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Exercise 4.14

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$E (X) = c \sum_{i = 1}^{\infty} p^{k} = c (\frac{1}{1 - p} - 1) = - \frac{1}{\log (1 - p)} \frac{p}{1 - p}$

$E (X) = E (X^{2}) - {(E (X))}^{2} = c \sum_{i = 1}^{\infty} i p^{i} - {(- \frac{1}{\log (1 - p)} \frac{p}{1 - p})}^{2} = - \frac{1}{\log (1 - p)} \frac{p}{{(p - 1)}^{2}} - {(- \frac{1}{\log (1 - p)} \frac{p}{1 - p})}^{2}$

fifthist

2021-12-05 00:00

Exercise 4.14

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