 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/i/solution

B_{n} = {N = n}

for

n \geq 0

, omitting any null cells. For

0 \leq s \leq 1

s^{X}

is bounded, so the law of total expectation applies without moment assumptions. Conditional on

N = n

, the sum contains exactly

n

terms. Independence of

N

from the whole sequence preserves their joint law under this conditioning; independence of the

Y_{k}

then gives

E (s^{X} ∣ N = n) = E (\prod_{k = 1}^{n} s^{Y_{k}}) = \prod_{k = 1}^{n} E (s^{Y_{k}}) = [G_{Y} (s)]^{n} .

For

n = 0

G_{X} (s) = n = 0 \sum \infty P (N = n) [G_{Y} (s)]^{n} = G_{N} (G_{Y} (s)) .

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