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Magazine Chapter 5: Problem 12
Show that the values of the Poisson distribution given in Equation 5.2 sum to \(1 .\)
Short Answer
Expert verified
The sum of the Poisson distribution probability values equals 1, as shown by the series expansion.
Step by step solution
01
Write the Poisson Probability Mass Function
The Poisson distribution provides the probability of a given number of events happening in a fixed interval of time or space. The probability mass function (PMF) of the Poisson distribution is given by the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \( \lambda \) is the average number of events in the interval, \( e \) is the base of the natural logarithm, and \( k \) is the number of events.
02
Express the Sum of Probabilities
Since the Poisson distribution needs to add to 1 over all possible values of \( k \), express the sum of probabilities for all \( k \) from 0 to infinity: \[ \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \] Our goal is to show that this sum equals 1.
03
Use Series Expansion to Simplify
Recognize that the series we have is of the exponential series form. Exponential functions have the series expansion given by:\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]If we set \( x = \lambda \), then:\[ e^\lambda = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \]
04
Relate the Series to Exponential Function
Since the series expansion of the exponential function \( e^\lambda \) matches the form of our sum in Step 2 (multiplied by \( e^{-\lambda} \)), we can multiply both sides of the exponential expansion by \( e^{-\lambda} \):\[ e^{-\lambda} \times e^{\lambda} = e^{0} = 1 \]
05
Conclude That the Sum Equals 1
From Step 4, we have shown that multiplying the exponential expansion on both sides by \( e^{-\lambda} \) results in 1. Therefore, the sum of the probabilities of a Poisson distribution indeed sums to 1: \[ \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} = 1 \] This confirms that the distribution is valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Mass Function
The probability mass function (PMF) is a fundamental concept to grasp when dealing with discrete probability distributions such as the Poisson distribution. It essentially describes the probability that a discrete random variable is exactly equal to some value. For the Poisson distribution, the PMF is given by:
- \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]
- \( X \) represents the random variable indicating the number of events occurring.
- \( k \) is the specific number of events that we want to find the probability for.
- \( \lambda \) is the average or expected number of events in a given interval.
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
- \( k! \) is the factorial of \( k \) (e.g., \( 3! = 3 \times 2 \times 1 = 6 \)).
Exponential Series
Understanding the exponential series is crucial when proving properties related to the Poisson distribution. An exponential series is an infinite sum that represents the exponential function. This is given by:
- \[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]
- For \( e^\lambda \), substitute \( x = \lambda \) in the series: \[ e^\lambda = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \]
- This indicates that the equation of the exponential and our Poisson PMF share the same structure except for the front multiplication by \( e^{-\lambda} \).
Mathematical Proof
A mathematical proof provides a logical reasoning that confirms a concept's validity using established rules. In our case, we need to show how the sum of probabilities in a Poisson distribution equals one. The formulation is:
- \[ \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} = 1 \]
- Recognize that \( \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \) is the series expansion for \( e^\lambda \).
- The expression \( e^{-\lambda} \times \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \) simplifies because multiplying by \( e^{-\lambda} \) results in: \[ e^{-\lambda} e^\lambda = e^0 = 1 \]
Sum of Probabilities
The sum of probabilities is a central concept when dealing with any probability distribution. For the Poisson distribution, it is necessary for the probabilities of all possible event counts (from 0 to infinity) to add up to 1. This confirms that the total probability accounts for all potential outcomes.
- The expression \( \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \) represents this cumulative sum for the Poisson distribution.
- Thanks to the exponential series \[ e^\lambda = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \], and multiplying it by \( e^{-\lambda} \), we smoothly reach the conclusion: \[ e^{-\lambda} e^\lambda = 1 \]
