Question

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

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

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.

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 \(k = 0, 1, 2, ...\)

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 \(\lambda\).