The Poisson Process

Let $\left\{ X_1(t), t \geq 0\right\}$ and $\left\{ X_2(t), t \geq 0\right\}$ be independent Poisson processes with common intensity $\lambda$. Suppose that $X_1(3) = 9$ and $X_2(3) = 5$. What is the probability that the $X_1$-process reaches level 10 before the $X_2$-process does?

Solve the same problem under the assumption that the processes have intensities $\lambda_1$ and $\lambda_2$, respectively.

Consider two independent Poisson processes $\left\{ X_1(t), t \geq 0\right\}$ and $\left\{ X_2(t), t \geq 0\right\}$ with common intensity. What is the probability that the two-dimensional process $\left\{ (X_1(t), X_2(t)), t \geq 0\right\}$ passes through the point

Susan likes pancakes very much, but Tom does not. The time to eat a pancake can be assumed to be exponential. Their mother has studied them over the years and estimates the parameters ($=1/\text{expected time}$) to be 7 and 2, respectively. Compute the probability that Susan finishes 10 pancakes before Tom has finished his first one.

Suppose that customers arrive at a counter or server according to a Poisson process with intensity $\lambda$ and that the service times are independent, $\mathrm{Exp}(1/\mu )$-distributed random variables. Suppose also that a customer arrives at time zero and finds the counter free.

Fredrik and Ulrich both received soap bubble machines for Christmas. The machines emit bubbles according to independent Poisson processes with intensities 3 and 2 (bubbles per minute), respectively. Suppose they turn them on at the same time.

At the center of espionage in Kznatropsk one is thinking of a new method for sending Morse telegrams. Instead of using the traditional method, that is, to send letters in groups of 5 according to a Poisson process with intensity 1, one might send them one by one according to a Poisson process with intensity 5. Before deciding which method to use one would like to know the following: What is the probability that it takes less time to send one group of 5 letters the traditional way than to send 5 letters the new way (the actual transmission time can be neglected).

Consider a Poisson process with intensity $\lambda$. We start observing at time $t=0$. Let $T$ be the time that has elapsed at the first occurrence. Continue to observe the process $T$ further units of time. Let $N(T)$ be the number of occurrences during the latter period (i.e., during $(T,2T]$). Determine the distribution of $N(T)$.

A particle source $A$ emits one particle at a time, according to a Poisson process with an intensity of two particles a minute. Another particle source $B$ emits two particles at a time, according to a Poisson process with an intensity of one pair of particles a minute. The sources are independent of each other. We begin to observe the sources at time zero. Compute the probability that source $A$ has emitted two particles before source $B$ has done so.

A specific component in a cryptometer has an $\mathrm{Exp}(\mu )$-distributed lifetime, $\mu > 0$. If replacement is made as soon as a component fails, and if $X(t) = \#$ failures during $(0,t] = \#$ replacements during $(0,t]$, then $\left\{ X(t), t \geq 0\right\}$ is, of course, a Poisson process. Let $\left\{ V_n, n \geq 1\right\}$ be these usual interreplacement times, and suppose, instead, that the $n$th component is replaced:

Karin arrives at the post office, which opens at 9:00 a.m., at 9:05 a.m. She finds two cashiers at work, both serving one customer each. The customers started being served at 9:00 and 9:01, respectively. The service times are independent and $\mathrm{Exp}(8)$-distributed. Let $T_k$ be the time from 9:05 until service has been completed for $k$ of the two customers, $k=1,2$. Find $\mathbb {E}\left[T_k\right]$ for $k=1$ and $2$.

Mns waits for the bus. The waiting time, $T$, is $\mathrm{U}(0,a)$-distributed. While he waits he tries to get a ride from cars that pass by according to a Poisson process with intensity $\lambda$. The probability of a passing car picking him up is $p$. Determine the probability that Mns is picked up by some car before the bus arrives.

