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I quote Life Insurance Mathematics (Gerber, 1997).


Let $i$ be an annual effective interest rate and $d$ an annual effective discount rate.

In case of interests-in-advance, a person investing an amount of $C$ will be credited interest equal to $dC$ immediately and the invested capital $C$ will be returned at the end of the period. Investing the interest $dC$ at the same conditions, the investor will receive additional interest of $d(dC)=d^2C$, and the additional invested amount will be returned at the end of the year; reinvesting the interest yields additional interest of $d(d^2C)=d^3C$, and so on.

$\color{red}{\text{Repeating this process ad infinitum,}}$ we find that the investor will receive the total sum of: \begin{equation} C+dC+d^2C+d^3C+\cdots=\frac{1}{1-d}C\tag{1} \end{equation} $\color{red}{\text{The equivalent effective interest rate } i \text{ is given by the equation:}}$

\begin{equation} \color{red} {\frac{1}{1-d}=1+i\tag{2}} \end{equation}



Could you please help me understand the logic underlying the passages in $\color{red}{\text{red}}$ above? Specifically (in particular, my main doubt is the one in bold below):

  1. Why is it needed to "repeat that process ad infinitum"? Does the aim correspond to get to $(2)$? If so, how could one justify that while in the "simple" case, there is no assumption of reinvesting interests (i.e. one invests $C$ at time $t=0$ and gets an interest of $C(1+i)$ at time $t+1$), in the other case assumption of reinvesting interests "infinitely many times" is made? How could one, starting from two different assumptions, get to the equality $(2)$?
  2. Why does $(2)$ hold true? Why does the equivalent effective interest rate $i$ is given by $\frac{1}{1-d}-1$?
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    $\begingroup$ Normally if we invest C we will have $(1+i)C$ after one period. In (1) we showed that with interest in advance we will have $\frac{1}{1-d}C$ after one period. Comparing these two we see that $(1+i)$ plays the same role as $\frac{1}{1-d}$. So to find the equivalent $i$ we set these equal: $\frac{1}{1-d}=1+i$ $\endgroup$
    – nbbo2
    Commented Oct 29, 2021 at 14:39
  • $\begingroup$ The terms $C,dC,d^2C,\cdots$ in (1) represent the principal, the interest, the interest on the interest, etc. all of which you are entitled to collect at the end of the period. $\endgroup$
    – nbbo2
    Commented Oct 29, 2021 at 14:46
  • $\begingroup$ Ok, but why do we need to assume that we repeat that process of investing infinitely many times so as to get to $(2)$? I cannot understand the logic underlying that infinite repetition of the process @noob2 $\endgroup$ Commented Oct 29, 2021 at 14:49
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    $\begingroup$ I think the author is alluding to summing the geometric series: en.wikipedia.org/wiki/Geometric_series#Sum $\endgroup$
    – Bob Jansen
    Commented Oct 29, 2021 at 16:03
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    $\begingroup$ There are two people, Ann and Bob. Ann invests C with simple interest at rate i. Bob also invests C amount but at rate d under assumption of interests-in-advance. What is required for both Ann and Bob to end the period with the same amount of money? The answer is (2). $\endgroup$
    – nbbo2
    Commented Oct 29, 2021 at 21:52

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Let $S=C+dC+d^2C+\dots+d^nC$ be your sum; multiply it by $d$ to get $dS=dC+d^2C+d^3C+\dots+d^{n+1}C$; subtract; $S-dS=C-d^{n+1}C$; it can be transformed to; $[1-d]S=C-d^{n+1}C$; divide both sides by $1-d$; $S=\frac{C-d^{n+1}C}{1-d}$ note that $\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}$ so; $S=\frac{C}{1-d}-\frac{d^{n+1}C}{1-d}$ Now if $-1<d<1$, $d^{n+1}$ can be made as small as we want by making $n$ large enough for example $d=\frac{1}{2}$ ,$d^2=1/4$,$d^3=1/8$, for $n=0 ,n=1, n=2$ ; But $\frac{d^{n+1}C}{1-d}$ is just $d^{n+1}$ multiplied by a constant term $\frac{C}{1-d}$ so it also can be made as small as we want; so equation $S=\frac{C}{1-d}+\frac{d^{n+1}C}{1-d}$ can be written as:

$S= $ constant term + something which can be made as small as we want by making some variable big enough.



Now let me made some definition;

For sum denoted by $A$, which depends of variable denoted by $M$, and whose value can be written as

$A= $ constant term + something which can be made as small as we want by making $ M $ big enough

we will from now call the value of constant term of the sum $A$ , "The limit of $A$ at infinity'.

So the limit of $S$ at infinity is $\frac{C}{1-d}$.

So back to your questions

  1. Why is it needed to "repeat that process ad infinitum"?

The more times you repeat the process the more money you get.In practice what you copy-pasted from book is a example of lending money to someone. When you lend money for profit you expect to get more money back from someone than you give them,usually as percent from lent sum, for example you can lend someone 100 dollars and after a month he will give you 150 dollars you have 50 dollars more, half of lent money.Interests-in-advance differ in that you receive 50 dollars surplus of money immediately and 100 dollars after a month so you still have 50 dollars to use. If you lend this money one more time with the same conditions you will get 25 dollars immediately and 50 dollars after a month, because you lend second time you receive 25 dollars more than you would get if you would only lend once. The more times you can repeat this process the more money you will get.But there is a limit of how much money you can get.You can be as close as you want to that limit but you would never surpass it.In our example you will never achieve 200 dollars but you can earn arbitrarily close sum.I written about this few lines above.

  1. Does the aim correspond to get to (2)?''

No, equation (2) is defintion, you define $1+i$ by $\frac{1}{1-d}$.

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  • $\begingroup$ First of all, thank you a lot for your explanation. Still I can't exactly get why in case of interests-in-advance we assume that we would like to make "profits on profits" (as much as possible), while in case of interest in arrears we do not start from such an assumption and we just get the interest $C(1+i)$ at the end of the period considered. Despite of that different assumption, we get to the definition $(2)$. I cannot really understand why such two situations are "comparable". Could you please help me understand that? $\endgroup$ Commented Nov 1, 2021 at 9:14

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