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ANSWERS TO PRACTICE QUESTIONS 6 – ACTSC 431/831, FALL 2011 1. (a) The probability is Pr { N 3 12 = 1 , N 5 12 = 2 } = Pr { N 3 12 = 1 } Pr { N 2 12 = 1 } = 0 . 02747 . (b) The probability is Pr { N 1 12 = 0 | N 1 2 = 3 } = Pr { N 1 12 =0 } Pr { N 5 12 =3 } Pr { N 1 2 =3 } = 0 . 5787 . (c) The probability is 5 2 ! (Pr { X i > 1 } ) 2 (Pr { X i 1 } ) 3 = 0 . 34183 . (d) The probability is Pr { N 1 = 2 , X 1 > 1 , X 2 > 1 } = Pr { N 1 = 2 } Pr { X 1 > 1 } Pr { X 2 > 1 } = 0 . 02489 . (e) The expected time of the 2nd claim is E T 2 | T 6 = 20 12 = E W 1 + W 2 | W 1 + · · · + W 6 = 20 12 = 2 E W 1 | W 1 + · · · + W 6 = 20 12 = 2 × 20 12 × 1 6 = 5 9 . (f) Let ψ n ( u ) be the probability that ruin occurs on or before the n th claim, n = 1 , 2 .... For any u 0, ψ 1 ( u ) = R 0 e - ( u +1 . 2 t ) e - t dt = e - u 2 . 2 . Furthermore, for any u 0, ψ 2 ( u ) = Z 0 " Z u +1 . 2 t 0 e - ( u +1 . 2 t - x ) 2 . 2 e - x dx + e - ( u +1 . 2 t ) # e - t dt = 1 . 2 2 . 2 3 + 1 2 . 2 + u 2 . 2 2 e - u = (0 . 567243 + 0 . 206612 u ) e - u . 2. (a) The expected surplus is E ( U 4 ) = 5 . 8 - E N 1 4 i =1 X i - E N 2 4 i =1 Y i = 1 . 8 . (b) The variance of the surplus is V ar ( U 4 ) = V ar N 1 4 i =1 X i + V ar N 2 4 i =1 Y i = 9 . 778 . (c) Note that { U t , t 0 } has the same distribution as a surplus process { U * t = u + 1 . 2 t - N t i =1 Z i , t 0 } , where N t is a Poisson process with rate 1 / 3 + 2 / 3 = 1 and Z has a two-point mixture distribution F Z ( x ) = (1 / 3)(1 - e - 3 x ) + (2 / 3)(1 - e - 3 x/ 4 ) , x 0. Thus, ψ ( u ) is equal to the infinite-time ruin probability in the surplus process
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