Monthly mortgage payment
The formula of the monthly mortgage payments for a fixed-rate loan can be found online as \begin{equation} M = P\frac{r(1+r)^n}{(1+r)^n-1}\,,\tag{1}\end{equation} where
- M is the mortgage payment.
- P is the principal, i.e., the initial amount borrowed.
- r is the monthly interest rate. For annual interest rate 2.5\%, r=2.5\% / 12.
- n is the number of payments. For 30-years fixed rate mortgage, n=30 * 12 = 360.
Here we provide a derivation of the above formula. Let b_i be the owned balance in the i-th month. Initially, b_0=P and we choose the constant monthly payment M such that b_n=0. We then have the recursion relation between two consecutive month as \begin{equation}b_{i+1}=b_i(1+r) - M\,.\tag{2}\end{equation}
A trick to solve (2) is to write Eq. (2) into the following form \begin{equation}b_{i+1}+C = \left(b_i+C\right)(1+r)\,,\tag{3}\end{equation} so that b_i+C is a geometric sequence: \begin{equation}b_i + C = (b_0 + C) (1+r)^i\,.\end{equation} Note that b_0=P and b_n=0, we thus have the equation \begin{equation} C = (P+C) (1+r)^n\,,\end{equation} from which we can solve \begin{equation}C= \frac{P(1+r)^n}{(1+r)^n-1}\,.\end{equation}
Finally, by comparing Eq. (2) and (3), we know that M=Cr, which then has the same form as Eq. (1).
After proving the formula (1), let's do a simple exercise. Suppose the 30-years fixed rate mortgage rate is 2.5\% and the loan amount is P_1. For exactly the same monthly mortgage payments, what is the new loan amount P_2 when the 30-years fixed rate mortgage rate is increased to 6.0\%?
Using Eq. (1), for the same M, we have the relation \begin{equation} \frac{P_2}{P_1} = \frac{r_1 (1 + r_1)^n}{r_2 (1 + r_2)^n}\frac{(1 + r_2)^n-1}{ (1 + r_1)^n-1}\,.\end{equation} The surprise is the final numerical result: after submitting r_1=2.5\%/12 and r_2=6.0\%/12, we can calculate \begin{equation} \frac{P_2}{P_1} = 0.659\,.\end{equation} The new loan amount is only two thirds of the old loan amount after the interest rate increases!
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