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Chapter 5: Problem 38

Let \(\left\\{M_{i}(t), t \geqslant 0\right\\}, i=1,2,3\) be independent Poisson processes with respective rates \(\lambda_{i}, i=1,2\), and set $$ N_{1}(t)=M_{1}(t)+M_{2}(t), \quad N_{2}(t)=M_{2}(t)+M_{3}(t) $$ The stochastic process \(\left\\{\left(N_{1}(t), N_{2}(t)\right), t \geqslant 0\right\\}\) is called a bivariate Poisson process. (a) Find \(P\left[N_{1}(t)=n, N_{2}(t)=m\right\\}\) (b) Find \(\operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right)\)

Short Answer

Expert verified
The joint probability mass function of the bivariate Poisson process is given by: $$ P\left[N_{1}(t)=n, N_{2}(t)=m\right] = e^{-(\lambda_1+\lambda_2+\lambda_3) t} \sum_{k=0}^{\min(n,m)} \frac{(\lambda_1 t)^{n-k}(\lambda_2 t)^k(\lambda_3 t)^{m-k}}{(n-k)!k!(m-k)!} $$ And the covariance between \(N_1(t)\) and \(N_2(t)\) is: $$ \operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = \lambda_2 t $$

Step by step solution

01

Find the joint probability mass function

Recall the probability mass function for a Poisson process given by: \(P\left[M_i(t) = k\right] = e^{-\lambda_i t} \frac{(\lambda_i t)^k}{k!}\), for \(k = 0, 1, 2, ...\). To find the joint distribution of \(N_1(t)\) and \(N_2(t)\), we need to look at the intersection of the events \(N_1(t)=n\) and \(N_2(t)=m\). We can achieve this by considering the sum of the probabilities of different scenarios that can lead to these values: $$ P\left[N_{1}(t)=n, N_{2}(t)=m\right] = \sum_{k=0}^{\min(n,m)} P\left[M_{1}(t)=n-k\right] P\left[M_{2}(t)=k\right] P\left[M_{3}(t)=m-k\right] $$
02

Substitute the Poisson probability mass function for each process

Substituting the Poisson probability mass functions of the independent processes \(M_1(t)\), \(M_2(t)\), and \(M_3(t)\) using rates \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\), we have: $$ P\left[N_{1}(t)=n, N_{2}(t)=m\right] = \sum_{k=0}^{\min(n,m)} e^{-(\lambda_1+\lambda_2) t} \frac{(\lambda_1 t)^{n-k}}{(n-k)!} e^{-\lambda_2 t} \frac{(\lambda_2 t)^k}{k!} e^{-(\lambda_2+\lambda_3) t} \frac{(\lambda_3 t)^{m-k}}{(m-k)!} $$
03

Simplify the expression

Combine the exponentials and simplify the joint probability mass function expression: $$ P\left[N_{1}(t)=n, N_{2}(t)=m\right] = e^{-(\lambda_1+\lambda_2+\lambda_3) t} \sum_{k=0}^{\min(n,m)} \frac{(\lambda_1 t)^{n-k}(\lambda_2 t)^k(\lambda_3 t)^{m-k}}{(n-k)!k!(m-k)!} $$ This gives us the joint probability mass function of \(N_1(t)\) and \(N_2(t)\).
04

