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Chapter 5: Problem 69

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). For \(sN(s))\) (b) \(P(N(s)=0, N(t)=3)\); (c) \(E[N(t) \mid N(s)=4] ;\) (d) \(E[N(s) \mid N(t)=4]\).

Short Answer

Expert verified
(a) \(P(N(t)>N(s)) = 1 - e^{-\lambda(t-s)}\) (b) \(P(N(s)=0, N(t)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\) (c) \(E[N(t) | N(s)=4] = 4 + \lambda(t-s)\) (d) \(E[N(s) | N(t)=4] = 4 - \lambda(t-s)\)

Step by step solution

01

Recall the Poisson Process properties

A Poisson process with rate 位 has the following properties: 1. N(0) = 0 2. It has independent increments. That is, for t >= s, the random variable N(t) - N(s) is independent of N(u) for u 鈮� s. 3. For any interval [s, t), the number of events in the interval has a Poisson distribution with parameter 位(t - s). Now we'll go through each part of the question.
02

Find (a) P(N(t) > N(s))

Let's find the probability \(P(N(t) - N(s) > 0)\). Since the increments are independent, we know that \(N(t) - N(s)\) has a Poisson distribution with parameter \(\lambda(t-s)\). So, \[P(N(t) > N(s)) = P(N(t) - N(s) > 0) = 1 - P(N(t) - N(s) = 0)\] Now, we know that a Poisson distribution with parameter \(\lambda(t-s)\) will have the probability mass function given by: \[P(N(t) - N(s) = k) = \frac{(\lambda(t - s))^k e^{-\lambda(t-s)}}{k!}\] So, for k = 0: \[P(N(t) - N(s) = 0) = \frac{(\lambda(t - s))^0 e^{-\lambda(t-s)}}{0!} = e^{-\lambda(t-s)}\] Thus, \[P(N(t) > N(s)) = 1 - e^{-\lambda(t-s)}\]
03

Find (b) P(N(s)=0, N(t)=3)

We know that \(N(t) - N(s)\) is independent of \(N(s)\) and has a Poisson distribution with parameter \(\lambda(t-s)\). Therefore, \[P(N(t) - N(s) = 3) = \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\] Now, we'll find the probability of \(N(s) = 0\): \[P(N(s) = 0) = \frac{(\lambda s)^0 e^{-\lambda s }}{0!} = e^{-\lambda s}\] And so, the joint probability is given by multiplying the two probabilities, since they are independent: \[P(N(s)=0, N(t)=3) = P(N(s)=0)P(N(t) - N(s)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\]
04

Find (c) E[N(t) | N(s)=4]

Given that the Poisson process has independent increments, we know that \(N(t) - N(s)\) is independent of \(N(s)\). Since \(N(t) - N(s)\) has a Poisson distribution with parameter \(\lambda(t-s)\), the expected value is given by \(E[N(t) - N(s)] = \lambda(t-s)\). Now, conditioning on \(N(s) = 4\), we have: \[E[N(t) | N(s) = 4] = E[N(s) + (N(t) - N(s)) | N(s) = 4] = E[4 + N(t) - N(s)] = 4 + E[N(t) - N(s)] = 4 + \lambda(t-s)\]
05

Find (d) E[N(s) | N(t)=4]

Similarly, since \(N(t) - N(s)\) is independent of \(N(s)\), it's also independent of \(N(t)\). We can write \(N(s) = N(t) - (N(t) - N(s))\). Now, conditioning on \(N(t) = 4\), we have: \[E[N(s) | N(t) = 4] = E[N(t) - (N(t) - N(s)) | N(t) = 4] = E[4 - (N(t) - N(s))] = 4 - E[N(t) - N(s)] = 4 - \lambda(t-s)\] So, we have the final answers for each part of the question: (a) \(P(N(t)>N(s)) = 1 - e^{-\lambda(t-s)}\) (b) \(P(N(s)=0, N(t)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\) (c) \(E[N(t) | N(s)=4] = 4 + \lambda(t-s)\) (d) \(E[N(s) | N(t)=4] = 4 - \lambda(t-s)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Increments
Understanding the concept of independent increments is critical when studying random processes like the Poisson process. In this context, independent increments signify that the number of events occurring in disjoint time intervals is independent of one another. This implies that what happens in one period does not influence the probability of events in another.

