Mortgage Formula Derivation

Geometry Series

$$ S_n = 1 + a + a^2 + \cdots + a^{n-1} + a^n$$

$$ S_n a = a + a^2 + a^3 + \cdots + a^n+ a^{n+1} $$

$$ S_n (a-1) = a^{n+1}-1 $$

$$ S_n = \frac{1-a^{n+1}}{(1-a)} $$

Derivation

$n$ number of months $m$ monthly payment $a$ annualized interest rate $r=a/12$ monthly interest rate (anuualized interest rate divided by 12 months) $p$ principal

First Month $n=1$

starting debt = principal $p$

current interest = starting debt * monthly interest rate $$pr$$ actual principal payment = ( monthly payment - current interest ) $$ m-pr $$ debt left = ( starting debt - actual principal payment ) $$ p-(m-pr)=(1+r)p-m $$

Second Month $n=2$

starting debt = last debt $(1+r)p-m$

current interest = starting debt * monthly interest rate $$ ( (1+r)p-m ) r= (1+r)rp-mr $$ actual principal payment = ( monthly payment - current interest ) $$ m-[(1+r)rp-mr] $$ debt left = ( starting debt - actual principal payment ) $$ \begin{aligned} & (1+r)p-m-[m-((1+r)rp-mr)]\
= & (1+r)p-2m+((1+r)rp-mr)\
= & (1+r)p+(1+r)rp-2m-mr\
= & (r+1)^2p-(1+r)m-m \end{aligned} $$

Third Month $n=3$

starting debt = last debt $(r+1)^2p-(1+r)m-m$

current interest = starting debt * monthly interest rate $$ \begin{aligned} & [ (r+1)^2p-(1+r)m-m ] r\
= & r(r+1)^2p-(1+r)rm-rm \end{aligned} $$ actual principal payment = ( monthly payment - current interest ) $$ m-[r(r+1)^2p-(1+r)rm-rm] $$ debt left = ( starting debt - actual principal payment ) $$ \begin{aligned} & (r+1)^2p-(1+r)m-rm - \left[ m-[r(r+1)^2p-(1+r)rm-rm] \right]\
= & (r+1)^3p -(1+r)^2m -(1+r)m - m \end{aligned} $$

Now the formula for debt left at the $n$th term seems to be $$(r+1)^np -m[ (1+r)^{n-1}+\cdots+(1+r)^2 +(1+r) + 1] $$. We use the fomula for the sum of geometric series $$ S_n = \frac{1-a^{n+1}}{(1-a)} $$. So the final formula becomes $$ (r+1)^{n}p + m\frac{ 1-(r+1)^{n} }{ r } $$.

We can use math induction to prove it.

the case of $n$th term

debt left $ (r+1)^{n}p + m\frac{ 1-(r+1)^{n} }{ r } $.

the case of $(n+1)$th term

starting debt = last debt $$ (r+1)^{n}p + m\frac{ 1-(r+1)^{n} }{ r } $$ current interest = starting debt * monthly interest rate $$ r(r+1)^{n}p + rm\frac{ 1-(r+1)^{n} }{ r } $$ actual principal payment = ( monthly payment - current interest ) $$ m - [ r(r+1)^{n}p + rm\frac{ 1-(r+1)^{n} }{ r } ] $$

debt left = ( starting debt - actual principal payment ) $$ \begin{aligned} & (r+1)^{n}p + m\frac{ 1-(r+1)^{n} }{ r } - \left [ m - [ r(r+1)^{n}p + rm\frac{ 1-(r+1)^{n} }{ r } ] \right ] \
= & (r+1)^{n+1}p + m\frac{ 1-(r+1)^{n} }{ r } - m + rm\frac{ 1-(r+1)^{n} }{ r } ]\
= & (r+1)^{n+1}p + m [ \frac{ 1-(r+1)^{n} }{ r } - \frac{r}{r} + \frac{ r-r(r+1)^{n} }{ r } ] \
= & (r+1)^{n+1}p + m [ \frac{ 1-(r+1)(r+1)^{n} }{ r } ] \
= & (r+1)^{n+1}p + m [ \frac{ 1-(r+1)^{n+1} }{ r } ] \end{aligned} $$

Cosistent with the formula for $n$th term. So it proves.

To calculate the monthly payment, we let the final debt of the $n$th term be zero. $$ (r+1)^np + m\frac{ 1-(r+1)^n }{ r } = 0 $$ So we have the formula for monthly payment $$ m = \frac{ (r+1)^np }{ \frac{ (r+1)^n - 1 }{ r } } = p r \frac{ (r+1)^n }{ (r+1)^n - 1 } = \frac{ p r }{ 1 - (r+1)^{-n} }$$

What is the Total Interest?

Total Interest = number of months * number of terms - principal $$ \begin{aligned} m*n - p & = \frac{ p r }{ 1 - (r+1)^{-n} }*n - p \
& = [ \frac{ n r }{ 1 - (r+1)^{-n} } - 1 ]p \end{aligned} $$

An Example

Suppose we are getting the mortgage loan 350K with interest rate 3%. Then $p=350000$ and fixed anualized interest rate $a=0.03$ for 30 years. Plug them in the formula. The monthly interest rate $r=a/12=0.03/12=0.0025$. Number of terms $ n = 30\times 12=360 $.

Monthly payment $$ \frac{ p r }{ 1 - (r+1)^{-n} } = \frac{ 350000\times 0.03 }{ 1 - (0.0025+1)^{-360} } = 1475 $$

Total interest $$ mn-p = 1475360-350000 = 181221 $$

So with annualized interest rate 3%, you have to pay more than half of the principal as interest. The fraction of total interest over principal is $$ \frac{ n r }{ 1 - (r+1)^{-n} } - 1 $$. The following table shows the change of total interest fraction on various interest rates ($n=30\times 12=360$).

With 5% annualized interest rate, the total interest is almost the same as the principal.