Chapter 3: Problem 44
Let \(N_{1}\) and \(N_{2}\) be independent random variables following Poisson distributions with parameters \(\lambda_{1}\) and \(\lambda_{2} .\) Show that the distribution of \(N=N_{1}+N_{2}\) is Poisson with parameter \(\lambda_{1}+\lambda_{2}\)
Short Answer
Expert verified
The sum \(N = N_1 + N_2\) is Poisson with parameter \(\lambda_1 + \lambda_2\).
Step by step solution
01
Understand the Problem
We are given two independent Poisson random variables, \(N_1\) with parameter \(\lambda_1\) and \(N_2\) with parameter \(\lambda_2\). We need to show that their sum, \(N = N_1 + N_2\), is also a Poisson random variable with parameter \(\lambda_1 + \lambda_2\).
02
Characteristic Property of Poisson Distribution
A Poisson random variable \(N\) with parameter \(\lambda\) is defined by its probability mass function: \(P(N = k) = \frac{e^{-\lambda} \lambda^k}{k!}\) for \(k = 0, 1, 2, \ldots\). We aim to show \(P(N = k)\) for \(N = N_1 + N_2\) adheres to this form with \(\lambda = \lambda_1 + \lambda_2\).
03
Use the Sum of Independent Poisson Variables Property
One key property of Poisson distributions is that the sum of two independent Poisson random variables is itself distributed as a Poisson random variable. This means if \(N_1 \sim \text{Poisson}(\lambda_1)\) and \(N_2 \sim \text{Poisson}(\lambda_2)\), then \(N_1 + N_2 \sim \text{Poisson}(\lambda_1 + \lambda_2)\), which completes the proof.
04
Formal Proof Using Generating Functions
The probability generating function for a Poisson random variable \(N\) with parameter \(\lambda\) is given by \(G_N(s) = e^{-\lambda(1-s)}\). For two independent variables \(N_1\) and \(N_2\), the generating function of the sum is the product of their generating functions: \(G(s) = G_{N_1}(s) \cdot G_{N_2}(s) = e^{-\lambda_1(1-s)} \cdot e^{-\lambda_2(1-s)} = e^{-((\lambda_1 + \lambda_2)(1-s))}\), which is the generating function of a Poisson distribution with parameter \(\lambda_1 + \lambda_2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Independent Random Variables
Independent random variables are a key concept in understanding how different probabilistic outcomes do not influence each other. When we have two random variables, say \(N_1\) and \(N_2\), they are independent if the outcome of one does not affect the outcome of the other.
For instance, in a Poisson distribution, if \(N_1\) represents the number of events occurring in one interval and \(N_2\) in another, these counts are independent if events in one interval do not influence the number in the other.
For instance, in a Poisson distribution, if \(N_1\) represents the number of events occurring in one interval and \(N_2\) in another, these counts are independent if events in one interval do not influence the number in the other.
- Formally, independence is expressed as \(P(N_1 = k, N_2 = j) = P(N_1 = k) \cdot P(N_2 = j)\) for all \(k\) and \(j\).
- This multiplication property reflects that knowing the occurrence of \(N_1\) does not change the probability distribution of \(N_2\).
Probability Mass Function
The probability mass function (PMF) is a fundamental tool in understanding discrete random variables, such as those following a Poisson distribution. The PMF gives the probability that a discrete random variable is exactly equal to some value.
In the context of Poisson distribution, the PMF for a random variable \(N\) with parameter \(\lambda\) is given by:
When dealing with the sum of two independent Poisson variables, \(N_1\) and \(N_2\), the resulting random variable \(N = N_1 + N_2\) also follows a Poisson distribution with parameter \(\lambda_1 + \lambda_2\). The PMF remains a crucial aspect to determine how often different event counts occur for \(N\).
In the context of Poisson distribution, the PMF for a random variable \(N\) with parameter \(\lambda\) is given by:
- \(P(N = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
When dealing with the sum of two independent Poisson variables, \(N_1\) and \(N_2\), the resulting random variable \(N = N_1 + N_2\) also follows a Poisson distribution with parameter \(\lambda_1 + \lambda_2\). The PMF remains a crucial aspect to determine how often different event counts occur for \(N\).
Generating Functions
Generating functions offer a powerful way to study random variables and their distributions, providing insights into their properties and behavior. For Poisson random variables, the probability generating function (PGF) is a particularly useful tool.
The PGF of a Poisson random variable \(N\) with parameter \(\lambda\) is expressed as:
The PGF of a Poisson random variable \(N\) with parameter \(\lambda\) is expressed as:
- \(G_N(s) = e^{-\lambda(1-s)}\)
- \(G(s) = G_{N_1}(s) \cdot G_{N_2}(s) = e^{-\lambda_1(1-s)} \cdot e^{-\lambda_2(1-s)}\)
- This simplifies to \(e^{-((\lambda_1 + \lambda_2)(1-s))}\), indicating that \(N = N_1 + N_2\) also follows a Poisson distribution, now with the parameter \(\lambda_1 + \lambda_2\). Generating functions, therefore, not only verify the resulting distribution but also provide an elegant way to handle sums of random variables.