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Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 2 / 4F / Solution

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 2 4F
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Writing pn​=P(X=n), the ordinary probability generating function is
PX​(s)=E[sX]=∑n=1K​pn​sn.
(1)
Its finite sum can be differentiated termwise, giving
EX=PX′​(1).​
(2)
The Dirichlet probability generating function is the different transform q(z)=E[X−z]. Evaluating at z=−1 gives the first ordinary moment, whereas differentiation gives q′(z)=−∑n​pn​(logn)n−z. Therefore
EX=q(−1),E[logX]=−q′(0).​
(3)
There is no convergence issue because the support is finite and all its values are positive.

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