Actuarial Outpost Lesson 20 proof of the alternative formula for EPV of continuous annuity
 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

 Enter your email to subscribe to DW Simpson weekly actuarial job updates. li.signup { display: block; text-align: center; text-size: .8; padding: 0px; margin: 8px; float: left; } Entry Level Casualty Health Life Pension All Jobs

 Long-Term Actuarial Math Old Exam MLC Forum

#1
01-12-2019, 08:11 AM
 ingenting Non-Actuary Join Date: Jan 2014 College: University of Washington Posts: 17
Lesson 20 proof of the alternative formula for EPV of continuous annuity

In lesson 20, we have direct way for EPV of continuous whole life annuity:
a_x=∫ a_t*t_p_x*μ_x+t dt, integral from 0 to infinite
The textbook says to plug a_t=(1-v^t)/ δ into the equation above and apply integration by parts, hence the new equation:
a_x=∫ (1-v^t)/δ*fx(t) dt since t_p_x*μ_x+t = fx(t)
= 1/δ * [∫fx(t) dt - ∫ v^t*fx(t) dt]
v^t is the discount factor, which equals to e^(-δt)，the first integral ∫fx(t) dt equals to 1 since I could split the integral into [0,t] and [t,∞], which gives me Fx(t) and Sx(t) that sum up to 1, so the equation becomes:
a_x=1/δ * [1- ∫ v^t*fx(t) dt]
now integration by parts,
u = e^(-δt) v= 1 as shown above
du=-δ*e^(-δt)dt dv=fx(t) dt
so ∫ v^t*fx(t) dt = e^(-δt) - ∫-δ*e^(-δt)dt
I realize that the solution is not the same as the alternative formula:
a_x = ∫ v^t*t_p_x dt
Any help on this one?
#2
01-12-2019, 08:43 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,047

Quote:
 Originally Posted by ingenting so ∫ v^t*fx(t) dt = e^(-δt) - ∫-δ*e^(-δt)dt Any help on this one?
I didn't follow all the steps, but ∫ v^t*fx(t) dt depends on mortality and
e^(-δt) - ∫-δ*e^(-δt)dt does not depend on mortality so something must be wrong there.
#3
01-12-2019, 09:21 AM
 ingenting Non-Actuary Join Date: Jan 2014 College: University of Washington Posts: 17

Quote:
 Originally Posted by Gandalf I didn't follow all the steps, but ∫ v^t*fx(t) dt depends on mortality and e^(-δt) - ∫-δ*e^(-δt)dt does not depend on mortality so something must be wrong there.
The mortality vanishes after integration by parts. I set
u = e^(-δt) v= 1
du=-δ*e^(-δt)dt dv=fx(t) dt
I set dv = fx(t) dt and when you go back from dv to v, you integrate dv from 0 to infinite to get 1. The formula for integration by parts is:
u*v - ∫v du
there is no mortality anywhere in equation, that's how the mortality just disappears. I agree that things get a little weird here but I couldn't find out what goes wrong.
#4
01-12-2019, 09:37 AM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,001

You are confusing definite integrals and anti-derivatives.

When you do integration by parts, you select u and dv. You then compute du and v. v is an antiderivative for dv.

You know that the definite integral of dv from 0 to infinity is one. That does not mean that v is one.

An example of such a dv(t) is exp(-t).
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"

"I think that probably clarifies things pretty good by itself."