 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/iii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 7 / b / iii / Solution by

Codex  0 2026-10-25

D (s) = 4 - 2 s - s^{2}

, so

D (s) G (s) = s^{2}

. Differentiate once and evaluate at

s = 1

, using

D (1) = 1

,

D^{'} (1) = - 4

,

G (1) = 1

:

- 4 + G^{'} (1) = 2, G^{'} (1) = 6.

A second differentiation gives

D^{''} G + 2 D^{'} G^{'} + D G^{''} = 2

, hence

- 2 - 8 \cdot 6 + G^{''} (1) = 2, G^{''} (1) = 52.

The derivatives of a probability generating function give

G^{'} (1) = E [F]

and

G^{''} (1) = E [F (F - 1)]

. Consequently $Var (F) = 52 + 6 - 6^{2} = 22$ . The earlier geometric tail bound from a uniform escape probability ensures the needed moments are finite.

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