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10\documentclass{article}
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12\author{ Sean E. O'Connor}
13\title{Mortgage Loan Derivation}
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46
47
48\begin{document}
49
50\maketitle
51
52
53
54\begin{abstract}
55We derive the equations for a \emph{mortgage} loan and give several useful formulas
56\footnote{Using using \LaTeX and \TeX Shop. See \url{https://www.ctan.org}}
57\end{abstract}
58
59
60
61\tableofcontents
62
63
64
65\pagebreak
66
67\section{The Repayment Structure of a Mortgage Loan}
68
69\subsection{Definition of Interest Rate}
70
71The interest rate on the loan note is quoted as a yearly amount but is applied periodically for $m$ periods per year, giving
72
73\begin{equation}
74i = {{Annual \: interest \: percent / 100} \over m}
75\end{equation}
76
77Most house and car loans have a monthly period, $m = 12$. % Inline math can use dollar sign delimiters like MathJax
78
79\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.
80
81\subsection{Periodic Payments of Principal + Interest}
82
83The $j^{th}$ payment PMT consists of interest $I_j$ on the previous unpaid balance $BAL_{j-1}$
84
85This payment is a fixed amount every month by design.
86
87Only the remainder of the payment after interest $R_j$ actually reduces the principal of the loan.
88
89Payments from the initial payment to the end payment have the form,
90
91\begin{tabular}{ clll }
92\toprule
93Payment Number & Payment Amount & Interest & Balance \\
94\midrule
95${1^{st}}$ & $PMT = R_1 + I_1$ & $I_1 = i \: BAL_0$ & $BAL_1 = BAL_0 - R_1$ \\
96$2^{nd}$ & $PMT = R_2 + I_2$ & $I_2 = i \: BAL_1 $ & $BAL_2 = BAL_1 - R_2$ \\
97\ldots & \ldots & \ldots & \ldots \\
98$j^{th}$ & $PMT = R_j + I_j$ & $I_j = i \: BAL_{j-1}$ & $BAL_j = BAL_{j-1} - R_j$ \\
99\ldots & \ldots & \ldots & \ldots \\
100$n^{th}$ & $PMT = R_n + I_n$ & $I_n = i \: BAL_{n-1}$ & $BAL_n= BAL_{n-1} - R_n = 0 $ \\
101\bottomrule
102\end{tabular}
103
104n = the number of periods of the loan
105
106\begin{remark}
107The loan amount is $PV = BAL_0$ which is the original balance before any payments are made.
108\end{remark}
109
110\begin{remark}
111The last balance is 0 because the loan is paid off, $BAL_n = 0$.
112\end{remark}
113
114We could write an Excel, Perl or Python script to compute this, but there are closed form solutions as we'll show.
115
116\pagebreak
117
118\section{Closed Form Solution}
119
120\subsection{Difference Equation}
121
122Substituting we get a difference equation for the balances in terms of interest i and constant periodic payment PMT,
123
124\begin{equation}
125BAL_j = BAL_{j-1} - PMT + I_j
126\end{equation}
127
128\begin{equation}
129BAL_j = BAL_{j-1} - PMT + i BAL_{j-1}
130\end{equation}
131
132\begin{equation}
133BAL_j = (1+i) BAL_{j-1} - PMT
134\end{equation}
135
136Let's solve this difference equation. But first, some lemmas,
137
138\begin{lemma}{}
139The geometric progression
140
141\begin{equation}
142g = 1 + b + b^2 + \ldots + b^{n-1}
143\end{equation}
144
145has the formula,
146
147\begin{equation}
148g = {{1 - b^n} \over {1 - b}}
149\end{equation}
150
151\end{lemma}
152
153\begin{proof}
154
155Multiply by b and subtract,
156
157\begin{equation}
158g - 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
159\end{equation}
160
161to get
162
163\begin{equation}
164g = {{1 - b^n} \over {1 - b}}
165\end{equation}
166
167\end{proof}
168
169\begin{lemma}{}
170 If
171
172\begin{equation}
173u_k = a + b u_{k-1}
174\end{equation}
175
176then
177
178\begin{equation}
179u_n = a {1 - b^n \over 1-b } + b^n u_0
180\end{equation}
181
182\end{lemma}
183
184\begin{proof} Expand out terms to see the pattern and use mathematical induction if you like,
185
186\begin{equation}
187u_1 = a + b u_0
188\end{equation}
189
190\begin{equation}
191u_2 = a + b u_1 = a + b(a + b u_0) = a (1 + b) + b^2 u_0
192\end{equation}
193
194\begin{equation}
195u_3 = a + b u_2 = a + b [ (1+b) + b^2 u_0] = a (1 + b + b^2) + b^3 u_0
196\end{equation}
197
198The general formula is
199
200\begin{equation}
201u_n = a (1 + b + b^2 + \ldots + b^{n-1}) + b^n u_0
202\end{equation}
203
204Now use the geometric progression lemma above.
