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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 --------------------
58
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\end{abstract}
66
67\footnote{I'm using \LaTeX and \TeX Shop. See \url{https://www.ctan.org}. See source code for more documentation.}
68
69\tableofcontents
70
71\bigskip % Put extra line space after the contents.
72
73% ------------------------------------------ First section begins --------------------------------------
74
75\pagebreak
76
77\section{The Repayment Structure of a Mortgage Loan}
78
79\subsection{Definition of Interest Rate}
80
81The interest rate on the loan note is quoted as a yearly amount but is applied periodically for $m$ periods per year, giving
82
83\begin{equation}
84i = {{Annual \: interest \: percent / 100} \over m}
85\end{equation}
86
87Most house and car loans have a monthly period, $m = 12$. % Inline math can use dollar sign delimiters like MathJax
88
89\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.
90
91\subsection{Periodic Payments of Principal + Interest}
92
93The $j^{th}$ payment PMT consists of interest $I_j$ on the previous unpaid balance $BAL_{j-1}$
94
95This payment is a fixed amount every month by design.
96
97Only the remainder of the payment after interest $R_j$ actually reduces the principal of the loan.
98
99Payments from the initial payment to the end payment have the form,
100
101\bigskip % Put extra line space before and after the table.
102
103\begin{tabular}{ clll }
104\toprule
105Payment Number & Payment Amount & Interest & Balance \\
106\midrule
107${1^{st}}$ & $PMT = R_1 + I_1$ & $I_1 = i \: BAL_0$ & $BAL_1 = BAL_0 - R_1$ \\
108$2^{nd}$ & $PMT = R_2 + I_2$ & $I_2 = i \: BAL_1 $ & $BAL_2 = BAL_1 - R_2$ \\
109\ldots & \ldots & \ldots & \ldots \\
110$j^{th}$ & $PMT = R_j + I_j$ & $I_j = i \: BAL_{j-1}$ & $BAL_j = BAL_{j-1} - R_j$ \\
111\ldots & \ldots & \ldots & \ldots \\
112$n^{th}$ & $PMT = R_n + I_n$ & $I_n = i \: BAL_{n-1}$ & $BAL_n= BAL_{n-1} - R_n = 0 $ \\
113\bottomrule
114\end{tabular}
115
116\bigskip
117
118n = the number of periods of the loan
119
120\begin{remark}
121The loan amount is $PV = BAL_0$ which is the original balance before any payments are made.
122\end{remark}
123
124\begin{remark}
125The last balance is 0 because the loan is paid off, $BAL_n = 0$.
126\end{remark}
127
128We could write an Excel, Perl or Python script to compute this, but there are closed form solutions as we'll show.
129
130\pagebreak
131
132\section{Closed Form Solution}
133
134\subsection{Difference Equation}
135
136Substituting we get a difference equation for the balances in terms of interest i and constant periodic payment PMT,
137
138\begin{equation}
139BAL_j = BAL_{j-1} - PMT + I_j
140\end{equation}
141
142\begin{equation}
143BAL_j = BAL_{j-1} - PMT + i BAL_{j-1}
144\end{equation}
145
146\begin{equation}
147BAL_j = (1+i) BAL_{j-1} - PMT
148\end{equation}
149
150Let's solve this difference equation. But first, some lemmas,
151
152\begin{lemma}{}
153The geometric progression
154
155\begin{equation}
156g = 1 + b + b^2 + \ldots + b^{n-1}
157\end{equation}
158
159has the formula,
160
161\begin{equation}
162g = {{1 - b^n} \over {1 - b}}
163\end{equation}
164
165\end{lemma}
166
167\begin{proof}
168
169Multiply by b and subtract,
170
171\begin{equation}
172g - 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
173\end{equation}
174
175to get
176
177\begin{equation}
178g = {{1 - b^n} \over {1 - b}}
179\end{equation}
180
181\end{proof}
182
183\begin{lemma}{}
184 If
185
186\begin{equation}
187u_k = a + b u_{k-1}
188\end{equation}
189
190then
191
192\begin{equation}
193u_n = a {1 - b^n \over 1-b } + b^n u_0
194\end{equation}
195
196\end{lemma}
197
198\begin{proof} Expand out terms to see the pattern and use mathematical induction if you like,
199
200\begin{equation}
201u_1 = a + b u_0
202\end{equation}
203
204\begin{equation}
205u_2 = a + b u_1 = a + b(a + b u_0) = a (1 + b) + b^2 u_0
206\end{equation}
207
208\begin{equation}
209u_3 = a + b u_2 = a + b [ (1+b) + b^2 u_0] = a (1 + b + b^2) + b^3 u_0
210\end{equation}
211
212The general formula is
213
214\begin{equation}
215u_n = a (1 + b + b^2 + \ldots + b^{n-1}) + b^n u_0
216\end{equation}
217
218Now use the geometric progression lemma above.
