problem 29

4.2.5 problem 29

Random variable $f (X)$ takes values that are the probabilities of a random value taken by $X$ . Since $X \sim Geom (p)$ , $f (X) \in {{(1 - p)}^{k} p ∣ ∣ k \in Z_{\geq 0}}$ , and each value ${(1 - p)}^{k} p$ of $f (X)$ occurs with probability ${(1 - p)}^{k} p$ . Thus,

E (X) = \infty \sum k = 0 {({(1 - p)}^{k} p)}^{2} = - \frac{p}{p - 2}

for $∣ ∣ p - 1 {∣ ∣}^{2} < 1$ .

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