Consider a sender that transmits signals according to a Poisson process with intensity $\lambda$. The signals are received by a receiver, however, in such a way that every signal is registered with probability $p$, $0 < p < 1$, and ``missed'' with probability $q = 1-p$. Registrations are independent. Let $X(t)$ be the number of transmitted signals during $(0,t]$, let $Y(t)$ be the number of registered signals, and let $Z(t)$ be the number of nonregistered signals during this period, where $t \geq 0$.

We have seen that a thinned Poisson process is, again, a Poisson process. Prove the following analog for a ``geometric process.'' More precisely:

A radio amateur wishes to transmit a message. The frequency on which she sends the Morse signals is subject to random disturbances according to a Poisson process with intensity $\lambda$ per second. In order to succeed with the transmission, she needs a time period of $a$ seconds without disturbances. She stops as soon as she is done. Let $T$ be the total time required to finish. Determine $\mathbb {E}\left[T\right]$.

Peter wishes to take a picture of his girlfriend Sheila. Since they are in a rather dark room, he needs a rather long exposure time, during which Sheila must not move. The following model can be used to describe the situation. The success of a photo is called an $A$-event.'' Each time Sheila moves, she causes a disturbance called a $D$-event.'' $A$-events and $D$-events occur according to independent Poisson processes with intensities $\lambda_A$ and $\lambda_D$, respectively. The experiment is started at time $t=0$. Let $T$ be the time of the first $A$-occurrence. The experiment is deemed successful if $T \geq 1$ and if no $D$-event occurs during the time interval $(0,T+2]$. What is the probability of a successful photo?

People arrive at an automatic teller machine (ATM) according to a Poisson process with intensity $\lambda$. The service time required at the ATM is constant, $a$ seconds. Unfortunately, this machine does not allow for any waiting customers (i.e., no queue is allowed), which means that persons who arrive while the ATM is busy have to leave. When the $a$ seconds of a customer have elapsed, the ATM is free to serve again, and so on. Suppose that the ATM is free at time zero, and let $T_n$ be the time of the arrival of the $n$th customer. Find the distribution of $T_n$, and compute $\mathbb {E}\left[T_n\right]$ and $\operatorname {Var}\left[T_n\right]$.

Suppose that we are at time zero. Passengers arrive at a train station according to a Poisson process with intensity $\lambda$. Compute the expected value of the total waiting time of all passengers who have come to the station in order to catch a train that leaves at time $t$.

Suppose that electrical pulses having i.i.d. random amplitudes $A_1, A_2, \ldots$ arrive at a counter in accordance with a Poisson process with intensity $\lambda$. The amplitude of a pulse is assumed to decrease exponentially, that is, if a pulse has amplitude $A$ upon its arrival, then its amplitude at time $t$ is $Ae^{-\alpha t}$, where $\alpha$ is some positive parameter. We finally assume that the initial amplitudes of the pulses are independent of the Poisson process. Compute the expected value of the total amplitude at time $t$.

Customers arrive at a computer center at time points generated by a Poisson process with intensity $\lambda$. The number of jobs brought to the center by the customers are independent random variables whose common generating function is $g(u)$. Compute the generating function of the number of jobs brought to the computer center during the time interval $(s,t]$.

We have seen that if we superposition a fixed number of Poisson processes we obtain a new Poisson process. This need however not be true if we superposition a random number of such processes. More precisely, let us superposition $N \in \mathrm{Fs}(p)$ independent Poisson processes, each with the same intensity $\lambda$, where $N$ is independent of the Poisson processes.

Let $X_1, X_2, \ldots$ be the i.i.d. lifetimes of some component in some large machine. The simplest replacement policy is to change a component as soon as it fails. In this case it may be necessary to call a repairman at night, which might be costly. Another policy, called replacement based on age, is to replace at failure or at some given age, $a$, say, whichever comes first, in which case the interreplacement times are

Suppose that $c_1$ is the cost for replacements due to failure and that $c_2$ is the cost for replacements due to age. In addition, let $Y_k$ be the cost attached to replacement $k$, $k \geq 1$, and let $N(t)$ be the number of replacements made in the time interval $(0,t]$, where $\left\{ N(t), t \geq 0\right\}$ is a Poisson process, which is independent of $X_1, X_2, \ldots$. This means that

is the total cost caused by the replacements in the time interval $(0,t]$ (with $Z(t)=0$ whenever $N(t)=0$).

