Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 38 3 iv Solution Created 2026-10-03 Updated 2026-10-07
For , the corresponding smaller root is . The conditional probability of selecting the larger root isIt is the same weight as a function of the actual selected value . HenceCombining the two open intervals proves the output has the Inverse Gaussian distribution . The event has probability zero, so no atom occurs at ; the limiting density there is finite. The algorithm always selects one of its positive roots, so these branch densities exhaust its probability distribution.
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 5J Solution Created 2026-09-24 Updated 2026-09-25
Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, . This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 218 1 a Solution Created 2026-09-24 Updated 2026-10-03
Expanding the exponent of the Inverse Gaussian distribution givesThus the exponential dispersion family representationhasandThe standard cumulant identities yieldHence the variance function is , and the canonical link function is .