# Interest Rate Math Basics

Suppose you deposit $1000 in a bank account that ears 5% a year. What is your balance at the end of five years? The answer to this question depends on the type of rate that’s being quoted here and can actually range between$1276.28 and $1284.03. When banks quote interest rates on credit card debt, home and auto loans, or savings accounts, they take advantage of this ambiguity to make the rates seem higher or lower to their benefit. Annual Percentage Rate (APR) Most interest rates are quoted as an APR or Annual Percentage Rate. You’ll see APRs quoted when opening a credit card, getting a home or auto loan, as well as investing in bonds. For one of their basic credit cards, Bank of America quotes an APR of 13.49%. If you had$1000 of credit card debt, how much would your debt grow in five years?

When given an APR, you also need to know the compounding frequency, i.e. how many times a year interest is added to your balance. If interest is simply added to your account as a lump sum at the end of every year, your ending balance would be:

$\displaystyle \1000 \times (1+13.49\%)^5=\ 1,882.73$

If interest added semiannually, instead of adding 13.49% at the end of the year, your debt will grow 6.75% every six months. In this case your balance will be

$\displaystyle \1000 \times \left(1+\frac{13.49\%}{2}\right)^{5\times 2} = \1,920.77$

This is slightly higher since interest added earlier accrues more interest itself. Credit card debt usually actually compounds interest monthly, giving us a balance of

$\displaystyle \1000 \times \left(1+\frac{5\%}{12}\right)^{5\times 12} = \1,955.68$

As we continue to increase the compounding frequency indefinitely, the ending balance approaches $1963.05. Using an infinite compounding frequency is called continuous compounding. The balance in an account that earns r% a year compounded continuously after t years is given by the formula $\displaystyle B_t = B_0 e^{rt}$ While continuous compounding is not found in real life, it is helpful in simplifying the math of compounding interest as we’ll see in later on. Annual Percentage Yield (APY) Banks usually advertise the interest rates they offer for their savings accounts as an Annual Percentage Yield (APY), also known as an Effective Annual Rate (EAR). An APY describes the interest accrued over one year regardless of how often it is compounded over the year. One Bank of America savings account offers an APY of .06% with interest added to your account daily. If you deposited$1000 in this account, after five years, your balance would simply be

$\displaystyle \1000 \times (1+.06\%)^5 = \ 1003.00$

However, if we were to rewrite this interest rate as an APR instead of an APY, we would find that the APR is slightly smaller than the APY.

$\displaystyle \left(1+\frac{APR}{365}\right)^{365}=1+.06\%$

$\displaystyle APR = .05998\%$

Alternatively, if we were to rewrite the 13.49% monthly compounded APR as an APY, we would get a much larger number.

$\left(1+\frac{13.49\%}{12}\right)^{12} = 1+APY$

$APY = 14.36\%$

In general, representing an interest rate as an APR makes interest rate look smaller, which is why APRs are used in quoting interest rates on loans or credit card debt. APYs, on the other hand, make interest rates look larger, and are often used in quoting interest earned in savings accounts.

A short proof of the continuous compounding formula

Using binomial expansion we can expand the compounded interest formula in the following way.

$\displaystyle \left(1+\frac{x}{n}\right)^n = 1+\binom{n}{1}\left(\frac{x}{n}\right)+\binom{n}{2}\left(\frac{x}{n}\right)^2 + \binom{n}{3}\left(\frac{x}{n}\right)^3+\cdots$

This is simply an n-degree polynomial on $x$ where the polynomial coefficient of $x^k$ is

$\displaystyle \binom{n}{k}\frac{1}{n^k}=\frac{n!}{(n-k)!k!n^k}$

Using the bounds $(n-k)!(n-k)^k \leq n! \leq (n-k)!n^k$ we can bound this coefficient above and below.

$\displaystyle \frac{(n-k)!(n-k)^k}{(n-k)!k!n^k} \leq \frac{n!}{(n-k)!k!n^k} \leq \frac{(n-k)!n^k}{(n-k)!k!n^k}$

$\displaystyle \left(\frac{n-k}{n}\right)^k\frac{1}{k!} \leq \frac{n!}{(n-k)!k!n^k} \leq \frac{1}{k!}$

Letting $n$ go to infinity, since both the upper and lower bound approach \$latex \frac{1}{k!}, so must the coefficient.

$\displaystyle \lim_{n \to \infty} \left(\frac{n-k}{n}\right)^k\frac{1}{k!} = \frac{1}{k!}$

$\displaystyle \lim_{n \to \infty} \frac{n!}{(n-k)!k!n^k} = \frac{1}{k!}$

Therefore

$\displaystyle \lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n = \sum_{k=1}^\infty \frac{1}{k!}x^k=e^x$

Finally,

$\displaystyle \lim_{n \to \infty} B_0\left(1+\frac{r}{n}\right)^{nt}=B_0\left(\lim_{n\to \infty}\left(1+\frac{r}{n}\right)^n\right)^t=B_0e^{rt}$