Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 8 / a

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

i

Solution

Put

q = 1 - p

. Summing the geometric series in the definition of the probability generating function gives

G_{X} (s) = E [s^{X}] = p k = 0 \sum \infty (q s)^{k} = \frac{p}{1 - q s}, ∣ s ∣ < \frac{1}{q} .

This is the zero-based geometric distribution, counting failures before the first success.

ii

Solution

Differentiate the probability generating function to obtain

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

The first derivative is the expected value, and the second gives the factorial moment

E [X (X - 1)]

. Hence

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

In particular,

E [X^{2}] = q (2 - p) / p^{2}

.

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