Midterm-Test-2-S-331-2011-F

# Midterm-Test-2-S-331-2011-F - SOLUTIONS TO MIDTERM TEST #2...

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SOLUTIONS TO MIDTERM TEST #2 – ACTSC 331, FALL 2011 1. (a) π ¯ a 45: 15 = 1000 ¯ A 1 45: 25 + 800 25 E 45 . (b) 20 V = 1000 ¯ A 1 65: 5 + 800 5 E 65 . (c) 20 V = ( π ¯ a 45: 15 - 1000 ¯ A 1 45: 20 ) 1 20 E 45 . (d) i. By (a), we get π = 55 . 17. ii. By (b), we get 20 V = 756 . 41. iii. 20 L = 1000 v T 45 - 20 for T 45 25 and 800 v 5 for T 45 > 25. iv. We have V ar ( 20 L | T 45 > 20) = E ( 20 L 2 | T 45 > 20) - ( 20 V ) 2 . By the aggregate mortality law, we get E ( 20 L 2 | T 45 > 20) = Z 5 0 1000 2 v 2 t f T 65 ( t ) dt + 800 2 v 2 × 5 5 p 65 = 578610 . 3323 . Hence, by d(ii), we get V ar ( 20 L | T 45 > 20) = 6454 . 24. 2. (a) Thiele’s diﬀerential equation for the policy value t V is ( t V ) 0 = 0 . 06 t V + π - 600 for 0 < t < 20. The boundary conditions are 0 V = 0 and 20 V = 20000. (b) Using the boundary condition 0 V = 0 and the general solution to the diﬀerential equation, we get t V = e 0 . 06 t ( π - 600) Z t 0 e - 0 . 06 x dx = ( π - 600) × e 0 . 06 t - 1 0 . 06 . Thus, using the boundary condition 20 V = 20000, we get π = 1117 . 22. 3. (a) Kolmogorov’s forward equation for t p 01 x is d dtt p 01 x = - μ 12 x + t t p 01 x + μ 01 x + t t p 00 x . (b) Using the boundary condition 0 p 01 x = 0 and the general solution to the diﬀerential equation, we get,
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## This note was uploaded on 01/04/2012 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.

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