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Introduction to Probability

Exercise 3.5

Answers

(a)

p (n)

is clearly non-negative. Also,

\sum_{n = 0}^{\infty} p (n) = \frac{1}{2} \sum_{n = 0}^{\infty} \frac{1}{2^{n}} = \frac{1}{2} * \frac{1}{1 - \frac{1}{2}} = 1 .

Thus, $p (n)$ is a valid PMF.

(b)

F (x) = \sum_{n = 0}^{⌊ x ⌋} p (n) = \frac{1}{2} \sum_{n = 0}^{⌊ x ⌋} \frac{1}{2^{n}} = \frac{1}{2} * \frac{1 - \frac{1}{2^{⌊ x ⌋ + 1}}}{1 - \frac{1}{2}} = 1 - \frac{1}{2^{⌊ x ⌋ + 1}}

for $x \geq 0$ and $0$ for $x < 0$ .

fifthist

2021-12-05 00:00

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