time to } 111\ldots 1\right] = 1+p_{1} \mathbb {E}\left[\text{time to } 111\ldots 1\right]+p_{0} \mathbb {E}\left[\text{time to } 111\ldots 10\right], applied with a run of six 1s) to check the relationship between these answers.

Problem 5.36

A large system is controlled by nn identical computers. Each computer independently alternates between an operational state and a repair state. The duration of the operational state, from completion of one repair until the next need for repair, is a random variable XX with finite expected duration E[X]\mathbb {E}\left[X\right]. The time required to repair a computer is an exponentially distributed random variable with density λe−λt\lambda e^{-\lambda t}. All operating durations and repair durations are independent. Assume that all computers are in the repair state at time 0.

?
(a)

For a single computer, say the iith, do the epochs at which the computer enters the repair state form a renewal process? If so, find the expected inter-renewal interval.

(b)

Do the epochs at which it enters the operational state form a renewal process?

(c)

Find the fraction of time over which the iith computer is operational and explain what you mean by fraction of time.

(d)

Let Qi(t)Q_{i}(t) be the probability that the iith computer is operational at time tt and find lim⁡t→∞Qi(t)\lim_{t \rightarrow \infty } Q_{i}(t).

(e)

The system is in failure mode at a given time if all computers are in the repair state at that time. Do the epochs at which system failure modes begin form a renewal process?

(f)

Let P(t)\mathbb {P}\left(t\right) be the probability that the the system is in failure mode at time tt. Find lim⁡t→∞P(t)\lim_{t \rightarrow \infty } \mathbb {P}\left(t\right).

(g)

For δ\delta small, find the probability that the system enters failure mode in the interval (t,t+δ](t, t+\delta ] in the limit as t→∞t \rightarrow \infty.

(h)

Find the expected time between successive entries into failure mode.

(i)

Next assume that the repair time of each computer has an arbitrary density rather than exponential, but has a mean repair time of 1/λ1/\lambda. Do the epochs at which system failure modes begin form a renewal process?

(j)

Repeat (f) for the assumption in (i).

Problem 5.37

Let {N1(t);t>0}\left\{ N_{1}(t); t>0\right\} and {N2(t);t>0}\left\{ N_{2}(t); t>0\right\} be independent renewal counting processes. Assume that each has the same CDF F(x)\mathsf{F}(x) for interarrival intervals and assume that a density f(x)f(x) exists for the interarrival intervals.

?
(a)

Is the counting process {N1(t)+N2(t);t>0}\left\{ N_{1}(t)+N_{2}(t); t>0\right\} a renewal counting process? Explain.

(b)

Let Y(t)Y(t) be the interval from tt until the first arrival (from either process) after tt. Find an expression for the CDF of Y(t)Y(t) in the limit t→∞t \rightarrow \infty (you may assume that time averages and ensemble averages are the same).

(c)

Assume that a reward RR of rate 1 unit per second starts to be earned whenever an arrival from process 1 occurs and ceases to be earned whenever an arrival from process 2 occurs. Assume that lim⁡t→∞(1/t)∫0tR(τ) dτ\lim_{t \rightarrow \infty }(1/t) \int_{0}^{t} R(\tau ) \, d\tau exists with probability 1 and find its numerical value.

(d)

Let Z(t)Z(t) be the interval from tt until the first time after tt that R(t)R(t) (as in (c)) changes value. Find an expression for E[Z(t)]\mathbb {E}\left[Z(t)\right] in the limit t→∞t \rightarrow \infty.

Problem 5.38

This problem provides another way of treating ensemble averages for renewal-reward problems. Assume for notational simplicity that XX is a continuous random variable.

?
(a)

Show that P(one or more arrivals in (τ,τ+δ))=m(τ+δ)−m(τ)−o(δ)\mathbb {P}\left(\text{one or more arrivals in } (\tau , \tau +\delta )\right) = m(\tau +\delta )-m(\tau )-\mathrm{o}(\delta ) where o(δ)≥0\mathrm{o}(\delta ) \geq 0 and lim⁡δ→0o(δ)/δ=0\lim_{\delta \rightarrow 0} \mathrm{o}(\delta )/\delta = 0.

(b)

Show that P(Z(t)∈[z,z+δ),X~(t)∈(x,x+δ))\mathbb {P}\left(Z(t) \in [z, z+\delta ), \widetilde{X}(t) \in (x, x+\delta )\right) is equal to [m(t−z)−m(t−z−δ)−o(δ)][FX(x+δ)−FX(x)][m(t-z)-m(t-z-\delta )-\mathrm{o}(\delta )][\mathsf{F}_{X}(x+\delta )-\mathsf{F}_{X}(x)] for x≥z+δx \geq z+\delta.

