Under de Moivre, the solution has calculated t/q(x)=1/(80-x). How did
they come up with this. Isn't de Moivre s(x)=w-x/w which would lead
to tpx of w-x-t/w-x.
In order to find t/q(x). We need tpx*qx+t. Am I wrong?
Under de Moivre, the probability of a life (x) dying in a given year is equal to
the probability of dying in any other given year. Assuming of course that the
life is younger than the terminal age. de Moivre's law assumes that deaths occur
uniformly over the future lifetime.
0|qx = 1|qx = 2|qx =. . .= w-x-1|qx.
Let's take a look at what you brought up t|qx = tpx * qx+t
tpx for de Moivre is what you said it was w-x-t/w-x.
1px+t = w-(x+t)-1/w-(x+t)= 1-(1/(w-x-t)) and
1qx+t = 1- 1px+t = 1/(w-x-t)
Finally, the product tpx * qx+t = 1/w-x
James M. Smieszkal