{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AMS410_SOLFINAL

# AMS410_SOLFINAL - AMS 410 Actuarial Mathematics Fall 2009...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern