Actuarial Outpost SOA 3 Nov 2003 Questions
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 Fill in a brief DW Simpson Registration Form to be contacted when new jobs meet your criteria.

 Long-Term Actuarial Math Old Exam MLC Forum

#61
11-05-2003, 10:02 AM
 IA_Hawk Member Join Date: Nov 2002 Location: Chicago Posts: 99

I vaguely remember the LER question. Do you remember the details of the question??

Was it something like for deductible d the LER was .3 ??
The LER for exponential is F(d) ??

.3 = 1 - exp(-d/theta)
.7 = exp(-d/theta)
ln(.7) = -d/theta

for next year the deductible is increased by 4/3

so 4/3 ln(.7) = -(4/3)d/theta then LER = 1 - exp[-(4/3)d/theta]

= 1- .7^(4/3) = .378

Does this hold any water?
#62
11-05-2003, 10:05 AM
 asarfatti Member Join Date: Nov 2003 Posts: 35

I followed the same procedure as hawk, but can't recall the details of the question (eg, specific numbers). Intuitively, a declining LER cannot be correct -- when you raise the deductible the LER must also rise if the underlying distribution remains the same. Memorylessness of exponential made this problem much easier.
#63
11-05-2003, 10:08 AM
 DukeCrow Member CAS AAA Join Date: Jan 2003 Posts: 77

I thought the LER was 0.70, originally. Or am I thinking of another problem? My new LER with the 4/3 increase in d was 0.80.
#64
11-06-2003, 11:44 AM
 Tommy Vercetti Member CAS Join Date: Jan 2003 Posts: 988

#8 and #28.
#65
11-06-2003, 12:22 PM
 Msta Member Join Date: Nov 2002 Posts: 53

I agree, help with #8.

For 28:

E(K^3) = 0*qx + 1*qx+1 + 2*qx+2 + 3*3px

if you die in the first year, K=0
second year, K=1
third year, K=2.
K is at most 3, and is equal to 3 if you survive to 3 (3px)

Calculate E(K^3) squared and get the variance.

Didn't get this on the exam, but is clear as mud now.
#66
11-06-2003, 01:06 PM
 Becoming An Actuary Member Join Date: May 2003 Posts: 162

Quote:
 Originally Posted by Msta I agree, help with #8. For 28: E(K^3) = 0*qx + 1*qx+1 + 2*qx+2 + 3*3px if you die in the first year, K=0 second year, K=1 third year, K=2. K is at most 3, and is equal to 3 if you survive to 3 (3px) Calculate E(K^3) squared and get the variance. Didn't get this on the exam, but is clear as mud now.
Not quite....the probability of dying in the second year, for example, is p_x*p_x+1 (not q_x+1)...here's the correct solution

K^3
= 0 w/ pr. q_x=0.1
= 1 w/ pr. p_x*q_x+1=0.9*0.2=0.18
= 2 w/ pr. p_x*p_x+1*q_x+2=0.9*0.8*0.3=0.216
= 3 w/ pr. p_x*p_x+1*p_x+2=0.9*0.8*0.7=0.504

E[K^3]=2.124
E[(K^3)^2]=5.58
VAR[K^3]=1.068624 ie. A)1.1
#67
11-06-2003, 01:10 PM
 Msta Member Join Date: Nov 2002 Posts: 53

That's what I meant. Forgot the px and px+1
#68
11-06-2003, 01:56 PM
 DukeCrow Member CAS AAA Join Date: Jan 2003 Posts: 77

Quote:
#8
At age 60 and every subsequent year, the PV of benefits equals the PV of payments (equal to 1000qv for every year), so the reserve <sub>20</sub>V is equal to zero. Therefore, you can look at the premium π as the level benefit premium for a 20-yr term insurance on (40), since that's all the 20 payments are paying for.

π = (A<sub>40</sub> - <sub>20</sub>E<sub>40</sub> A<sub>60</sub>) / (ä<sub>40</sub> - <sub>20</sub>E<sub>40</sub> ä<sub>60</sub>) = 60.13 / 11.76 (from ILT) = 5.11 B
.
#69
11-06-2003, 02:03 PM
 Tommy Vercetti Member CAS Join Date: Jan 2003 Posts: 988

ah. chapter 8 crap...
thx.