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Relating Effective Interest and Discount Rates Through Reinvestment

Article Quant Q&A · Author: Strictly_increasing

Summary

The document explains how an annual effective discount rate can be related to an effective interest rate when interest received in advance is repeatedly reinvested. Starting with capital C, each advance interest payment is reinvested under the same terms, creating a geometric sequence of amounts. If the discount rate is between negative one and one, the accumulated series converges to C divided by one minus the discount rate.

To compare this with ordinary effective interest, the accumulated amount over a period is set equal to the principal grown at the effective interest rate. This yields the stated relationship between the two rates. The answer illustrates the series limit by considering finite partial sums and letting the number of reinvestments grow. Its explanation is informal, and the displayed algebra contains a sign inconsistency in one intermediate expression; the convergent geometric-series result and rate equivalence are the intended points. The treatment assumes repeated reinvestment at unchanged terms and does not address real-world constraints on that assumption.

Key ideas

  • Repeated reinvestment of interest received in advance forms a geometric series.
  • The series converges when the discount rate lies between negative one and one.
  • The accumulated series is used to relate the effective discount rate to the effective interest rate.
  • The equivalence assumes the interest payments can be reinvested repeatedly under identical terms.
  • One intermediate algebraic expression in the answer has a sign error, though its stated limit is the geometric-series result.

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Full text
# Intuition behind reasoning around interests-in-advance


# Intuition behind reasoning around interests-in-advance












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):

- 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 $(2)$?

- Why does $(2)$ hold true? Why does the equivalent effective interest rate $i$ is given by $\frac{1}{1-d}-1$?

## Answer by user59776 (score 1)

https://quant.stackexchange.com/a/68589

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

- 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.

- Does the aim correspond to get to (2)?''

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.