For example, if you're observing a process where you count the number of emails you receive per hour, the number of emails received in one hour is unaffected by the number received in the previous hour, assuming this is a Poisson process. This specific property is what allows us to solve various probability questions related to the Poisson process as we can treat segments of time as distinct and separate in our calculations.
Probability Mass Function
The probability mass function (PMF) is an essential concept for discrete random variables, like the number of occurrences in a Poisson process. A PMF assigns a probability to each possible outcome. For a Poisson process with a given rate \( \lambda \), the PMF tells us the likelihood of observing exactly \( k \) events in a set time frame. Formally, it's expressed as:
\[ P(N(t) - N(s) = k) = \frac{(\lambda(t - s))^k e^{-\lambda(t-s)}}{k!} \]
Understanding and using the PMF is pivotal in working out specific probabilities, as we see in the textbook exercise where it's used to calculate the probability of a certain number of events in different intervals. Having this function at hand enables students to compute probabilities swiftly for any given number of occurrences within the process.
Conditional Expectation
Conditional expectation is a profound concept in probability theory, which deals with the expected value of a random variable given that another random variable or an event has occurred. In the realm of the Poisson process, this allows us to determine the expected number of events in one time frame knowing the number of events in another.

For instance, if we want to find out the expected number of emails we will receive by the end of the day, given that we've already received a certain amount by lunchtime, we would use the concept of conditional expectation. The formula would look something like this: \[ E[N(t) | N(s) = x] = x + \lambda(t-s) \]
By incorporating the rate of the process and the known count of occurrences up to time \( s \) (in this case, lunchtime), we're able to predict the average total count by time \( t \) (end of the day).

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Most popular questions from this chapter

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Consider the coupon collecting problem where there are \(m\) distinct types of coupons, and each new coupon collected is type \(j\) with probability \(p_{j}, \sum_{j=1}^{m} p_{j}=1\). Suppose you stop collecting when you have a complete set of at least one of each type. Show that $$ P\\{i \text { is the last type collected }\\}=E\left[\prod_{j \neq i}\left(1-U^{\lambda_{i} / \lambda_{i}}\right)\right] $$ where \(U\) is a uniform random variable on \((0,1)\).

(a) Let \(\\{N(t), t \geqslant 0\\}\) be a nonhomogeneous Poisson process with mean value function \(m(t) .\) Given \(N(t)=n\), show that the unordered set of arrival times has the same distribution as \(n\) independent and identically distributed random variables having distribution function $$ F(x)=\left\\{\begin{array}{ll} \frac{m(x)}{m(t)}, & x \leqslant t \\ 1, & x \geqslant t \end{array}\right. $$ (b) Suppose that workmen incur accidents in accordance with a nonhomogeneous Poisson process with mean value function \(m(t) .\) Suppose further that each injured man is out of work for a random amount of time having distribution F. Let \(X(t)\) be the number of workers who are out of work at time \(t\). By using part (a), find \(E[X(t)]\)

There are three jobs that need to be processed, with the processing time of job \(i\) being exponential with rate \(\mu_{i} .\) There are two processors available, so processing on two of the jobs can immediately start, with processing on the final job to start when one of the initial ones is finished. (a) Let \(T_{i}\) denote the time at which the processing of job \(i\) is completed. If the objective is to minimize \(E\left[T_{1}+T_{2}+T_{3}\right]\), which jobs should be initially processed if \(\mu_{1}<\mu_{2}<\mu_{3} ?\) (b) Let \(M\), called the makespan, be the time until all three jobs have been processed. With \(S\) equal to the time that there is only a single processor working, show that $$ 2 E[M]=E[S]+\sum_{i=1}^{3} 1 / \mu_{i} $$ For the rest of this problem, suppose that \(\mu_{1}=\mu_{2}=\mu, \quad \mu_{3}=\lambda .\) Also, let \(P(\mu)\) be the probability that the last job to finish is either job 1 or job 2, and let \(P(\lambda)=1-P(\mu)\) be the probability that the last job to finish is job 3 . (c) Express \(E[S]\) in terms of \(P(\mu)\) and \(P(\lambda)\). Let \(P_{i, j}(\mu)\) be the value of \(P(\mu)\) when \(i\) and \(j\) are the jobs that are initially started. (d) Show that \(P_{1,2}(\mu) \leqslant P_{1,3}(\mu)\). (e) If \(\mu>\lambda\) show that \(E[M]\) is minimized when job 3 is one of the jobs that is initially started. (f) If \(\mu<\lambda\) show that \(E[M]\) is minimized when processing is initially started on jobs 1 and \(2 .\)

Customers arrive at a bank at a Poisson rate \(\lambda .\) Suppose two customers arrived during the first hour. What is the probability that (a) both arrived during the first 20 minutes? (b) at least one arrived during the first 20 minutes?
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