By Yiu-Kuen Tse
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Additional info for Nonlife Actuarial Models: Theory, Methods and Evaluation (International Series on Actuarial Science)
14) X for a general mixed distribution X . 15) g(xi )fX (xi ). 17) 0 where g (x) is the derivative of g(x) with respect to x. Deﬁning g(x) = x, so that g(0) = 0 and g (x) = 1, the mean of X can be evaluated by ∞ E(X ) = ∞ [1 − FX (x)] dx = 0 SX (x) dx. 2 Let X ∼ U(0, 100). 18). Deﬁne a random variable Y as follows Y = 0, X − 20, for X ≤ 20, for X > 20. Determine the df of Y , and its density and mass function. 3 for a proof. 01 0 1002 2 = 50. 01x dx = 50. 0 Thus, the result is veriﬁed. 2. 01y. 01y, ⎪ ⎩ 1, for y < 0, for y = 0, for 0 < y ≤ 80, for y > 80.
Thus, by the uniqueness of mgf, X ∼ PN (λ). 3 Some discrete distributions for claim frequency 13 It turns out that the converse of the above result is also true, as summarized in the following theorem. 2 Suppose an event A can be partitioned into m mutually exclusive and exhaustive events Ai , for i = 1, · · · , m. Let X be the number of occurrences of A, and Xi be the number of occurrences of Ai , so that X = X1 + · · · + Xm . e. Pr(Ai | A) = pi , with m i=1 pi = 1. If X ∼ PN (λ), then Xi ∼ PN (λi ), where λi = λpi .
1). If these two blocks of policies are independent, what is the probability that their aggregate claim frequency is less than 3? 30 Claim-frequency distribution Solution Note that the aggregate claim frequency S of the two blocks of policies is the sum of two compound Poisson distributions. Thus, S is itself a compound Poisson distribution with λ = 2 + 4 = 6. We shall apply the Panjer recursion to compute the distribution of S. To do this, we ﬁrst consider the pf of the secondary distributions of the blocks of policies.
Nonlife Actuarial Models: Theory, Methods and Evaluation (International Series on Actuarial Science) by Yiu-Kuen Tse