(a) Give an alternative proof (without m.g.f’s) of the result of Example 2.3-2 using Example 6.4.2-2.

(b) Generalize the result of Example 2.3-2 to the case where the Xi’s are still exponential but the distribution of N is the second version of the negative binomial distribution with parameters (p, ν) (the distribution of K_ν in Section 1.2.2) where ν is a positive integer.

Transcribed Image Text:

EXAMPLE 2. This beautiful example is close to Example 6.4.2-2. Now, we use m.g.f.'s. Let X; be standard exponential, and N have the second version of the geometric distribution with parameter p; that is, P(N = k) = pq, where k = 0, 1, ... and q = 1 - p. We have Then, by (2.3.2), Ms(z): - 1 MX (2) = 1/ - + P p(1-z) 1-qMx(z) 1-q(1 z)- 1-z-q 1 P(pz) p(1-P) =p+ =p+q. P-Z P-Z P MN (2) = q - 1-qe - - pq P-z P-pz p- - pz P-Z P-Z 1 1-z/p -= + P-p P-z (2.3.5) The function is the m.g.f. of the exponential distribution with parameter a = p. Let a (non-random) variable Y = 0. The m.g.f. My(z) = 1. We see that Ms(z) is the linear combination (with the coefficients p and q) of two m.g.f.'s: My(z) and the exponential m.g.f. with parameter p.