Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 8 / a

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

i

Solution

In independent Bernoulli trials with success probability

p

,

N

counts failures before the first success. The event

N = n

requires

n

failures followed by one success, giving

(1 - p)^{n} p

. This is the zero-based geometric distribution, rather than the count of trials including the success.

ii

Solution

Let

q = 1 - p

. By the definition of a probability generating function and the geometric series,

G_{N} (s) = E [s^{N}] = \sum_{n = 0}^{\infty} p q^{n} s^{n} = \frac{p}{1 - q s} .

The series converges absolutely when

∣ q s ∣ < 1

, equivalently

∣ s ∣ < 1/ (1 - p)

.

iii

Solution

Differentiate the probability generating function inside its radius of convergence:

G_{N}^{'} (s) = \frac{pq}{( 1 - q s ) ^{2}}, G_{N}^{''} (s) = \frac{2 p q ^{2}}{( 1 - q s ) ^{3}} .

The first moment and second factorial moment are

E [N] = G_{N}^{'} (1) = q / p

and

E [N (N - 1)] = G_{N}^{''} (1) = 2 q^{2} / p^{2}

. Consequently

E [N] = \frac{1 - p}{p}, Var (N) = \frac{2 q ^{2}}{p ^{2}} + \frac{q}{p} - \frac{q ^{2}}{p ^{2}} = \frac{1 - p}{p ^{2}} .

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