Let $\left\{ X(t), t \geq 0\right\}$ be a Poisson process with random intensity $\Lambda \in \Gamma (m,\theta )$.

Let $\left\{ X(t), t \geq 0\right\}$ be a Poisson process with intensity $\lambda$ that is run $N$ time units, where $N \in \mathrm{Fs}(p)$.

A Poisson process is observed during $n$ days. The intensity is, however, not constant, but varies randomly day by day, so that we may consider the intensities during the $n$ days as $n$ independent, $\mathrm{Exp}(1/\alpha )$-distributed random variables. Determine the distribution of the total number of occurrences during the $n$ days.

Let $\left\{ X(t), t \geq 0\right\}$ be a Poisson process and let $\left\{ T_k, k \geq 1\right\}$ be the occurrence times. Suppose that we know that $T_3=1$ and that $T_1=x$, where $0 < x < 1$. Our intuition then tells us that the conditional distribution of $T_2$ should be $\mathrm{U}(x,1)$-distributed. Prove that this is indeed the case, i.e., show that

Suppose that $X_1$, $X_2$, and $X_3$ are independent, $\mathrm{Exp}(1)$-distributed random variables, and let $X_{(1)}, X_{(2)}, X_{(3)}$ be the order variables. Determine $\mathbb {E}_{}\left[X_{(3)} \mid X_{(1)}=x\right]$.

Example 4.2.3: Let $X_1$, $X_2$, and $X_3$ be independent, $\mathrm{Exp}(1)$-distributed random variables. Compute $\mathbb {E}_{}\left[X_{(3)} \mid X_{(1)}=x\right]$. This is solved there by first finding the joint density $f_{X_{(1)},X_{(3)}}(x,y) = 3 \cdot 2(e^{-x}-e^{-y})e^{-(x+y)}$ for $0<x<y$ (a special case of the joint density of two order statistics), from which the conditional density $f_{X_{(3)} \mid X_{(1)}=x}(y) = 2(e^{-x}-e^{-y})e^{2x-y}$ for $0<x<y$ is obtained, and then the conditional expectation is computed by direct integration.

Consider a queueing system where customers arrive according to a Poisson process with intensity $\lambda$ customers per minute. Let $X(t)$ be the total number of customers that arrive during $(0,t]$. Compute the correlation coefficient of $X(t)$ and $X(t+s)$.

A particle is subject to hits at time points generated by a Poisson process with intensity $\lambda$. Every hit moves the particle a horizontal, $\mathrm{N}(0,\sigma^{2})$-distributed distance. The displacements are independent random variables, which, in addition, are independent of the Poisson process. Let $S_t$ be the location of the particle at time $t$ (we begin at time zero).

Consider a Poisson process with intensity $\lambda$, and let $T$ be the time of the first occurrence in the time interval $(0,t]$. If there is no occurrence during $(0,t]$, we set $T=t$. Compute $\mathbb {E}\left[T\right]$.

In the previous example, let, instead, $T$ be the time of the last occurrence in the time interval $(0,t]$. If there is no occurrence during $(0,t]$, we set $T=0$. Compute $\mathbb {E}\left[T\right]$.

A further (and final) definition of the Poisson process runs as follows (recall that, per the definition used throughout this chapter, a process $\left\{ X(t), t \geq 0\right\}$ is called a Poisson process with intensity $\lambda$ if it is nonnegative, integer-valued, with $X(0)=0$, has independent, stationary increments, and $X(t) \in \mathrm{Po}(\lambda t)$ for every $t > 0$): A nondecreasing stochastic process $\left\{ X(t), t \geq 0\right\}$ is a Poisson process iff