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

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)) .

ii

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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