(c)

Assuming that m′(τ)=dm(τ)/dτm'(\tau ) = dm(\tau )/d\tau exists for all τ\tau, show that the joint density of Z(t),X~(t)Z(t), \widetilde{X}(t) is fZ(t),X~(t)(z,x)=m′(t−z)fX(x)\mathsf{f}_{Z(t), \widetilde{X}(t)}(z, x) = m'(t-z) \mathsf{f}_{X}(x) for x>zx>z.

(d)

Show that E[R(t)]=∫z=0t∫x=z∞R(z,x)fX(x) dx m′(t−z) dz\mathbb {E}\left[R(t)\right] = \int_{z=0}^{t} \int_{x=z}^{\infty } \mathcal{R}(z, x) \mathsf{f}_{X}(x) \, dx \, m'(t-z) \, dz.

Problem 5.39

In this problem, we show how to calculate the distribution of the residual life Y(t)Y(t) as a transient in tt. Let μ(t)=dm(t)/dt\mu (t) = dm(t)/dt where m(t)=E[N(t)]m(t) = \mathbb {E}\left[N(t)\right], and let the interarrival distribution have the density fX(x)\mathsf{f}_{X}(x). Let Y(t)Y(t) have the density fY(t)(y)\mathsf{f}_{Y(t)}(y).

?
(a)

Show that these densities are related by the integral equation

μ(t+y)=fY(t)(y)+∫u=0yμ(t+u)fX(y−u) du.(A.109) \mu (t+y) = \mathsf{f}_{Y(t)}(y)+\int _{u=0}^{y} \mu (t+u) \mathsf{f}_{X}(y-u) \, du. \tag {A.109}
(b)

Let Lμ,t(r)=∫y≥0μ(t+y)e−ry dyL_{\mu , t}(r) = \int_{y \geq 0} \mu (t+y) e^{-ry} \, dy and let LY(t)(r)L_{Y(t)}(r) and LX(r)L_{X}(r) be the Laplace transforms of fY(t)(y)\mathsf{f}_{Y(t)}(y) and fX(x)\mathsf{f}_{X}(x) respectively. Find LY(t)(r)L_{Y(t)}(r) as a function of Lμ,tL_{\mu , t} and LXL_{X}.

(c)

Consider the inter-renewal density fX(x)=(1/2)e−x+e−2x\mathsf{f}_{X}(x) = (1/2) e^{-x}+e^{-2x} for x≥0x \geq 0 (as in Example 5.6.1, which derives its Laplace transform LX(r)=(3/2)r+2(r+1)(r+2)L_{X}(r) = \frac{(3/2)r+2}{(r+1)(r+2)} and the resulting Lm(s)=43r2+19r−19(r+3/2)L_{m}(s) = \frac{4}{3r^{2}}+\frac{1}{9r}-\frac{1}{9(r+3/2)}, m(t)=4t3+1−exp⁡[−(3/2)t]9m(t) = \frac{4t}{3}+\frac{1-\exp [-(3/2)t]}{9}). Find Lμ,t(r)L_{\mu , t}(r) and LY(t)(r)L_{Y(t)}(r) for this example.

(d)

Find fY(t)(y)\mathsf{f}_{Y(t)}(y). Show that your answer reduces to that of (5.30) in the limit as t→∞t \rightarrow \infty.

(e)

Explain how to go about finding fY(t)(y)\mathsf{f}_{Y(t)}(y) in general, assuming that fX\mathsf{f}_{X} has a rational Laplace transform.

Problem 5.40

This problem is designed to give you an alternate way of looking at ensemble averages for renewal-reward problems. First we find an exact expression for P(SN(t)>s)\mathbb {P}\left(S_{N(t)}>s\right). We find this for arbitrary ss and tt, 0<s<t0<s<t.

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(a)

By breaking the event {SN(t)>s}\left\{ S_{N(t)}>s\right\} into subevents {SN(t)>s,N(t)=n}\left\{ S_{N(t)}>s, N(t) = n\right\}, explain each of the following steps:

