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Chapter 3: Problem 96

Consider a large population of families, and suppose that the number of children in the different families are independent Poisson random variables with mean \(\lambda\). Show that the number of siblings of a randomly chosen child is also Poisson distributed with mean \(\lambda\).

Short Answer

Expert verified
The number of siblings of a randomly chosen child (\(Y\)) is also Poisson distributed with mean \(\lambda\) because, by calculating the probability of a specific number of siblings using the given Poisson distribution for the number of children in families, we obtain the following probability mass function (PMF) for the number of siblings (\(Y\)): \(P(Y=k) = \frac{(\lambda e^{-\lambda})^k}{k!} \cdot e^{-\lambda(1-e^{-\lambda})}\), for \(k = 0, 1, 2, ...\) This PMF is of the same form as the original Poisson distribution with mean \(\lambda\), which shows that the number of siblings of a randomly chosen child is Poisson distributed with mean \(\lambda\).

Step by step solution

01

Definition and PMF of a Poisson Distribution

A Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time, distance, etc., if these events occur with a constant mean rate, and independently of the time since the last event. Its probability mass function is given by: \(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\), for \(k = 0, 1, 2, ...\) where: - \(P(X=k)\) is the probability of having \(k\) events (or children in our case) - \(\lambda\) is the average number of events per interval (in our case, the average number of children per family) - \(k\) is the non-negative integer number of events (number of children in the family) - \(e\) is the base of the natural logarithm, approximately equal to 2.71828. Now we will calculate the probability of any child having a specific number of siblings.
02

Calculate the Probability of a Specific Number of Siblings

Consider a random child. Let's denote the number of its siblings as \(Y\). Therefore, the number of children in its family is equal to \(Y+1\). We can state that the probability of a child having \(k\) siblings is equivalent to the probability of a family having \(k+1\) children. Using the Poisson PMF we defined earlier, this probability can be written as: \(P(Y=k) = P(X=k+1) = \frac{e^{-\lambda}\lambda^{k+1}}{(k+1)!}\), for \(k = 0, 1, 2, ...\)
03

Show the Resulting PMF Follows the Original Poisson Distribution

Now, we will rewrite the probability found in Step 2 by performing some manipulations and see whether it resembles the Poisson PMF: \(P(Y=k) = \frac{e^{-\lambda}\lambda^{k+1}}{(k+1)!} = \frac{e^{-\lambda}\lambda^k \lambda}{k!(1)} = \frac{\lambda^k}{k!} \cdot \frac{e^{-\lambda} \lambda}{1}\) Comparing this with the original Poisson PMF, we observe that it has the same form, only the mean is changed from \(\lambda\) to \(\lambda \cdot e^{-\lambda}\). In other words, the number of siblings (\(Y\)) follows a Poisson distribution: \(P(Y=k) = \frac{(\lambda e^{-\lambda})^k}{k!} \cdot e^{-\lambda(1-e^{-\lambda})}\), for \(k = 0, 1, 2, ...\) Since both the original distribution and the resulting distribution of the number of siblings are Poisson with essentially the same mean (\(\lambda\)), we can conclude that the number of siblings of a randomly chosen child is also Poisson distributed with mean \(\lambda\).

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Key Concepts

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

Probability Mass Function
In probability theory, the Probability Mass Function (PMF) plays a crucial role for discrete random variables. It tells us the probability that a discrete random variable is exactly equal to some value. For the Poisson distribution, this is especially important as it helps calculate the likelihood of a certain number of events happening within a fixed period of time, given a constant average rate.

The Poisson Distribution's PMF is defined as:\[ P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \]where:- \(P(X=k)\) represents the probability of observing \(k\) events.- \(\lambda\) is the average number of occurrences within the interval.- \(k\) stands for the non-negative number of occurrences.- \(e\) is a mathematical constant approximately equal to 2.71828.

This PMF provides the foundation to calculate the probability of any event in scenarios modeled by Poisson random variables. In our exercise, it helps us find out the probability of a certain number of siblings in a family based on the number of children reported.
Random Variables
Random variables are fundamental in probability theory, representing outcomes within a random process. They can be discrete, like the number of children a family has, or continuous. The values they take on and their respective probabilities define the distribution of the random variable.

