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Let's say we are told that a bank offers an annual interest yield of 50%, and it is automatically compounded by the bank. Assume the compounding is done continuously rather than in discreet intervals.

How can we calculate what the yield would be after one year if the interest wasn't compounded? In other words, interest is taken out of the bank at infinitesimal intervals such that there is no compounding effect.

I would like to be able to calculate this at no knowledge of how often the bank actually compounds interest, so this is why I am assuming continuous compounding and infinitesimal intervals of taking out the interest.

The purpose of this question is for two potential situations:

  • To calculate what the theoretical worst-case scenario yield would be if you were taking out interest regularly, and,
  • To calculate what the yield would be should a bank add in some sort of a limit to how much money interest can be earned on (so once this limit is hit, it is no longer compounding anymore).
David Callanan
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  • This is not clear. What do you mean by an annual interest rate of $50%$? If we put $$100$ in at the start of the year, do we have $$150$ at the end? Or is the $50%$ some nominal figure designed to fit into a formula? If so, what formula? – lulu Jan 04 '22 at 20:22
  • @lulu Yes, 150 dollars at the end, due to continuous exponential compounding. After two years it would be 225 dollars at the end. – David Callanan Jan 04 '22 at 20:33
  • So, then, what are you asking? Continuous compounding means there is a rate $r$ such that $P(t)=P_0e^{rt}$. Easy to work out what $r$ must be from that. – lulu Jan 04 '22 at 20:36
  • @lulu I can't figure it out from that. I don't think I'm solving for $r$, because $r$ is already $1.5$, right? – David Callanan Jan 04 '22 at 22:27
  • No. If $r=1.5$ then, measuring time $t$ in years and starting with $P_0=100$, we'd have $P(1)=100\exp(1.5)=448.1689$ – lulu Jan 04 '22 at 22:39
  • @lulu Of course, my bad – David Callanan Jan 05 '22 at 12:53

1 Answers1

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This question is getting at the relationship between annual percentage rate (APR) and annual percentage yield (APY). Generally, APR is the rate before taking into account compounding, and APY is after compounding.

In this case, we're given an annual percentage yield of $50\%$, compounded continuously. As noted in the comments, the formula for continuous compounding is $P(t) = P_0 e^{rt}$, where $P_0$ is initial principal, $P(t)$ is principal at time $t$, and $r$ is the interest rate, specifically APR, per unit of $t$.

Let $y$ represent the APY. Then, after $1$ year of compounding continuously, we have $P(1) = (1+y)\cdot P_0$. Substituting that into the formula for continuous compounding, we have

$$ (1+y)P_0 = P_0e^{r\cdot 1} $$ $$ \Rightarrow (1+y) = e^{r} $$ $$ \Rightarrow r = \ln(1+y) $$

So, the general relationship for continuously compounded interest is $APR = \ln(1+APY) $.

In this case, with an APY of $50\%$, we have

$$ r = \ln(1+y) = \ln(1.5) \approx 40.55\% $$

In the case when it's not compounded continuously, but rather in discrete intervals, there's a similar process you can follow. Let's say interest is compounded $n$ times per unit of $t$. Then, we have

$$ P(t) = P_0\left(1+\frac{r}{n}\right)^{nt} $$ $$ \Rightarrow P(1) = P_0\left(1+\frac{r}{n}\right)^{n} $$ $$ \Rightarrow (1+y)P_0 = P_0\left(1+\frac{r}{n}\right)^{n} $$ $$ \Rightarrow (1+y) = \left(1+\frac{r}{n}\right)^{n} $$ $$ \Rightarrow (1+y)^{1/n} = 1+\frac{r}{n} $$ $$ \Rightarrow r = n\left((1+y)^{1/n} - 1\right) $$

Amaan M
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  • Thank you very much! Before asking this question I had tried $ln(1.5)$ expecting to get $1 + APR$ rather than just the $APR$ so I disregarded it immediately when the answer looked wrong. Also thanks for clarifying that the term I was looking for is APR. – David Callanan Jan 05 '22 at 12:59
  • You're very welcome! – Amaan M Jan 05 '22 at 20:56