Exercise 6. If X is a geometric random variable, show that P(X > n + k − 1|X > n − 1) = P(X > k) In light of the construction of a geometric distribution from a sequence of independent Bernoulli trials, how can this be interpreted so that it is "obvious”?
Exercise 6. If X is a geometric random variable, show that P(X > n + k − 1|X > n − 1) = P(X > k) In light of the construction of a geometric distribution from a sequence of independent Bernoulli trials, how can this be interpreted so that it is "obvious”?
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 31E
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