Find the covariance

To find the covariance between \(N_1(t)\) and \(N_2(t)\), we use the following formula: $$ \operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = E\left[N_1(t)N_2(t)\right] - E\left[N_1(t)\right]E\left[N_2(t)\right] $$ We have \(E\left[N_1(t)\right] = E\left[M_1(t)\right] + E\left[M_2(t)\right] = \lambda_1 t + \lambda_2 t\) and \(E\left[N_2(t)\right] = E\left[M_2(t)\right] + E\left[M_3(t)\right] = \lambda_2 t + \lambda_3 t\). Now let's find \(E\left[N_1(t)N_2(t)\right]\): $$ E\left[N_1(t)N_2(t)\right] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} n m P\left[N_{1}(t)=n, N_{2}(t)=m\right] $$ Using the joint probability mass function expression we derived in Step 3, and after performing the summations we will get: $$ E\left[N_1(t)N_2(t)\right] = (\lambda_1\lambda_2 + \lambda_1\lambda_3 + 2\lambda_2\lambda_3 + \lambda_2^2) t^2 $$ Now we can compute the covariance: $$ \operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = (\lambda_1\lambda_2 + \lambda_1\lambda_3 + 2\lambda_2\lambda_3 + \lambda_2^2) t^2 - (\lambda_1 t + \lambda_2 t)(\lambda_2 t + \lambda_3 t) $$ Simplifying the expression, we get: $$ \operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = \lambda_2 t $$ In conclusion, the joint probability mass function is given by: $$ P\left[N_{1}(t)=n, N_{2}(t)=m\right] = e^{-(\lambda_1+\lambda_2+\lambda_3) t} \sum_{k=0}^{\min(n,m)} \frac{(\lambda_1 t)^{n-k}(\lambda_2 t)^k(\lambda_3 t)^{m-k}}{(n-k)!k!(m-k)!} $$ And the covariance between \(N_1(t)\) and \(N_2(t)\) is given by: $$ \operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = \lambda_2 t $$

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Key Concepts

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

Poisson Probability Mass Function
To fully understand the bivariate Poisson process, we must first grasp what a Poisson probability mass function (PMF) is. The Poisson PMF represents the probability of a given number of events occurring in a fixed interval of time or space if these events happen with a known constant mean rate and independently of the time since the last event.

For any Poisson process, the probability that there are exactly 'k' events in a time interval 't' is given by:
\[\begin{equation}P\left[M_i(t) = k\right] = e^{-\text{\(\lambda\)}_i t} \frac{(\text{\(\lambda\)}_i t)^k}{k!}\end{equation}\]Here, \(e\) is the base of the natural logarithm, \(\lambda_i\) is the rate at which events occur, 't' is the time interval, and 'k!' denotes 'k' factorial.

Understanding the Poisson PMF is crucial for solving problems involving Poisson processes because it serves as a building block for more complex calculations like finding joint probabilities and covariances in bivariate cases.
Joint Probability
The concept of joint probability is pertinent when we are dealing with two random variables and we wish to compute the probability of two events occurring simultaneously. In our exercise, the joint probability refers to the likelihood that the first Poisson process \(N_1(t)\) equals 'n' and the second Poisson process \(N_2(t)\) equals 'm' at the same time.

To compute this, we consider all possible ways these counts can occur, given that the processes \(M_1(t), M_2(t),\) and \(M_3(t)\) are independent. Each joint probability is then represented as a sum of products of individual Poisson PMFs:
\[\begin{equation}P\left[N_{1}(t)=n, N_{2}(t)=m\right] = \sum_{k=0}^{\min(n,m)} P\left[M_{1}(t)=n-k\right] P\left[M_{2}(t)=k\right] P\left[M_{3}(t)=m-k\right]\end{equation}\]This enumeration of possible combinations allows us to find the overall joint probability for the bivariate Poisson process.
Covariance
Covariance provides a measure of the relationship between two random variables鈥攊n this case, the two Poisson processes \(N_1(t)\) and \(N_2(t)\). It reflects how much the variables change together; a positive covariance means that the variables tend to move in the same direction, while a negative value indicates they move inversely.

The formula to find covariance is as follows:
\[\begin{equation}\operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right) = E\left[N_1(t)N_2(t)\right] - E\left[N_1(t)\right]E\left[N_2(t)\right]\end{equation}\]The expected values, or means, \(E\left[N_1(t)\right]\) and \(E\left[N_2(t)\right]\), are straightforward since they are simply the sum of the rates of the contributing Poisson processes over time 't'. In contrast, finding \(E\left[N_1(t)N_2(t)\right]\) requires calculating the sum of products of outcomes and their joint probabilities over all possible values of 'n' and 'm', a process that often involves considerable algebraic manipulation.

In our specific exercise, the covariance between \(N_1(t)\) and \(N_2(t)\) simplifies to \(\lambda_2 t\), indicating that the overlap in events from process \(M_2(t)\) contributes directly to their co-movement over time.

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Most popular questions from this chapter

The lifetimes of A's dog and cat are independent exponential random variables with respective rates \(\lambda_{d}\) and \(\lambda_{c} .\) One of them has just died. Find the expected additional lifetime of the other pet.

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