Homepage › Solution manuals › Joseph Blitzstein › Introduction to Probability › Exercise 4.29

Exercise 4.29

Answers

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) = \sum_{k = 0}^{\infty} {({(1 - p)}^{k} p)}^{2} = - \frac{p}{p - 2}

for $| p - 1 |^{2} < 1$ .

fifthist

2021-12-05 00:00

Exercise 4.29

Answers

Comments

Add answer