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Mortality of (50) follows De Moivre's law with w=100 and i=0.06. I find it hard to evaluate the value of $(IA)_{50}$. Any tip will be much appreciated

Heisenberg
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  • I would like to comment, before any one gives a bad comment or downvote, that everything in this question is perfectly clear if you have studied actuarial science, which is the subject of this question. – GeoffDS Jul 12 '13 at 13:23
  • If this was a continuous situation I find it simple to just do it by integration but i simply cannot solve it when using the summation – Heisenberg Jul 13 '13 at 04:19
  • Then your question has nothing to do with actuarial science in reality. It is just how to do a certain sum. So, ask that question instead. A very small percentage of the people on here know what De Moivre's law is and what the IA symbol means. Everyone on here will understand a sum and some people may know how to do it. Feel free to edit this question, especially with no answers or comments other than mine. You can justify why you care about the sum by leaving what you already have. – GeoffDS Jul 15 '13 at 01:32

1 Answers1

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So, you have a constant increasing stream of revenue starting at 1 and increasing by 1 each year.

The probability to survive $t$ years from age $x$ is according to De Moivre's law

$${}_{t}p_{x} = \frac{1-\frac{x+t}{w}}{1-\frac{x}{w}} = \frac{w-x-t}{w-x}=1-\frac{t}{w-x}$$

Thus

$$(IA)_{50} = \sum_{k=0}^{50} (k+1)\; {}_{k}p_{50} \; v^k = \sum_{k=0}^{50} \; (k+1)v^k - \frac{1}{50}\sum_{k=0}^{50} \; (k+1)k \;v^k$$

with $v=(1+i)^{-1}$.

We are left with evaluating the sums $\sum_{k=0}^{50} \; (k+1)v^k$ and $\sum_{k=0}^{50} \; (k+1)k \;v^k$. There are several tricks that work for tackling those sums. Take for instance the first sum. Let's call it $S$. Then

$$vS = \sum_{k=0}^{50} \; (k+1)v^{k+1} = \sum_{k=1}^{51} \; k \; v^{k} = \sum_{k=0}^{50} \; k \; v^{k} + 51v^{51} = S - \sum_{k=0}^{50} \; v^{k} + 51\; v^{51} \; ,$$

thus,

$$S = \frac{1}{v-1}\left(51\; v^{51} - \sum_{k=0}^{50} \; v^{k} \right) \; .$$

The same trick can be applied to $\sum_{k=0}^{50} \; v^{k}$. But maybe you recognize this as a partial sum of a geometric series. Thus,

$$S = \frac{1}{v-1}\left(51\; v^{51} - \frac{v^{51}-1}{v-1} \right) \; .$$

You should now be able to apply the same technique to the second sum.

Raskolnikov
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