205
206\end{proof}
207
208\subsection{Closed Form Solution for $BAL_k$}
209
210To solve the balance difference equation let $u_k = BAL_k$, $a = -PMT$, and $b = i+1$
211
212The balance AFTER payment k is then
213
214\begin{equation}
215BAL_k = {(-PMT) {1 - (1+i)^k \over 1 - (1+i) } + (1+i)^k BAL_0}
216\end{equation}
217
218rearranging,
219
220\begin{equation}
221BAL_k = {1 \over (1+i)^{-k} } \left( PV + PMT { (1+i)^{-k} -1 \over i } \right)
222\end{equation}
223
224Now set $BAL_n = 0$ because the last nth payment PMT finishes the mortgage.
225Check: $BAL_0 = PV$
226
227\pagebreak
228
229\section{Other Useful Closed Form Solutions}
230
231\subsection{Periodic Payment PMT From PV, n and i}
232
233The constant periodic payment is then
234
235\begin{equation}
236PMT = {i PV \over 1-(1+i)^{-n}}
237\end{equation}
238
239\subsection{Number of Periods n from PV, i and PMT}
240
241The number of periods (months) comes from the loan amount PV and the periodic payment PMT after inverting,
242
243\begin{equation}
244n = -{ ln \left( {1 - i {PV \over PMT}} \right) \over { ln( 1 + i ) } }
245\end{equation}
246
247where we're using the natural logarithm, but any log will work.
248
249\subsection{Reduced Number of Periods $n'$ from Increasing the PMT}
250
251You can use this to determine the reduced number of periods if you prepay principal. Change $PMT$ to $(PMT + additional\: principal \: prepayment)$
252
253\begin{equation}
254n' = -{ ln \left( {1 - i {PV \over {PMT + additional\: principal \: prepayment}}} \right) \over { ln( 1 + i ) } }
255\end{equation}
256
257\subsection{Equivalent Simple Interest $i'$}
258
259Equivalent simple interest $i'$ on the loan is how much interest would have been had it been payable up front immediately.
260
261We use this formula to deduce it,
262
263\begin{equation}
264PV + i' PV = n PMT
265\end{equation}
266
267Equivalent simple interest is then
268
269\begin{equation}
270i' = {n i \over 1-(1+i)^{-n}} -1
271\end{equation}
272
273\subsection{Accumulated Interest Payments j though k}
274
275What is the accumulated interest for payments j though k inclusive?
276
277\begin{equation}
278Int_{j \rightarrow k} = I_j + \ldots + I_k
279\end{equation}
280
281\begin{equation}
282Int_{j \rightarrow k} = \left( k - j + 1 \right) PMT - \sum_{l=j}^k R_l
283\end{equation}
284
285but
286
287\begin{equation}
288BAL_l - BAL_{l-1} = -R_l
289\end{equation}
290
291We get a telescoping sum
292
293\begin{equation}
294- \sum_{l=j}^k R_l = BAL_k - BAL_{j-1}
295\end{equation}
296
297Interest AFTER the kth payment is
298
299\begin{equation}
300Int_{1 \rightarrow k} = k PMT + BAL_k - PV
301\end{equation}
302
303Check: When $k=n$,
304
305\begin{equation}
306Int_{1 \rightarrow n} = n PMT + BAL_n - PV = n PMT - PV
307\end{equation}
308
309as expected because the total n payments of PMT include repaying PV plus interest.
310
311\end{document}
