/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 If \(A(t)\) and \(Y(t)\) are, re... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 31

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

Short Answer

Expert verified
The short answer for the given problem is: $$ P\{Y(t) > x \mid A(t) = s\} = \frac{P\{X_{N(t)+1} > s + x\}}{P\{X_{N(t)+1} > s\}} $$ where \(X_{N(t)+1}\) is the remaining time till the next event in the renewal process.

Step by step solution

01

Understand the problem

We need to find $$ P\{Y(t) > x \mid A(t) = s\} $$ where \(A(t)\) and \(Y(t)\) are the age and excess of a renewal process at time \(t\), which has interarrival distribution \(F\).
02

Use the definition of conditional probability

By definition, the conditional probability is given by $$ P\{Y(t) > x \mid A(t) = s\} = \frac{P\{Y(t) > x \cap A(t) = s\}}{P\{A(t) = s\}} $$
03

Determine the joint probability of \(Y(t) > x\) and \(A(t) = s\)

\(A(t) = s\) means that the renewal process has just completed its \(s\)th arrival at time \(t\). For \(Y(t) > x\), we have the remaining time until the next arrival is greater than \(x\) with respect to that. So the joint probability is $$ P\{Y(t) > x \cap A(t) = s\} = P\{A(t) = s, N(t+s) - N(t) = 0\} $$ where \(N(t)\) is the number of arrivals in the process up to time \(t\).
04

Determine the probability of \(A(t) = s\)

The probability of \(A(t) = s\) can be determined by examining the time since the last event. Specifically, it is the probability of having an arrival at time \(t-s\) and no arrivals within the interval \([t-s,t)\): $$ P\{A(t) = s\} = P\{N(t-s) - N(t) = -1\} $$
05

Use renewal process properties

We know the renewal process has interarrival distribution \(F\). We can rewrite the joint probability as $$ P\{Y(t) > x \cap A(t) = s\} = P\{A(t) = s, N(t+s) - N(t) = 0\} = P\{X_{N(t)+1} > s+x\} $$ and we can rewrite the probability of \(A(t) = s\) as $$ P\{A(t) = s\} = P\{N(t-s) - N(t) = -1\} = P\{X_{N(t)+1} > s\} $$
06

Calculate the conditional probability

Now we have everything we need to calculate the conditional probability. Substitute the obtained results back into the equation obtained in Step 2: $$ P\{Y(t) > x \mid A(t) = s\} = \frac{P\{X_{N(t)+1} > s + x\}}{P\{X_{N(t)+1} > s\}} $$ Note that the conditional probability does not depend on \(t\).

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Most popular questions from this chapter

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

In Example \(7.7\), suppose that potential customers arrive in accordance with a renewal process having interarrival distribution \(F\). Would the number of events by time \(t\) constitute a (possibly delayed) renewal process if an event corresponds to a customer (a) entering the bank? (b) leaving the bank? What if \(F\) were exponential?

Wald's equation can be used as the basis of a proof of the elementary renewal theorem. Let \(X_{1}, X_{2}, \ldots\) denote the interarrival times of a renewal process and let \(N(t)\) be the number of renewals by time \(t\). (a) Show that whereas \(N(t)\) is not a stopping time, \(N(t)+1\) is. Hint: Note that $$ N(t)=n \Leftrightarrow X_{1}+\cdots+X_{n} \leqslant t \quad \text { and } \quad X_{1}+\cdots+X_{n+1}>t $$ (b) Argue that $$ E\left[\sum_{i=1}^{N(t)+1} X_{i}\right]=\mu[m(t)+1] $$ (c) Suppose that the \(X_{i}\) are bounded random variables. That is, suppose there is a constant \(M\) such that \(P\left\\{X_{i}

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 \(1 ?\)

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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