Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 7 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii
- iv a
  - Solution iv

i

Solution

The variable has positive-integer support, so

P (X > 0) = 1

. Subtract consecutive values of the survival function:

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

This is the positive-support convention for the geometric distribution.

ii

Solution

Summing a geometric series gives the probability generating function

G_{X} (s) = k \geq 1 \sum p (1 - p)^{k - 1} s^{k} = \frac{p s}{1 - ( 1 - p ) s} .

The series converges absolutely for

∣ s ∣ < 1/ (1 - p)

. Differentiating the rational expression yields

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

iii

Solution

The expected value is the derivative of the probability generating function at one. The expression from part (ii) gives

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

Its finiteness also follows directly from the exponentially decaying geometric probabilities.

iv

Solution

By independence, for

k \geq 0

,

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

This is the geometric minimum law, giving

W \sim Geom (1 - (1 - p)^{2}) = Geom (2 p - p^{2}) .

The support remains the positive integers; this parameter is a success probability, not a mean.

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