 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/probability/8/b/ii/solution

For

0 \leq s \leq 1

, condition on

N = k

. Independence of the descendant total progeny variables gives

E [s^{\sum_{i = 1}^{N} Z_{i}} ∣ N = k] = \prod_{i = 1}^{k} E [s^{Z_{i}}] = G_{Z} (s)^{k},

1

when

k = 0

. Taking expectation over

N

E [s^{\sum_{i = 1}^{N} Z_{i}}] = k = 0 \sum \infty P (N = k) G_{Z} (s)^{k} = G_{N} (G_{Z} (s)) .

The same identity holds for complex

∣ s ∣ < 1

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