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#1
07-19-2012, 03:23 PM
 kangh4 Member Join Date: Mar 2010 Posts: 132
ASM 11th edition second printing page 89

For example 5G, i'm wondering why the p_x calculation has the curt ate future lifetime in the integrate in the first line of the calculation instead of the force of mortality. Can someone explain why that would be or if this is a typo? It works out if we substitute u_x+s instead.
#2
07-19-2012, 05:52 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,197

It's a typo.
#3
07-20-2012, 10:32 AM

For page 269, problem 13.1, the question asked Pr(Z<=400). The solution provided is probablilty from 0 to t where z will be 400. Should it be 1 - the probability from 0 to t where at that T, z will be 400? Because present value decrease as t increases right?
#4
07-20-2012, 11:33 AM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,197

The solution is correct. Pr(Z<=400)=Pr(T>=t)=e^{-0.04t}, because for constant force of mortality, $\Pr(T\ge t)=e^{-\mu t}$.
#5
07-20-2012, 12:09 PM

Im sorry I am still confused. Pr(T>t) - probabillity T will be greater than given t - should that be the second half of the graph where T ( Times) > t (given time). also that means 1 - e-(ut) = 1 - the probability to the left of given t.
#6
07-20-2012, 12:46 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,197

Well, Pr(T<t) is the probability that T is less than t, and on the usual left-to-right graph, that means T will be some number to the left of t. Is that what you're asking?
#7
07-20-2012, 02:43 PM

yes. that is Pr(T<t). So is Pr(T<t) = e^-ut? which makes Pr(T>t) = 1- e^-ut.

thanks
#8
07-20-2012, 03:17 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,197

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