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2% MortgageLoanDerivation.tex
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16\documentclass{article}
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18\author{ Sean E. O'Connor}
19\title{Mortgage Loan Derivation}
20
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57% ----------------- Document begins: title, abstract, table of contents --------------------
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59\begin{document}
60
61\maketitle
62
63\begin{abstract}
64We derive the equations for a \emph{mortgage} loan and give several useful formulas
65\footnote{I'm using \LaTeX and \TeX Shop. See \url{https://www.ctan.org}. See source code for more documentation.}
66\end{abstract}
67
68\tableofcontents
69
70% ------------------------------------------ First section begins --------------------------------------
71
72\pagebreak
73
74\section{The Repayment Structure of a Mortgage Loan}
75
76\subsection{Definition of Interest Rate}
77
78The interest rate on the loan note is quoted as a yearly amount but is applied periodically for $m$ periods per year, giving
79
80\begin{equation}
81i = {{Annual \: interest \: percent / 100} \over m}
82\end{equation}
83
84Most house and car loans have a monthly period, $m = 12$. % Inline math can use dollar sign delimiters like MathJax
85
86\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.
87
88\subsection{Periodic Payments of Principal + Interest}
89
90The $j^{th}$ payment PMT consists of interest $I_j$ on the previous unpaid balance $BAL_{j-1}$
91
92This payment is a fixed amount every month by design.
93
94Only the remainder of the payment after interest $R_j$ actually reduces the principal of the loan.
95
96Payments from the initial payment to the end payment have the form,
97
98\bigskip % Put extra line space before and after the table.
99
100\begin{tabular}{ clll }
101\toprule
102Payment Number & Payment Amount & Interest & Balance \\
103\midrule
104${1^{st}}$ & $PMT = R_1 + I_1$ & $I_1 = i \: BAL_0$ & $BAL_1 = BAL_0 - R_1$ \\
105$2^{nd}$ & $PMT = R_2 + I_2$ & $I_2 = i \: BAL_1 $ & $BAL_2 = BAL_1 - R_2$ \\
106\ldots & \ldots & \ldots & \ldots \\
107$j^{th}$ & $PMT = R_j + I_j$ & $I_j = i \: BAL_{j-1}$ & $BAL_j = BAL_{j-1} - R_j$ \\
108\ldots & \ldots & \ldots & \ldots \\
109$n^{th}$ & $PMT = R_n + I_n$ & $I_n = i \: BAL_{n-1}$ & $BAL_n= BAL_{n-1} - R_n = 0 $ \\
110\bottomrule
111\end{tabular}
112
113\bigskip
114
115n = the number of periods of the loan
116
117\begin{remark}
118The loan amount is $PV = BAL_0$ which is the original balance before any payments are made.
119\end{remark}
120
121\begin{remark}
122The last balance is 0 because the loan is paid off, $BAL_n = 0$.
123\end{remark}
124
125We could write an Excel, Perl or Python script to compute this, but there are closed form solutions as we'll show.
126
127\pagebreak
128
129\section{Closed Form Solution}
130
131\subsection{Difference Equation}
132
133Substituting we get a difference equation for the balances in terms of interest i and constant periodic payment PMT,
134
135\begin{equation}
136BAL_j = BAL_{j-1} - PMT + I_j
137\end{equation}
138
139\begin{equation}
140BAL_j = BAL_{j-1} - PMT + i BAL_{j-1}
141\end{equation}
142
143\begin{equation}
144BAL_j = (1+i) BAL_{j-1} - PMT
145\end{equation}
146
147Let's solve this difference equation. But first, some lemmas,
148
149\begin{lemma}{}
150The geometric progression
151
152\begin{equation}
153g = 1 + b + b^2 + \ldots + b^{n-1}
154\end{equation}
155
156has the formula,
157
158\begin{equation}
159g = {{1 - b^n} \over {1 - b}}
160\end{equation}
161
162\end{lemma}
163
164\begin{proof}
165
166Multiply by b and subtract,
167
168\begin{equation}
169g - 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
170\end{equation}
171
172to get
173
174\begin{equation}
175g = {{1 - b^n} \over {1 - b}}
176\end{equation}
177
178\end{proof}
179
180\begin{lemma}{}
181 If
182
183\begin{equation}
184u_k = a + b u_{k-1}
185\end{equation}
186
187then
188
189\begin{equation}
190u_n = a {1 - b^n \over 1-b } + b^n u_0
191\end{equation}
192
193\end{lemma}
194
195\begin{proof} Expand out terms to see the pattern and use mathematical induction if you like,
196
197\begin{equation}
198u_1 = a + b u_0
199\end{equation}
200
201\begin{equation}
202u_2 = a + b u_1 = a + b(a + b u_0) = a (1 + b) + b^2 u_0
203\end{equation}
204
205\begin{equation}
206u_3 = a + b u_2 = a + b [ (1+b) + b^2 u_0] = a (1 + b + b^2) + b^3 u_0
207\end{equation}
208
209The general formula is
210
211\begin{equation}
212u_n = a (1 + b + b^2 + \ldots + b^{n-1}) + b^n u_0
213\end{equation}
214
215Now use the geometric progression lemma above.
