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#61




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




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.

#65




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




Quote:
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




That's what I meant. Forgot the px and px+1

#68




Quote:
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 20yr 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 . 
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