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Chapter 7: Problem 20

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 \(n \rightarrow \infty\)

Short Answer

Expert verified
As \(n\) approaches infinity, \(W_n\) converges to \(\frac{E[R]}{E[X]}\), which is the average reward earned during a single cycle. This result is obtained by calculating the expected value of \(W_n\), applying the law of large numbers, and simplifying the resulting expressions.

Step by step solution

01

Rewrite the expression for \(W_n\)

First, let us rewrite the expression for \(W_n\) in a more compact form using the sum notation: \[W_n = \frac{\sum_{i=1}^{n}R_i}{\sum_{i=1}^{n}X_i'}\]
02

Calculate the expected value of \(W_n\)

Now, we want to find the expected value of \(W_n\). To do this, note that the expected value of the sum of random variables is equal to the sum of their expected values: \[E\left[\frac{\sum_{i=1}^{n}R_i}{\sum_{i=1}^{n}X_i'}\right] = \frac{E\left[\sum_{i=1}^{n}R_i\right]}{E\left[\sum_{i=1}^{n}X_i\right]} = \frac{\sum_{i=1}^{n}E[R_i]}{\sum_{i=1}^{n}E[X_i]}\]
03

Apply the law of large numbers

Now, the law of large numbers tells us that, as \(n\) approaches infinity, the average of a sequence of independent and identically distributed random variables converges to their expected value. So we have, \[\lim_{n\rightarrow\infty} \frac{\sum_{i=1}^{n}E[R_i]}{\sum_{i=1}^{n}E[X_i]} = \lim_{n\rightarrow\infty} \frac{nE[R]}{nE[X]}\]
04

Simplify and finalize the result

Simplify the expression to obtain the required result: \[\lim_{n\rightarrow\infty} \frac{nE[R]}{nE[X]} = \frac{E[R]}{E[X]}\] Therefore, we have shown that: \[W_n \rightarrow \frac{E[R]}{E[X]} \text{ as } n \rightarrow \infty\]

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Most popular questions from this chapter

Each of \(n\) skiers continually, and independently, climbs up and then skis down a particular slope. The time it takes skier \(i\) to climb up has distribution \(F_{i}\), and it is independent of her time to ski down, which has distribution \(H_{i}, i=1, \ldots, n\). Let \(N(t)\) denote the total number of times members of this group have skied down the slope by time \(t .\) Also, let \(U(t)\) denote the number of skiers climbing up the hill at time \(t\). (a) What is \(\lim _{t \rightarrow \infty} N(t) / t\) ? (b) Find \(\lim _{t \rightarrow \infty} E[U(t)]\). (c) If all \(F_{i}\) are exponential with rate \(\lambda\) and all \(G_{i}\) are exponential with rate \(\mu\), what is \(P\\{U(t)=k\\} ?\)

A truck driver regularly drives round trips from \(\mathrm{A}\) to \(\mathrm{B}\) and then back to \(\mathrm{A}\). Each time he drives from \(A\) to \(B\), he drives at a fixed speed that (in miles per hour) is uniformly distributed between 40 and \(60 ;\) each time he drives from \(\mathrm{B}\) to \(\mathrm{A}\), he drives at a fixed speed that is equally likely to be either 40 or 60 . (a) In the long run, what proportion of his driving time is spent going to \(\mathrm{B}\) ? (b) In the long run, for what proportion of his driving time is he driving at a speed of 40 miles per hour?

An airport shuttle bus picks up all passengers waiting at a bus stop and drops them off at the airport terminal; it then returns to the stop and repeats the process. The times between returns to the stop are independent random variables with distribution \(F\), mean \(\mu\), and variance \(\sigma^{2} .\) Passengers arrive at the bus stop in accordance with a Poisson process with rate \(\lambda\). Suppose the bus has just left the stop, and let \(X\) denote the number of passengers it picks up when it returns. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\). (c) At what rate does the shuttle bus arrive at the terminal without any passengers? Suppose that each passenger that has to wait at the bus stop more than \(c\) time units writes an angry letter to the shuttle bus manager. (d) What proportion of passengers write angry letters? (e) How does your answer in part (d) relate to \(F_{e}(x) ?\)

Consider a single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G\). The server serves one customer at a time. However, the server processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are three other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service. If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(i\) spends in the system. Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{i}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geqslant 1\), are independent random variables having distribution \(G\). (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t .\) Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 13\()\) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G .\) That is, the average time that customers spend in the system is equal to the average work they bring to the system.

Consider 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)
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