How to convert continuous interest compounding rate to non-compounding rate?

Solution 1:

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

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