Exercise 4.17

Answers

(a): The expected number of children in a randomly selected family during a particular era is $E (X) = \sum_{k = 0}^{\infty} k \frac{n_{k}}{\sum_{k = 0}^{\infty} n_{k}} = \frac{m_{1}}{m_{0}}$ .
(b): The expected number of children in the family of a randomly selected child is $E (X) = \sum_{k = 0}^{\infty} k \frac{k n_{k}}{\sum_{k = 0}^{\infty} k n_{k}} = \frac{m_{2}}{m_{1}}$ .
(c): answer in part $b$ is larger than the answer in part $a$ . Since the average in part $a$ is taken over randomly selected families, families with fewer children are weighted the same as families with more children. The average in part $b$ , on the other hand, is taken over individual children, skewing the weights in favor of families with more children.

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2021-12-05 00:00