# Life Contingencies

• May 1st 2009, 07:21 AM
plm2e
Life Contingencies
please see attached word doc for problem, I could not get all the symbols to properly display on the thread.

there are 2 problems, 1st uses the survival function and the second is using de moivre's law. any help would be appreciated
• May 1st 2009, 01:50 PM
mr fantastic
Quote:

Originally Posted by plm2e
please see attached word doc for problem, I could not get all the symbols to properly display on the thread.

there are 2 problems, 1st uses the survival function and the second is using de moivre's law. any help would be appreciated

Many members do not open attached files of any sort - and with good reason.

It would be easier for everyone if you just typed your questions rather than attaching them as a word file. That way they can be easily seen and quoted.

If you want to help, it's in your own self-interest to make things as easy as possible for people.
• May 2nd 2009, 07:27 AM
plm2e
ok, sorry about that. I was trying to make it look neat.

You are given that
s(x) = 1 - .01x^2 for 0 ≤ x ≤ 10 and l0 = 1000.
Find a.) μ(4) b.) °e8 (the ° is supposed to be above the e)

• May 2nd 2009, 11:04 AM
plm2e
Quote:

Originally Posted by plm2e
ok, sorry about that. I was trying to make it look neat.

You are given that
s(x) = 1 - .01x^2 for 0 ≤ x ≤ 10 and l0 = 1000.
Find a.) μ(4) b.) °e8 (the ° is supposed to be above the e)

i think i figured it out if anyone can check my work.

a.
μ(4) = -s'(x)/s(x) = .08/.84 = .0952

b.
°e8 = E[T(8)] = (from 0 to infin) tp8 dt
= [1-.01(8+t)^2] / [1-.01(8^2)]
= .36 1-.64-.16t-.01t^2 dt =
= [.36t-.08t^2-(.01/3)t^3]l (from 0 to 2) *[(.36)]
= (.72-.32-.27)*.36 = .1344
• May 3rd 2009, 06:32 PM
TKHunny
a. I'd prefer the prettier 2/21.

b. Does that make sense? Curtate survival to age 9 is s(9)/s(8) = 0.528