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I'm unsure how to calculate a continuously varying, continuously paid annuity. I'll write up my solution (which I suspect is wrong) to one, sample question, and I would greatly appreciate any correction.

The question (problem 29.13 on page 273):

Payments are made to an account at a continuous rate of $(8k+tk)$, where $0\le t\le10$. Interest is credited at a force of interest $\delta_t=\frac1{8+t}$. After 10 years, the account is worth 20,000. Calculate $k$.

My attempt at a solution:

The accumulation function is $e^{\int_0^t(8+s)^{-1}ds}$, which comes to $e^{\left.\ln\left|s+8\right|\right|_0^t}=\frac{t+8}8$, so that the discount (inverse-accumulation) function (to evaluate the present ($t=0$) value) is $\frac8{t+8}$.

Then the present value is the limit of sums of $(\textrm{discount})\times(\textrm{payment})$ i.e. $\int_0^{10}(8k+tk)\frac8{t+8}dt=80k$.

The accumulated value is then $80k\times(\textrm{accumulation(10)})=80k\frac{18}8$; since that's given as $20000$, we have $k=20000/180\approx111.11$.

Could anyone post the right solution or explain what (if anything) is wrong with mine, please?

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  • $\begingroup$ I can tell you that Finan has an answer key (no solutions, just answers) and his answer is 111.11. I don't have much time these days to look into it further! $\endgroup$ – Graphth Jan 12 '13 at 13:58
  • $\begingroup$ @Graphth Thank you! I didn't know about his key. Where is it? $\endgroup$ – user55618 Jan 13 '13 at 0:20
  • $\begingroup$ It's not publicly available. I have it because I taught a class from his book. If you are taking a class that uses his book, then your teacher may not want you to have it. So, he doesn't give it out to everyone. $\endgroup$ – Graphth Jan 13 '13 at 12:45
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As Finan said, the PV of an $n$ period continuously varying annuity with the continuous rate of payments $f(t)$ and with force of interest $\delta_t$ is

$$\int_0^n f(t) e^{-\int_0^t\delta_r dr}dt.$$

Your answer mimics this formula and is correct.

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