P(SN(t)>s)=∑n=1∞P(t≥Sn>s,Sn+1>t)=∑n=1∞∫y=stP(Sn+1>t∣Sn=y) dFSn(y)=∫y=stFXc(t−y) d∑n=1∞FSn(y)=∫y=stFXc(t−y) dm(y)where m(y)=E[N(y)]. \begin{aligned} \mathbb {P}\left(S_{N(t)}>s\right) & = \sum _{n=1}^{\infty } \mathbb {P}\left(t \geq S_{n}>s, S_{n+1}>t\right) \\ & = \sum _{n=1}^{\infty } \int _{y=s}^{t} \mathbb {P}\left(S_{n+1}>t \mid S_{n} = y\right) \, d\mathsf{F}_{S_{n}}(y) \\ & = \int _{y=s}^{t} \mathsf{F}_{X}^{c}(t-y) \, d\sum _{n=1}^{\infty } \mathsf{F}_{S_{n}}(y) \\ & = \int _{y=s}^{t} \mathsf{F}_{X}^{c}(t-y) \, dm(y) \qquad \text{where } m(y) = \mathbb {E}\left[N(y)\right]. \end{aligned}
(b)

Show that for 0<s<t<u0<s<t<u,

P(SN(t)>s,SN(t)+1>u)=∫y=stFXc(u−y) dm(y).(A.116) \mathbb {P}\left(S_{N(t)}>s, S_{N(t)+1}>u\right) = \int _{y=s}^{t} \mathsf{F}_{X}^{c}(u-y) \, dm(y). \tag {A.116}
(c)

Draw a two dimensional sketch, with age and duration as the axes, and show the region of (age, duration) values corresponding to the event {SN(t)>s,SN(t)+1>u}\left\{ S_{N(t)}>s, S_{N(t)+1}>u\right\}.

(d)

Assume that for large tt, dm(y)dm(y) can be approximated (according to Blackwell) as (1/X‾) dy(1/\overline{X}) \, dy, where X‾=E[X]\overline{X} = \mathbb {E}\left[X\right]. Assuming that XX also has a density, use the result in parts (b) and (c) to find the joint density of age and duration.

Problem 5.41

Show that for a G/G/1 queue, the time-average wait in the system is the same as lim⁡n→∞E[Wn]\lim_{n \rightarrow \infty } \mathbb {E}\left[W_{n}\right].

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Problem 5.42

If one extends the definition of renewal processes to include inter-renewal intervals of duration 0, with P(X=0)=α\mathbb {P}\left(X = 0\right) = \alpha, show that the expected number of simultaneous renewals at a renewal epoch is 1/(1−α)1/(1-\alpha ), and that, for a non-arithmetic process, the probability of 1 or more renewals in the interval (t,t+δ](t, t+\delta ] tends to (1−α)δ/E[X]+o(δ)(1-\alpha ) \delta /\mathbb {E}\left[X\right]+\mathrm{o}(\delta ) as t→∞t \rightarrow \infty.

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Problem 5.43

The purpose of this exercise is to show why the interchange of expectation and sum in the proof of Wald's equality is justified when E[J]<∞\mathbb {E}\left[J\right]<\infty but not otherwise. Let X1,X2,…X_{1}, X_{2}, \ldots, be a sequence of IID rv's, each with the CDF FX\mathsf{F}_{X}. Assume that E[∣X∣]<∞\mathbb {E}\left[\left|X\right|\right]<\infty.

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(a)

Show that Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} is a rv for each integer n>0n>0. Note: SnS_{n} is obviously a mapping from the sample space to the real numbers, but you must show that it is finite with probability 1. Hint: Recall the additivity axiom for the real numbers.

(b)

Let JJ be a stopping trial for X1,X2,…X_{1}, X_{2}, \ldots. Show that SJ=X1+⋯XJS_{J} = X_{1}+\cdots X_{J} is a rv. Hint: Represent P(SJ≤A)\mathbb {P}\left(S_{J} \leq A\right) as ∑n=1∞P(J=n,Sn≤A)\sum_{n=1}^{\infty } \mathbb {P}\left(J = n, S_{n} \leq A\right).

(c)

For the stopping trial JJ above, let J(k)=min⁡(J,k)J^{(k)} = \min (J, k) be the stopping trial JJ truncated to integer kk. Explain why the interchange of sum and expectation in the proof of Wald's equality is justified in this case, so E[SJ(k)]=X‾E[J(k)]\mathbb {E}\left[S_{J^{(k)}}\right] = \overline{X} \mathbb {E}\left[J^{(k)}\right].

(d)

In parts (d), (e), and (f), assume, in addition to the assumptions above, that FX(0)=0\mathsf{F}_{X}(0) = 0, i.e., that the XiX_{i} are positive rv's. Show that lim⁡k→∞E[SJ(k)]<∞\lim_{k \rightarrow \infty } \mathbb {E}\left[S_{J^{(k)}}\right]<\infty if E[J]<∞\mathbb {E}\left[J\right]<\infty and lim⁡k→∞E[SJ(k)]=∞\lim_{k \rightarrow \infty } \mathbb {E}\left[S_{J^{(k)}}\right] = \infty if E[J]=∞\mathbb {E}\left[J\right] = \infty.

