Re: May 1996 Actex #9

james.smieszkal@zurich.com
Fri, 17 Sep 1999 10:28:10 -0500

NSP for endowment=1000( NSP for 3 year term + NSP for 3 year pure endowment)=
=1000(q/(q+i)times(1- exponent(-3(0.12+0.09)) + exponent(-3(0.12+0.09))=
=1000times0.787614=787.61396 is your answer of course if my arithmetic is
correct
q=0.1130796 and i=0.0941743
P.S. This is exponential function. So NSP for 3 year term=NSP for whole life
times(1-NSP for 3year pure endowment)
And as you know: NSP for whole life for exp for discrete case is (q/q+i)

I think you are confusing the discrete case with the continuous case.

Ax:3=v qx + v^2 * px * q(x+1) + v^3 * 2px (Note payment is made at end of the
3rd
year
regardless of whether he lives or dies
in the
third year.)

You are correct that the term plus the pure endowment equals the endowment.

Ax1:3+Ax:13= v qx + v^2 * px * q(x+1) + v^3 * 2px * q(x+2) + v^3 * 3px

3px = 2px * p(x+2)

Ax1:3+Ax:13= v qx + v^2 * px * q(x+1) + v^3 * 2px * q(x+2) + v^3 * 2px * p(x+2)

Ax1:3+Ax:13= v qx + v^2 * px * q(x+1) + v^3 * 2px * (q(x+2) + p(x+2))

Ax1:3+Ax:13= v qx + v^2 * px * q(x+1) + v^3 * 2px

Regarding the NSP for a whole life insurance payable at the end
of the year of death given constant force:

d=delta, u=force of mort.

u/(u+d) is the NSP for the continuous case.

The NSP for the discrete case is derived as follows:

NSP = e^-d * (1-e^-u) + e^-2d * e^-u * (1-e^-u) + e^-3d * e^-2u * (1-e^-u). . .
.. . . .

NSP = [e^-d * (1-e^-u)] * [1+ e^-d * e^-u + e^-2d * e^-2u. . . . . . . . . .

The second portion is a infinite geometric series.

NSP = [e^-d * (1-e^-u)] * [1/(1- e^-d * e^-u)]