Let X have the moment generating function of Example 3.29 and let Y = X − 1. Recall that X is the number of people who need to be checked to get someone who is Rh+, so Y is the number of people checked before the first Rh+ person is found. Find MY(t) using the last proposition in this section.

Example 3.29

Example 3.10

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Example 3.29 Consider testing individuals' blood samples one by one in order to find someone whose blood type is Rh+. The rv X = the number of tested samples should follow the pmf specified in Example 3.10 with p = .85: p(x) = .85(.15)-1 for x = 1,2,3,.... Determining the moment generating function here requires using the formula for the sum of a geometric series: 1+r+²+...= = 1/(1-r) for r] < 1. The moment generating function is ∞ ∞ Mx (t) = E(e¹x) = exp(x) = e.85(.15)*-¹ = .85e¹e¹(x-1)(.15)*-1 XED ∞ x=1 x=1 = .85e¹(.15e¹)-¹ = .85e¹[1 +.15e¹ + (.15e¹)² + -] - x=1 .85et 1.15et The condition on r requires .15e| < 1. Dividing by .15 and taking logs gives t <-In(.15)~ 1.90; i.e., this function is defined in the interval (-∞, 1.90). The result is an interval of values that includes 0 in its interior, so the mgf exists. As a check, Mx(0) = .85/(1.15) = 1, as required.