219
220\end{proof}
221
222\subsection{Closed Form Solution for $BAL_k$}
223
224To solve the balance difference equation let $u_k = BAL_k$, $a = -PMT$, and $b = i+1$
225
226The balance AFTER payment k is then
227
228\begin{equation}
229BAL_k = {(-PMT) {1 - (1+i)^k \over 1 - (1+i) } + (1+i)^k BAL_0}
230\end{equation}
231
232rearranging,
233
234\begin{equation}
235BAL_k = {1 \over (1+i)^{-k} } \left( PV + PMT { (1+i)^{-k} -1 \over i } \right)
236\end{equation}
237
238Now set $BAL_n = 0$ because the last nth payment PMT finishes the mortgage.
239Check: $BAL_0 = PV$
240
241\pagebreak
242
243\section{Other Useful Closed Form Solutions}
244
245\subsection{Periodic Payment PMT From PV, n and i}
246
247The constant periodic payment is then
248
249\begin{equation}
250PMT = {i PV \over 1-(1+i)^{-n}}
251\end{equation}
252
253\subsection{Number of Periods n from PV, i and PMT}
254
255The number of periods (months) comes from the loan amount PV and the periodic payment PMT after inverting,
256
257\begin{equation}
258n = -{ ln \left( {1 - i {PV \over PMT}} \right) \over { ln( 1 + i ) } }
259\end{equation}
260
261where we're using the natural logarithm, but any log will work.
262
263\subsection{Reduced Number of Periods $n'$ from Increasing the PMT}
264
265You can use this to determine the reduced number of periods if you prepay principal. Change $PMT$ to $(PMT + additional\: principal \: prepayment)$
266
267\begin{equation}
268n' = -{ ln \left( {1 - i {PV \over {PMT + additional\: principal \: prepayment}}} \right) \over { ln( 1 + i ) } }
269\end{equation}
270
271\subsection{Equivalent Simple Interest $i'$}
272
273Equivalent simple interest $i'$ on the loan is how much interest would have been had it been payable up front immediately.
274
275We use this formula to deduce it,
276
277\begin{equation}
278PV + i' PV = n PMT
279\end{equation}
280
281Equivalent simple interest is then
282
283\begin{equation}
284i' = {n i \over 1-(1+i)^{-n}} -1
285\end{equation}
286
287\subsection{Accumulated Interest Payments j though k}
288
289What is the accumulated interest for payments j though k inclusive?
290
291\begin{equation}
292Int_{j \rightarrow k} = I_j + \ldots + I_k
293\end{equation}
294
295\begin{equation}
296Int_{j \rightarrow k} = \left( k - j + 1 \right) PMT - \sum_{l=j}^k R_l
297\end{equation}
298
299but
300
301\begin{equation}
302BAL_l - BAL_{l-1} = -R_l
303\end{equation}
304
305We get a telescoping sum
306
307\begin{equation}
308- \sum_{l=j}^k R_l = BAL_k - BAL_{j-1}
309\end{equation}
310
311Interest AFTER the kth payment is
312
313\begin{equation}
314Int_{1 \rightarrow k} = k PMT + BAL_k - PV
315\end{equation}
316
317Check: When $k=n$,
318
319\begin{equation}
320Int_{1 \rightarrow n} = n PMT + BAL_n - PV = n PMT - PV
321\end{equation}
322
323as expected because the total n payments of PMT include repaying PV plus interest.
324
325\end{document}
