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Let \(X\) have a Poisson distribution with parameter \(\mu\). Show that \(E(X)=\mu\) directly from the definition of expected value. [Hint: The first term in the sum equals 0 , and then \(x\) can be canceled. Now factor out \(\mu\) and show that what is left sums to 1.]

Short Answer

Expert verified
The expected value of a Poisson distribution with parameter \( \mu \) is \( \mu \).

Step by step solution

01

Write the Definition of Expected Value for a Poisson Distribution

The expected value (mean) of a discrete random variable like a Poisson distribution is given by the formula: \[ E(X) = \sum_{x=0}^{\infty} x P(X = x) \]. For a Poisson distribution, the probability mass function is \( P(X = x) = \frac{e^{-\mu} \mu^x}{x!} \). Thus, the expected value becomes: \[ E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\mu} \mu^x}{x!} \].
02

Simplify the Series Expression

The first term when \( x = 0 \) is 0 because it is multiplied by \( x \). Therefore, we can start the sum from \( x = 1 \): \[ E(X) = \sum_{x=1}^{\infty} x \cdot \frac{e^{-\mu} \mu^x}{x!} = \sum_{x=1}^{\infty} \frac{e^{-\mu} \mu^x}{(x-1)!} \]. This transformation comes from canceling \( x \) with \( x! = x \cdot (x-1)! \).
03

Factor Out the Common Terms

Factor \( \mu \) out of the series:\[ E(X) = \mu \cdot \sum_{x=1}^{\infty} \frac{e^{-\mu} \mu^{x-1}}{(x-1)!} \]. Notice this resembles the sum of a Poisson distribution for \( x' = x-1 \), effectively resembling e raised to a power sum.
04

Simplify the Summation Expression

Let \( x' = x - 1 \), then we can rewrite the sum starting from \( x' = 0 \): \[ \sum_{x'=0}^{\infty} \frac{e^{-\mu} \mu^{x'}}{x'!} \]. This is the expansion of the exponential function \( e^{\mu} \), where \( e^{-\mu} \cdot e^{\mu} = 1 \). As a result, \( \sum_{x=0}^{\infty} \frac{e^{-\mu} \mu^x}{x!} = 1 \).
05

Conclude with the Expected Value

Since the transformed series equals 1 and the factor \( \mu \) was taken outside the summation, we have \( E(X) = \mu \cdot 1 = \mu \). Therefore, we conclude that the expected value of a Poisson distribution with parameter \( \mu \) is indeed \( \mu \).

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Key Concepts

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

Expected Value
The expected value, often referred to as the mean, is a central concept in probability and statistics. It represents the average value you would expect to see from a random variable over many trials. In the context of a Poisson distribution, the expected value is particularly useful. For a discrete random variable like one with a Poisson distribution, the expected value is determined by the formula:\[ E(X) = \sum_{x=0}^{\infty} x \cdot P(X = x) \]Here, \(P(X = x)\) is the probability mass function. This expression essentially finds a weighted average of all possible values, where each value is weighted by its probability of occurrence. When working with a Poisson distribution, you're dealing with a distribution that models the number of events happening in a fixed interval, each event happening independently of the time since the last event. As such, the expected value directly reflects the average rate of occurrence, simplified in our current example to \( \mu \), the parameter of the distribution.
Discrete Random Variable
A discrete random variable is a type of random variable that can only take on a countable number of distinct values. This is in contrast to continuous random variables, which can take on an infinite number of values within a given range. In the realm of Poisson distribution, the random variable typically represents counts of occurrences of an event within a fixed period of time or space. For example, counting the number of cars that pass through a toll booth in an hour, or the number of emails received in a day. Key characteristics of a discrete random variable include:
  • The ability to list all possible outcomes, each associated with a probability.
  • Probabilities that sum up to 1 across all possible values.
  • Discrete values, meaning there are gaps between possible values (e.g., 0, 1, 2, etc.).
Understanding that the Poisson distribution is a discrete type helps in calculating expected values and probabilities tied to events happening discretely rather than continuously.
Probability Mass Function
The probability mass function (PMF) is crucial for understanding discrete random variables like those in a Poisson distribution. The PMF gives you the probability that a discrete random variable is exactly equal to some value.For a Poisson distribution, the probability mass function is represented as:\[ P(X = x) = \frac{e^{-\mu} \mu^x}{x!} \]Here:
  • \( \mu \) is the average rate (the expected value).
  • \( x! \) is the factorial of \( x \), which is the product of all positive integers up to \( x \).
  • \( e \) is the base of the natural logarithm (approximately 2.71828).
This function is derived from the nature of Poisson events, accommodating the way occurrences happen independently and at a constant average rate. Using the PMF, students can determine the likelihood of observing a given number of events within a specified interval.
Exponential Function
The exponential function plays a crucial role in continuous growth models and also within the probability context, particularly within Poisson distributions through the PMF. The function \( e^{x} \) is defined for all real numbers, and you might frequently see it in the form \( e^{-\mu} \) when dealing with the Poisson PMF:\[ P(X = x) = \frac{e^{-\mu} \mu^x}{x!} \]In this expression, \( e^{-\mu} \) represents the probability of zero occurrences modified by the likelihood of a particular count \( x \). This portion of the PMF ensures that cumulatively, probabilities sum to 1 when considering all possibilities from zero to infinity.Understanding \( e^{-\mu} \) helps in demonstrating why certain sums, like ones involving the PMF, converge. It simplifies analyzing behavior across different \( \mu \) values. Thus, in the realm of probability, the exponential function is a tool for modeling situations where values grow or shrink at a consistent relative rate.

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Most popular questions from this chapter

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha=10\) per hour. Suppose that with probability \(.5\) an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed \(y \geq 10\), what is the probability that \(y\) arrive during the hour, of which ten have no violations? c. What is the probability that ten "no-violation" cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from \(y=10\) to \(\left.x_{-}\right]\)

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 25 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using. [Hint: Let \(h(X)=\) the revenue when \(X\) of the 25 purchasers want new copies. Express this as a linear function.]

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either \((1 / 3.5)\) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of five per hour. a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45 min period?

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