 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/8/c/ii/solution

1/2

has

G_{Y} (s) = (1 + s) /2

G_{X} (s) = \frac{1}{2 - ( 1 + s ) /2} = \frac{2}{3 - s} = \frac{2}{3} \sum_{j = 0}^{\infty} (\frac{s}{3})^{j} .

P (X = j) = \frac{2}{3} (\frac{1}{3})^{j}, j = 0, 1, \dots .

Thus $X$ has the zero-based geometric distribution with success parameter $2/3$ . The probabilities sum to one, and its mean

1/2

agrees with the random-sum mean

E (N) E (Y_{1}) = 1 \cdot 1/2

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