216
217\end{proof}
218
219\subsection{Closed Form Solution for $BAL_k$}
220
221To solve the balance difference equation let $u_k = BAL_k$, $a = -PMT$, and $b = i+1$
222
223The balance AFTER payment k is then
224
225\begin{equation}
226BAL_k = {(-PMT) {1 - (1+i)^k \over 1 - (1+i) } + (1+i)^k BAL_0}
227\end{equation}
228
229rearranging,
230
231\begin{equation}
232BAL_k = {1 \over (1+i)^{-k} } \left( PV + PMT { (1+i)^{-k} -1 \over i } \right)
233\end{equation}
234
235Now set $BAL_n = 0$ because the last nth payment PMT finishes the mortgage.
236Check: $BAL_0 = PV$
237
238\pagebreak
239
240\section{Other Useful Closed Form Solutions}
241
242\subsection{Periodic Payment PMT From PV, n and i}
243
244The constant periodic payment is then
245
246\begin{equation}
247PMT = {i PV \over 1-(1+i)^{-n}}
248\end{equation}
249
250\subsection{Number of Periods n from PV, i and PMT}
251
252The number of periods (months) comes from the loan amount PV and the periodic payment PMT after inverting,
253
254\begin{equation}
255n = -{ ln \left( {1 - i {PV \over PMT}} \right) \over { ln( 1 + i ) } }
256\end{equation}
257
258where we're using the natural logarithm, but any log will work.
259
260\subsection{Reduced Number of Periods $n'$ from Increasing the PMT}
261
262You can use this to determine the reduced number of periods if you prepay principal. Change $PMT$ to $(PMT + additional\: principal \: prepayment)$
263
264\begin{equation}
265n' = -{ ln \left( {1 - i {PV \over {PMT + additional\: principal \: prepayment}}} \right) \over { ln( 1 + i ) } }
266\end{equation}
267
268\subsection{Equivalent Simple Interest $i'$}
269
270Equivalent simple interest $i'$ on the loan is how much interest would have been had it been payable up front immediately.
271
272We use this formula to deduce it,
273
274\begin{equation}
275PV + i' PV = n PMT
276\end{equation}
277
278Equivalent simple interest is then
279
280\begin{equation}
281i' = {n i \over 1-(1+i)^{-n}} -1
282\end{equation}
283
284\subsection{Accumulated Interest Payments j though k}
285
286What is the accumulated interest for payments j though k inclusive?
287
288\begin{equation}
289Int_{j \rightarrow k} = I_j + \ldots + I_k
290\end{equation}
291
292\begin{equation}
293Int_{j \rightarrow k} = \left( k - j + 1 \right) PMT - \sum_{l=j}^k R_l
294\end{equation}
295
296but
297
298\begin{equation}
299BAL_l - BAL_{l-1} = -R_l
300\end{equation}
301
302We get a telescoping sum
303
304\begin{equation}
305- \sum_{l=j}^k R_l = BAL_k - BAL_{j-1}
306\end{equation}
307
308Interest AFTER the kth payment is
309
310\begin{equation}
311Int_{1 \rightarrow k} = k PMT + BAL_k - PV
312\end{equation}
313
314Check: When $k=n$,
315
316\begin{equation}
317Int_{1 \rightarrow n} = n PMT + BAL_n - PV = n PMT - PV
318\end{equation}
319
320as expected because the total n payments of PMT include repaying PV plus interest.
321
322\end{document}
