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Unformatted text preview: ACTSC 431/831: Loss Models 1 Spring 2009  TEST #2 1. (12 marks) Let S 1 ; S 2 and S 3 be three independent compound Poisson random variables: & S 1 has a Poisson parameter & 1 = 2 and secondary distribution q = 0 : 4 and q 2 = 0 : 6 ; & S 2 has a Poisson parameter & 2 = 3 and secondary distribution q = 0 : 2 and q 1 = 0 : 8 ; & S 3 has a Poisson parameter & 3 = 5 and secondary distribution q = 0 : 3 , q 1 = 0 : 3 and q 2 = 0 : 4 . (a) (5 marks) Show that S = S 1 + S 2 + S 3 is also compound Poisson. Identify its Poisson parameter & and its secondary distribution. Solution: Using probability generating functions, P S ( t ) = E & t S ¡ = E & t S 1 + S 2 + S 3 ¡ = E & t S 1 ¡ E & t S 2 ¡ E & t S 3 ¡ = e & 1 ( : 4+0 : 6 t 2 & 1 ) e & 2 (0 : 2+0 : 8 t & 1) e & 3 ( : 3+0 : 3 t +0 : 4 t 2 & 1 ) = e & ( P ( t ) & 1) , where & = & 1 + & 2 + & 3 = 2 + 3 + 5 = 10 , and P ( t ) = & 1 & ¢ : 4 + 0 : 6 t 2 £ + & 2 & (0 : 2 + 0 : 8 t ) + & 3 & ¢ : 3 + 0 : 3 t + 0 : 4 t 2 £ = 2 10 ¢ : 4 + 0 : 6 t 2 £ + 3 10 (0 : 2 + 0 : 8 t ) + 5 10 ¢ : 3 + 0 : 3 t + 0 : 4 t 2 £ = 2 10 (0 : 4) + 3 10 (0 : 2) + 5 10 (0 : 3) + ¤ 3 10 (0 : 8) + 5 10 (0 : 3) ¥ t + ¤ 2 10 (0 : 6) + 5 10 (0 : 4) ¥ t 2 = 0 : 29 + 0 : 39 t + 0 : 32 t 2 ....
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 Spring '09
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 Probability theory, Trigraph, Probabilitygenerating function, Generating function, 1 j

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