\documentclass[onecolumn]{article}
\author{ Sean E. O'Connor}
\title{Mortgage Loan Derivation}
\usepackage{amssymb}
\begin{document}
\maketitle
\begin{abstract}
We derive the equations for a mortgage loan and give several useful formulas.
\end{abstract}
\section{The Repayment Structure of a Mortgage Loan}
\subsection{Definition of Interest Rate}
The interest rate on the loan note is quoted as a yearly amount but is applied periodically for $m$ periods per year, giving
\begin{equation}
i = {{Annual \: interest \: percent / 100} \over m}
\end{equation}
Most house and car loans have a monthly period, $m = 12$.
\emph{NOTE:} This is NOT the APR, which is the interest percentage as if the loan were compounded yearly instead of monthly. I would ignore the APR.
\subsection{Periodic Payments of Principal + Interest}
The $j^{th}$ payment PMT consists of interest $I_j$ on the previous unpaid balance $BAL_{j-1}$
This payment is a fixed amount every month by design.
Only the remainder of the payment after interest $R_j$ actually reduces the principal of the loan.
Payments from the initial payment to the end payment have the form,
\begin{equation}
\left( \begin{array} { cccc }
Payment \: Num & Payment \: Amount & Interest & Balance \\
{1^{st}} & {PMT = R_1 + I_1} & {I_1 = i \: BAL_0} & {BAL_1 = BAL_0 - R_1} \\
2^{nd} & PMT = R_2 + I_2 & I_2 = i \: BAL_1 & BAL_2 = BAL_1 - R_2 \\
\ldots & \ldots & \ldots & \ldots \\
j^{th} & PMT = R_j + I_j & I_j = i \: BAL_{j-1} & BAL_j = BAL_{j-1} - R_j \\
\ldots & \ldots & \ldots & \ldots \\
n^{th} & PMT = R_n + I_n & I_n = i \: BAL_{n-1} & BAL_n= BAL_{n-1} - R_n = 0 \\
\end{array} \right)
\end{equation}
n = the number of periods of the loan
The loan amount is $PV = BAL_0$ which is the original balance before any payments are made.
The last balance is 0 because the loan is paid off, $BAL_n = 0$.
We could write a Perl or Python script to compute this, but there are closed form solutions as we'll show.
\section{Closed Form Solution}
\subsection{Difference Equation}
Substituting we get a difference equation for the balances in terms of interest i and constant periodic payment PMT,
\begin{equation}
BAL_j = BAL_{j-1} - PMT + I_j
\end{equation}
\begin{equation}
BAL_j = BAL_{j-1} - PMT + i BAL_{j-1}
\end{equation}
\begin{equation}
BAL_j = (1+i) BAL_{j-1} - PMT
\end{equation}
Let's solve this difference equation. But first, some lemmas,
\emph{Lemma} The geometric progression
\begin{equation}
g = 1 + b + b^2 + \ldots + b^{n-1}
\end{equation}
has the formula,
\begin{equation}
g = {{1 - b^n} \over {1 - b}}
\end{equation}
\emph{Proof}
Multiply by b and subtract,
\begin{equation}
g - b g = g(1 - b) = \left( 1 + b + b^2 + \ldots + b^{n-1} \right) - \left( b + b^2 + b^3 + \ldots + b^n \right) = 1 - b^n
\end{equation}
to get
\begin{equation}
g = {{1 - b^n} \over {1 - b}}
\end{equation}
\emph{QED.}
\emph{Lemma.} If
\begin{equation}
u_k = a + b u_{k-1}
\end{equation}
then
\begin{equation}
u_n = a {1 - b^n \over 1-b } + b^n u_0
\end{equation}
\emph{Proof.} Expand out terms to see the pattern and use mathematical induction if you like,
\begin{equation}
u_1 = a + b u_0
\end{equation}
\begin{equation}
u_2 = a + b u_1 = a + b(a + b u_0) = a (1 + b) + b^2 u_0
\end{equation}
\begin{equation}
u_3 = a + b u_2 = a + b [ (1+b) + b^2 u_0] = a (1 + b + b^2) + b^3 u_0
\end{equation}
The general formula is
\begin{equation}
u_n = a (1 + b + b^2 + \ldots + b^{n-1}) + b^n u_0
\end{equation}
Now use the geometric progression lemma above.
\emph{QED.}
\subsection{Closed Form Solution for $BAL_k$}
To solve the balance difference equation let $u_k = BAL_k$, $a = -PMT$, and $b = i+1$
The balance AFTER payment k is then
\begin{equation}
BAL_k = {(-PMT) {1 - (1+i)^k \over 1 - (1+i) } + (1+i)^k BAL_0}
\end{equation}
rearranging,
\begin{equation}
BAL_k = {1 \over (1+i)^{-k} } \left( PV + PMT { (1+i)^{-k} -1 \over i } \right)
\end{equation}
Now set $BAL_n = 0$ because the last nth payment PMT finishes the mortgage.
Check: $BAL_0 = PV$
\section{Other Useful Closed Form Solutions}
\subsection{Periodic Payment PMT From PV, n and i}
The constant periodic payment is then
\begin{equation}
PMT = {i PV \over 1-(1+i)^{-n}}
\end{equation}
\subsection{Number of Periods n from PV, i and PMT}
The number of periods (months) comes from the loan amount PV and the periodic payment PMT after inverting,
\begin{equation}
n = -{ ln \left( {1 - i {PV \over PMT}} \right) \over { ln( 1 + i ) } }
\end{equation}
where we're using the natural logarithm, but any log will work.
\subsection{Reduced Number of Periods n' from Increasing the PMT}
You can use this to determine the reduced number of periods if you prepay principal. Change $PMT$ to $(PMT + additional\: principal \: prepayment)$
\begin{equation}
n' = -{ ln \left( {1 - i {PV \over {PMT + additional\: principal \: prepayment}}} \right) \over { ln( 1 + i ) } }
\end{equation}
\subsection{Equivalent Simple Interest i'}
Equivalent simple interest $i'$ on the loan is how much interest would have been had it been payable up front immediately.
We use this formula to deduce it,
\begin{equation}
PV + i' PV = n PMT
\end{equation}
Equivalent simple interest is then
\begin{equation}
i' = {n i \over 1-(1+i)^{-n}} -1
\end{equation}
\subsection{Accumulated Interest Payments j though k}
What is the accumulated interest for payments j though k inclusive?
\begin{equation}
Int_{j \rightarrow k} = I_j + \ldots + I_k
\end{equation}
\begin{equation}
Int_{j \rightarrow k} = \left( k - j + 1 \right) PMT - \sum_{l=j}^k R_l
\end{equation}
but
\begin{equation}
BAL_l - BAL_{l-1} = -R_l
\end{equation}
We get a telescoping sum
\begin{equation}
- \sum_{l=j}^k R_l = BAL_k - BAL_{j-1}
\end{equation}
Interest AFTER the kth payment is
\begin{equation}
Int_{1 \rightarrow k} = k PMT + BAL_k - PV
\end{equation}
Check: When $k=n$,
\begin{equation}
Int_{1 \rightarrow n} = n PMT + BAL_n - PV = n PMT - PV
\end{equation}
as expected because the total n payments of PMT include repaying PV plus interest.
\end{document}