exam mlc nov 2005 solutions

exam mlc nov 2005 solutions - Fall 2005 Exam M Solutions...

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Fall 2005 Exam M Solutions Question #1 Key: C [] [] 2 2 Var Z E Z E Z ⎡⎤ =− ⎣⎦ [] () () ( ) 0.08 0.03 0.02 00 0.02 tt t t tt x x EZ vb p td t e e e d t µ ∞∞ −− == ∫∫ 0.07 0 0.02 2 0.02 7 0.07 t ed t = 2 2 2 0.05 0.02 0.02 tt t xx E Z v b p t dt e e dt 0.12 0 21 0.02 12 6 t x et d t = 2 11 4 2 0.08503 7 66 4 9 Var Z =− = Question #2 Key: C From 1 x Ad a x ±± we have 0.1 3 18 11 1.1 x A = 10 0.1 5 16 11 1.1 x A + = xx i AA δ 10 30 . 1 0.2861 11 ln 1.1 50 . 1 0.4769 11 ln 1.1 x x A A + = = 10 10 10 VA P Aa ++ × 0.2861 0.4769 6 8 ⎛⎞ ⎜⎟ ⎝⎠ = 0.2623 There are many other equivalent formulas that could be used.
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Question #3 Key: C Regular death benefit 0.06 0.001 0 100,000 0.001 tt ee d t −− × 0.001 100,000 0.06 0.001 ⎛⎞ = ⎜⎟ + ⎝⎠ = 1639 34 . Accidental death () 10 0.06 0.001 0 100,000 0.0002 d t = 10 0.061 0 20 t ed t = 0.61 1 20 149.72 0.061 e ⎡⎤ == ⎢⎥ ⎣⎦ Actuarial Present Value =+ = 1639 34 149 72 1789 06 .. . Question #4 Key: D Once you are dead, you are dead. Thus, you never leave state 2 or 3, and rows 2 and 3 of the matrix must be (0 1 0) and (0 0 1). Probability of dying from cause 1 within the year, given alive at age 61, is 160/800 = 0.20. Probability of dying from cause 2 within the year, given alive at age 61, is 80/800 = 0.10 Probability of surviving to 62, given alive at 61, is 560/800 = 0.70 (alternatively, 1 – 0.20 – 0.10), so correct answer is D.
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Question #5 Key: C This first solution uses the method on the top of page 9 of the study note. Note that if the species is it is not extinct after 3 Q it will never be extinct. This solution parallels the example at the top of page 9 of the Daniel study note. We want the second entry of the product () 123 3 QQQe ×× which is equal to ( ) ( ) 3 QQQ e ××× . 3 00 . 1 11 Q = 2 . 0 1 0.1 0.27 Q = 1 0.01 0.049 0.27 0.489 Q = The second entry is 0.489; that’s our answer. Alternatively, start with the row matrix (0 1 0) and project it forward 3 years. (0 1 0 ) 1 Q = (0.00 0.70 0.30) (0 0.70 0.30) 2 Q = (0.07 0.49 0.44) (0.07 0.49 0.44) 3 Q = (0.16 0.35 0.49) Thus, the probability that it is in state 3 after three transitions is 0.49. Yet another approach would be to multiply × × , and take the entry in row 2, column 3. That would work but it requires more effort.
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Question #6 Key: B Probabilities of being in each state at time t : t Active Disabled Dead Deaths 0 1.0 0.0 0.0 1 0.8 0.1 0.1 0.1 2 0.65 0.15 0.2 0.1 3 not needed not needed 0.295 0.095 We built the Active Disabled Dead columns of that table by multiplying each row times the transition matrix. E.g., to move from t = 1 to t = 2, (0.8 0.1 0.1) Q = (0.65 0.15 0.2) The deaths column is just the increase in Dead. E.g., for t = 2, 0.2 – 0.1 = 0.1. v = 0.9 APV of death benefits = () 23 100,000* 0.1 0.1 0.095 24,025.5 vv v ++ = APV of $1 of premium = 2 1 0.8 0.65 2.2465 = 24,025.5 Benefit premium = 10,695 2.2465 =
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Question #7 Key: A Split into three independent processes: Deposits, with () ( ) ( ) * 0.2 100 8 160 λ == per day Withdrawals, with ( ) ( ) * 0.3 100 8 240 per day Complaints. Ignore, no cash impact. For aggregate deposits, ()( ) ( ) 160 8000 1,280,000 ED ( ) ( ) 22 160 1000 160 8000 Var D =+ 10 1.04 10 For aggregate withdrawals 240 5000 1,200,000 EW ( ) ( ) ( ) 240 2000 240 5000 Var W 10 0.696 10 1,200,000 1,280,000 80,000 EW D −= = 10 10 10 0.696 10 1.04 10 1.736 10 Var W D × + × = × 131,757 SD W D ( ) 80,000 80,000 Pr Pr 0 Pr 131,757 131,757 WD −+ ⎛⎞ >= > = > ⎜⎟ ⎝⎠ ( ) 1 0.607 0.27 =−Φ =
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Question #8 Key: D Exponential inter-event times and independent implies Poisson process (imagine additional batteries being activated as necessary; we don’t care what happens after two have failed).
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This note was uploaded on 12/21/2010 for the course ACT act247 taught by Professor Act during the Spring '08 term at University of Toronto- Toronto.

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exam mlc nov 2005 solutions - Fall 2005 Exam M Solutions...

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