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Unformatted text preview: AMS 410 Actuarial Mathematics Fall 2009 Final Exam Solution 12/17/2009 1. Let X denote the annual numbers of tornadoes in counties P, Y denote the annual numbers of tornadoes in counties Q. An actuary determines that X and Y are jointly distributed as follows: Y 1 2 3 0.12 0.06 0.05 0.02 X 1 0.13 0.15 0.12 0.03 2 0.05 0.15 0.10 0.02 Calculate the conditional variance of the annual number of tornadoes in county Q, given that there are no tornadoes in county P. (A) 0.51 (B) 0.84 (C) 0.88 (D) 0.99 (E) 1.76 Solution : The requested is V ar ( Y  X = 0) . P ( X = 0) = 0 . 12 + 0 . 06 + 0 . 05 + 0 . 02 = 0 . 25 . Thus the conditional probabilities are: Y  X = 0 1 2 3 0.12/.25=.48 0.06/.25=.24 0.05/.25=.2 0.02/.25=.08 Thus E ( X  Y = 0) = . 48 × 0 + . 24 × 1 + . 2 × 2 + . 08 × 3 = . 88 , E ( X 2  Y = 0) = . 48 × 0 + . 24 × 1 + . 2 × 4 + . 08 × 9 = 1 . 76 . So V ar ( Y  X = 0) = 1 . 76 . 88 2 = 0 . 9856 . 2. Let N 1 and N 2 represent the numbers of claims led by a family in January and February, respectively. N 1 follows a Poisson distribution with mean 2. N 2 follows a Binomial distribution with parameters n = 5 and p = 0 . 2 . N 1 and N 2 are independent. Calculate the probability that in total there will be less than 2 claims submitted in the two months. (A) 0.188 (B) 0.144 (C) 0.133 (D) 0.089 (E) 0.055 Solution : The requested probability is P ( N 1 + N 2 < 2) = P ( N 1 = 0 ,N 2 = 0) + P ( N 1 = ,N 2 = 1) + P ( N 1 = 1 ,N 2 = 0) . By independence, the probability = P ( N 1 = 0) · P ( N 2 = 0) + P ( N 1 = 0) · P ( N 2 = 1) + P ( N 1 = 1) · P ( N 2 = 0) . N 1 follows a Poisson distribution with λ = 2 , P ( N 1 = 0) = e 2 = 0 . 135 , P ( N 1 = 1) = 2 · e 2 = 0 . 271 . N 2 follows Bin(5,0.2), P ( N 2 = 0) = ( 5 ) · . 2 · . 8 5 = 0 . 8 5 = 0 . 328 , P ( N 2 = 1) = ( 5 1 ) · . 2 1 · . 8 4 = 0 . 8 4 = 0 . 410 . Plug in those probabilities, P ( N 1 + N 2 < 2) = 0 . 188 . 3. The moment generating function of a random variable X is given by M X ( t ) = 1 1 100 t 2 for . 1 < t < . 1 . Determine the variance of X . 1 (A) 50 (B) 100 (C) 200 (D) 400 (E) 800 Solution : M X ( t ) = (1 100 t 2 ) 1 ; M X ( t ) = ( 1)(1 100 t 2 ) 2 ( 200 t ) = 200 t · (1 100 t 2 ) 2 , E ( X ) = M X (0) = 0 ; M 00 X ( t ) = 200 · (1 100 t 2 ) 2 +200 t · ( 2) · (1 100 t 2 ) 3 ( 200 t ) = 200 · (1 100 t 2 ) 2 +80000 t 2 · (1 100 t 2 ) 3 , E ( X 2 ) = M 00 X (0) = 200 . V ar ( X...
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 Spring '08
 Yang,Y
 Normal Distribution, Probability theory, probability density function

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