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

[ingenting]

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?

[Gandalf]

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.

[ingenting]

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.

[Colymbosathon ecplecticos]

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).