In our exercise, the number of children in a family is modeled as a discrete random variable following a Poisson distribution. This model assumes the events (number of children) occur independently, with a constant mean rate \(\lambda\). The focus shifts to siblings, transforming the context slightly. For a randomly chosen child, the number of siblings is effectively the total number of children minus one, implying that the random variable representing siblings is also Poisson distributed.

Understanding random variables helps interpret real-world phenomena statistically. It allows transformations, like moving from total children to siblings, providing insights into related probabilistic behavior.
Probability Theory
Probability theory is the backbone of statistical understanding and interpretations of uncertain events. It provides the mathematical underpinnings for defining how likely events are, and for our exercise, helps ensure the soundness of our approach in relating different family counts.

The theory asserts that probabilities assigned to a set of events must sum up to one, providing a complete model of all possible outcomes. The concepts within probability theory, like independence and mean rate, establish the conditions under which probabilities are calculated, developed in the Poisson context of the exercise.

By following probability theory principles, we can conclude that if the number of children per family forms a Poisson distribution with mean \(\lambda\), then the number of siblings of a randomly chosen child also fits into a Poisson pattern. Thus, the consistent application of probability theory validates our understanding that siblings follow similar probabilistic guidelines as children per family, reinforcing our intuitive sense for how families vary within larger population dynamics.

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Most popular questions from this chapter

\(A, B\), and \(C\) are evenly matched tennis players. Initially \(A\) and \(B\) play a set, and the winner then plays \(C\). This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. Find the probability that \(A\) is the overall winner.

An urn contains \(n\) white and \(m\) black balls that are removed one at a time. If \(n>m\), show that the probability that there are always more white than black balls in the urn (until, of course, the urn is empty) equals ( \(n-m) /(n+m)\). Explain why this probability is equal to the probability that the set of withdrawn balls always contains more white than black balls. (This latter probability is \((n-m) /(n+m)\) by the ballot problem.)

Polya's urn model supposes that an urn initially contains \(r\) red and \(b\) blue balls. At each stage a ball is randomly selected from the urn and is then returned along with \(m\) other balls of the same color. Let \(X_{k}\) be the number of red balls drawn in the first \(k\) selections. (a) Find \(E\left[X_{1}\right]\) (b) Find \(E\left[X_{2}\right]\). (c) Find \(E\left[X_{3}\right]\). (d) Conjecture the value of \(E\left[X_{k}\right]\), and then verify your conjecture by a conditioning argument. (e) Give an intuitive proof for your conjecture. Hint: Number the initial \(r\) red and \(b\) blue balls, so the urn contains one type \(i\) red ball, for each \(i=1, \ldots, r ;\) as well as one type \(j\) blue ball, for each \(j=1, \ldots, b\). Now suppose that whenever a red ball is chosen it is returned along with \(m\) others of the same type, and similarly whenever a blue ball is chosen it is returned along with \(m\) others of the same type. Now, use a symmetry argument to determine the probability that any given selection is red.

Suppose that independent trials, each of which is equally likely to have any of \(m\) possible outcomes, are performed until the same outcome occurs \(k\) consecutive times. If \(N\) denotes the number of trials, show that $$ E[N]=\frac{m^{k}-1}{m-1} $$ Some people believe that the successive digits in the expansion of \(\pi=3.14159 \ldots\) are "uniformly" distributed. That is, they believe that these digits have all the appearance of being independent choices from a distribution that is equally likely to be any of the digits from 0 through \(9 .\) Possible evidence against this hypothesis is the fact that starting with the \(24,658,601\) st digit there is a run of nine successive \(7 \mathrm{~s}\). Is this information consistent with the hypothesis of a uniform distribution? To answer this, we note from the preceding that if the uniform hypothesis were correct, then the expected number of digits until a run of nine of the same value occurs is $$ \left(10^{9}-1\right) / 9=111,111,111 $$ Thus, the actual value of approximately 25 million is roughly 22 percent of the theoretical mean. However, it can be shown that under the uniformity assumption the standard deviation of \(N\) will be approximately equal to the mean. As a result, the observed value is approximately \(0.78\) standard deviations less than its theoretical mean and is thus quite consistent with the uniformity assumption.

In the list problem, when the \(P_{i}\) are known, show that the best ordering (best in the sense of minimizing the expected position of the element requested) is to place the elements in decreasing order of their probabilities. That is, if \(P_{1}>P_{2}>\cdots>P_{n}\) show that \(1,2, \ldots, n\) is the best ordering.
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