Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a

Put

q = 1 - p

. Since

X

P (X > 0) = 1 = q^{0}

. Taking differences of the given tail probabilities therefore yields, also for

k = 1

P (X = k) = P (X > k - 1) - P (X > k) = p q^{k - 1}, k \geq 1.

{1, 2, \dots}

ii

G_{X} (s) = E (s^{X}) = \sum_{k \geq 1} p q^{k - 1} s^{k} = \frac{p s}{1 - q s}, ∣ q s ∣ < 1.

G_{X}^{'} (s) = \frac{p ( 1 - q s ) + pq s}{( 1 - q s ) ^{2}} = \frac{p}{( 1 - s + p s ) ^{2}} .

E X = G_{X}^{'} (1) = \frac{p}{p ^{2}} = \frac{1}{p} .

1/ (1 - p) > 1

W > k

occurs exactly when both

X_{1} > k

and

X_{2} > k

P (W > k) = q^{k} q^{k} = q^{2 k} = (1 - (2 p - p^{2}))^{k} .

Thus

W \sim Geom (2 p - p^{2}) .

The new success probability

2 p - p^{2}

is the chance that at least one of the two trials succeeds.

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