(e)

Show that

P(SJ(k)>x)≤P(SJ>x)for all k,x.(A.117) \mathbb {P}\left(S_{J^{(k)}}>x\right) \leq \mathbb {P}\left(S_{J}>x\right) \qquad \text{for all } k, x. \tag {A.117}
(f)

Show that E[SJ]=X‾E[J]\mathbb {E}\left[S_{J}\right] = \overline{X} \mathbb {E}\left[J\right] if E[J]<∞\mathbb {E}\left[J\right]<\infty and E[SJ]=∞\mathbb {E}\left[S_{J}\right] = \infty if E[J]=∞\mathbb {E}\left[J\right] = \infty.

(g)

Now assume that XX has both negative and positive values with non-zero probability and let X+=max⁡(0,X)X^{+} = \max (0, X) and X−=min⁡(X,0)X^{-} = \min (X, 0). Express SJS_{J} as SJ++SJ−S_{J}^{+}+S_{J}^{-} where SJ+=∑i=1JXi+S_{J}^{+} = \sum_{i=1}^{J} X_{i}^{+} and SJ−=∑i=1JXi−S_{J}^{-} = \sum_{i=1}^{J} X_{i}^{-}. Show that E[SJ]=X‾E[J]\mathbb {E}\left[S_{J}\right] = \overline{X} \mathbb {E}\left[J\right] if E[J]<∞\mathbb {E}\left[J\right]<\infty and that E[Sj]\mathbb {E}\left[S_{j}\right] is undefined otherwise.

Problem 5.44

This is a very simple exercise designed to clarify confusion about the roles of past, present, and future in stopping rules. Let {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} be a sequence of IID binary rv's, each with the pmf pX(1)=1/2\mathsf{p}_{X}(1) = 1/2, pX(0)=1/2\mathsf{p}_{X}(0) = 1/2. Let JJ be a positive integer-valued rv that takes on the sample value nn of the first trial for which Xn=1X_{n} = 1. That is, for each n≥1n \geq 1,

{J=n}={X1=0,  X2=0,…,Xn−1=0,  Xn=1}. \left\{ J = n\right\} = \left\{ X_{1} = 0, \; X_{2} = 0, \ldots , X_{n-1} = 0, \; X_{n} = 1\right\} .
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(a)

Use the definition of stopping trial, Definition 5.5.1 (a stopping trial, or stopping time, JJ for a sequence of rv's X1,X2,…X_{1}, X_{2}, \ldots is a positive, integer-valued rv such that the indicator function of the event {J=n}\left\{ J = n\right\} is a function of X1,…,XnX_{1}, \ldots , X_{n} for each n≥1n \geq 1), to show that JJ is a stopping trial for {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\}.

(b)

Show that for any given nn, the rv's XnX_{n} and IJ=n\mathbb {I}_{J=n} are statistically dependent.

(c)

Show that for every m>nm>n, XnX_{n} and IJ=m\mathbb {I}_{J=m} are statistically dependent.

(d)

Show that for every m<nm<n, XnX_{n} and IJ=m\mathbb {I}_{J=m} are statistically independent.

(e)

Show that XnX_{n} and IJ≥n\mathbb {I}_{J \geq n} are statistically independent. Give the simplest characterization you can of the event {J≥n}\left\{ J \geq n\right\}.

(f)

Show that XnX_{n} and IJ>n\mathbb {I}_{J>n} are statistically dependent.

Problem 5.45

Assume a friend has developed an excellent program for finding the steady-state probabilities for finite-state Markov chains. More precisely, given the transition matrix [P][P], the program returns lim⁡n→∞Piin\lim_{n \rightarrow \infty } P_{ii}^{n} for each ii. Assume all chains are aperiodic.

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(a)

You want to find the expected time to first reach a given state kk starting from a different state mm for a Markov chain with transition matrix [P][P]. You modify the matrix to [P′][P'] where Pkm′=1P'_{km} = 1, Pkj′=0P'_{kj} = 0 for j≠mj \neq m, and Pij′=PijP'_{ij} = P_{ij} otherwise. How do you find the desired first-passage time from the program output given [P′][P'] as an input? (Hint: The times at which a Markov chain enters any given state can be considered as renewals in a (perhaps delayed) renewal process).

(b)

Using the same [P′][P'] as the program input, how can you find the expected number of returns to state mm before the first passage to state kk?

(c)

Suppose, for the same Markov chain [P][P] and the same starting state mm, you want to find the probability of reaching some given state nn before the first-passage to kk. Modify [P][P] to some [P′′][P''] so that the above program with P′′P'' as an input allows you to easily find the desired probability.

