Negative-binomial sum of exponential claims 2026-10-14
Let and let the original independent count have negative binomial distribution with probability mass function on . With unit-rate exponential distribution severities, its moment-generating function is for . Thus it also has a random sum of independent claims representation with independent binomial distribution count and rate- exponential distribution steps. For the law is . The hurdle decomposition of a positive random sum therefore has zero mass and positive density .
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 38 1 a Solution Created 2026-10-03 Updated 2026-10-14
Write , and suppose first that . Let be policy 's aggregate, and the moment-generating function of its claim sizes. The Poisson distribution count has probability generating function , so the random-sum transform identity and independence of the policies giveThe expression in the weighted sum is the moment-generating function of a mixture distribution with claim cumulative distribution functionThus the aggregate has a compound Poisson distribution with intensity and severity law . This identification remains valid without positive exponential moments, using the same calculation for Laplace transforms of nonnegative random variables and their uniqueness. It is the Poisson superposition of insurance portfolios.
For common unit-rate exponential distribution severities, the merged count is Poisson distributed with mean . Because each severity is strictly positive, exactly when , givingFor , the sum of unit-rate exponential distribution claims has Erlang distribution density . This density follows by induction: convolving the -claim density with gives . Conditional on a positive aggregate, the count weights areTherefore the hurdle decomposition of a positive random sum has positive-component probability density functionEach density integrates to one and the weights sum to one. Hence for , where is the conditional cumulative distribution function given . If , the aggregate is identically zero, and the positive component is undefined; the requested presupposes a positive total intensity.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 31 1 b Solution Created 2026-10-03 Updated 2026-10-06
The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies exactly when . ChooseIf , take to have the zero-truncated claim-count distribution, namely the conditional law of given . ThenChoose this count independently of a fresh independent claim-size sequence and put . Its probability distribution is that of conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable of success probability realizes . This establishes the distributional representation, including that is itself a positive random sum of independent claims. If , the aggregate is identically zero; set and choose any positive , for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.
For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value has transform . Substitution yieldsHence is exponential with rate and expected value . This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking .
The resulting distribution function isIts jump of size at zero is important: the aggregate law is not a purely continuous exponential distribution.