(d)

Let P(X(0)=i)=Qi\mathbb {P}\left(X(0) = i\right) = Q_{i}, 1≤i≤M1 \leq i \leq \mathsf{M} be an arbitrary set of initial probabilities for the same Markov chain [P][P] as above. Show how to modify [P][P] to some [P′′′][P'''] for which the steady-state probabilities allow you to easily find the expected time of the first-passage to state kk.

Problem 5.46

Consider a ferry that carries cars across a river. The ferry holds an integer number kk of cars and departs the dock when full. At that time, a new ferry immediately appears and begins loading newly arriving cars ad infinitum. The ferry business has been good, but customers complain about the long wait for the ferry to fill up.

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(a)

Assume that cars arrive according to a renewal process. The IID interarrival times have mean X‾\overline{X}, variance σ2\sigma^{2} and moment generating function gX(r)\mathsf{g}_{X}(r). Does the sequence of departure times of the ferries form a renewal process? Explain carefully.

(b)

Find the expected time that a customer waits, starting from its arrival at the ferry terminal and ending at the departure of its ferry. Note 1: Part of the problem here is to give a reasonable definition of the expected customer waiting time. Note 2: It might be useful to consider k=1k = 1 and k=2k = 2 first.

(c)

Is there a `slow truck' phenomenon (a dependence on E[X2]\mathbb {E}\left[X^{2}\right]) here? Give an intuitive explanation. Hint: Look at k=1k = 1 and k=2k = 2 again.

(d)

In an effort to decrease waiting, the ferry managers institute a policy where no customer ever has to wait more than one hour. Thus, the first customer to arrive after a ferry departure waits for either one hour or the time at which the ferry is full, whichever comes first, and then the ferry leaves and a new ferry starts to accumulate new customers. Does the sequence of ferry departures form a renewal process under this new system? Does the sequence of times at which each successive empty ferry is entered by its first customer form a renewal process? You can assume here that t=0t = 0 is the time of the first arrival to the first ferry. Explain carefully.

(e)

Give an expression for the expected waiting time of the first new customer to enter an empty ferry under this new strategy.

Problem 5.47

Consider a (G/G/∞\infty) `queueing' system. That is, the arriving customers form a renewal process, i.e., the interarrival intervals {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} are IID. You may assume throughout that E[X]<∞\mathbb {E}\left[X\right]<\infty. Each arrival immediately enters service; there are infinitely many servers, so one is immediately available for each arrival. The service time YiY_{i} of the iith customer is a rv of expected value Y‾<∞\overline{Y}<\infty and is IID with all other service times and independent of all inter-arrival times. There is no queueing in such a system, and one can easily intuit that the number in service never becomes infinite since there are always available servers.  {#fig-1 width="70%"}

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(a)

Give a simple example of distributions for XX and YY in which this system never becomes empty. Hint: Deterministic rv's are fair game.

(b)

We want to prove Little's theorem for this type of system, but there are no renewal instants for the entire system. As illustrated above, let N(t)N(t) be the renewal counting process for the arriving customers and let L(t)L(t) be the number of customers in the system (i.e., receiving service) at time tt. In distinction to our usual view of queueing systems, assume that there is no arrival at time 0 and the first arrival occurs at time S1=X1S_{1} = X_{1}. The nnth arrival occurs at time Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n}.

Carefully explain why, for each sample point ω\omega and each time t>0t>0,

∫0tL(t,ω) dt≤∑i=1N(t,ω)Yi(ω). \int _{0}^{t} L(t, \omega ) \, dt \leq \sum _{i=1}^{N(t, \omega )} Y_{i}(\omega ).
(c)

Find the limit as t→∞t \rightarrow \infty of 1t∑i=1N(t,ω)Yi(ω)\frac{1}{t} \sum_{i=1}^{N(t, \omega )} Y_{i}(\omega ) and show that this limit exists WP1.

(d)

Assume that the service time distribution is bounded between 0 and some b>0b>0, i.e., that FY(b)=1\mathsf{F}_{Y}(b) = 1. Carefully explain why

∫0t+bL(τ,ω) dτ≥∑i=1N(t,ω)Yi(ω). \int _{0}^{t+b} L(\tau , \omega ) \, d\tau \geq \sum _{i=1}^{N(t, \omega )} Y_{i}(\omega ).
(e)

Find the limit as t→∞t \rightarrow \infty of 1t∫0tL(τ,ω) dτ\frac{1}{t} \int_{0}^{t} L(\tau , \omega ) \, d\tau and indicate why this limit exists WP1.