<p>Note most of the argumentation adapts off the standard derivation for <a href="https://en.wikipedia.org/wiki/Mortgage_calculator#Monthly_payment_formula" rel="nofollow noreferrer">constant nominal payments from Wikipedia</a>.</p>

<p>Let us assume <span class="math-container">$c$</span> is a yearly payment for the sake of simplicity. Consider how much is left to be repaid after each year. The principal remaining after the first year is <span class="math-container">$P_1=(1+r)P-c$</span>. That is, the initial amount plus interest less the payment. After the second year <span class="math-container">$P_2=(1+r) P_1 - c$</span> is left, so <span class="math-container">$P_2=(1+r)((1+r)P-c)-c(1+\pi)$</span>. Note we need to pay <span class="math-container">$c(1+\pi)$</span> instead of <span class="math-container">$c$</span> so that we are still paying <span class="math-container">$c$</span> in real terms for the second year. If the whole loan was repaid after two years, <span class="math-container">$P_2=0$</span>, so <span class="math-container">$P = \frac{c}{1+r}+\frac{c(1+\pi)}{(1+r)^2}$</span>.</p>

<p>This equation generalizes for a term of <span class="math-container">$n$</span> years, <span class="math-container">$$P = \frac{c}{1+r} \sum_{k=0}^{n-1} \frac{(1+\pi)^k}{(1+r)^k}$$</span>. This is a <a href="https://en.wikipedia.org/wiki/Geometric_series" rel="nofollow noreferrer">geometric series</a> which has the sum
<span class="math-container">$P=\frac{c}{1+r}\left(\frac{1-\frac{1+\pi}{1+r}^n}{1-\frac{1+\pi}{1+r}}\right)$</span>
which can be rearranged to give <span class="math-container">$$c=\frac{P(1+r)}{\left(\frac{1-\frac{1+\pi}{1+r}^n}{1-\frac{1+\pi}{1+r}}\right)}$$</span>. The nominal payment for year <span class="math-container">$n$</span> will be <span class="math-container">$c(1+\pi)^{(n-1)}$</span>.</p>

CC BY-SA 4.0
2019-09-16T22:48:22.963Z
2019-09-16T22:48:22.963Z
Kent Shikama
<p>From the <a href="https://en.wikipedia.org/wiki/Mortgage_calculator#Monthly_payment_formula" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>The fixed monthly payment for a fixed rate mortgage is the amount paid
by the borrower every month that ensures that the loan is paid off in
full with interest at the end of its term. The monthly payment formula
is based on the annuity formula. The monthly payment c depends upon:</p>
<p>r - the monthly interest rate, expressed as a decimal, not a percentage. Since the quoted yearly percentage rate is not a
compounded rate, the monthly percentage rate is simply the yearly
percentage rate divided by 12; dividing the monthly percentage rate by 100 gives r, the monthly rate expressed as a decimal.</p>
<p>N - the number of monthly payments, called the loan's term, and</p>
<p>P - the amount borrowed, known as the loan's principal.</p>
<p>In the standardized calculations used in the United States, c is given by the formula:
<span class="math-container">$$c = \begin{cases}
  \frac{r P}{1-(1+r)^{-N}} = \frac {rP(1+r)^N}{(1+r)^N-1}, &amp; r\ne 0; \\
  \frac{P}{N}, &amp; r = 0.
\end{cases}$$</span></p>
</blockquote>
<p>The &quot;issue&quot; with this monthly payment schedule is that, if inflation is positive, the amount you pay in real terms the first year is more than the amount you pay the last year. Assuming constant inflation <span class="math-container">$\pi$</span>, what would the monthly payment formula look like if it had to be constant in real terms?</p>

CC BY-SA 4.0
2019-09-16T22:54:13.873Z
Note this question was originally created as a response for https://economics.stackexchange.com/q/31866/8387
Kent Shikama
<inflation><interest-rate>
What would mortgage payments be if monthly payments were constant in real terms?
Kent Shikama
# What would mortgage payments be if monthly payments were constant in real terms?

(Asked by: username_0 on 2019-09-16)

From the [Wikipedia](https://en.wikipedia.org/wiki/Mortgage_calculator#Monthly_payment_formula):

> The fixed monthly payment for a fixed rate mortgage is the amount paid
> by the borrower every month that ensures that the loan is paid off in
> full with interest at the end of its term. The monthly payment formula
> is based on the annuity formula. The monthly payment c depends upon:
>
> r - the monthly interest rate, expressed as a decimal, not a percentage. Since the quoted yearly percentage rate is not a
> compounded rate, the monthly percentage rate is simply the yearly
> percentage rate divided by 12; dividing the monthly percentage rate by 100 gives r, the monthly rate expressed as a decimal.
>
> N - the number of monthly payments, called the loan's term, and
>
> P - the amount borrowed, known as the loan's principal.
>
> In the standardized calculations used in the United States, c is given by the formula:
> $$c = \begin{cases}
> \frac{r P}{1-(1+r)^{-N}} = \frac {rP(1+r)^N}{(1+r)^N-1}, & r\ne 0; \\
> \frac{P}{N}, & r = 0.
> \end{cases}$$

The "issue" with this monthly payment schedule is that, if inflation is positive, the amount you pay in real terms the first year is more than the amount you pay the last year. Assuming constant inflation $\pi$, what would the monthly payment formula look like if it had to be constant in real terms?

**Tags:** inflation, interest-rate
> **Comments:**
> **username_0**: Note this question was originally created as a response for https://economics.stackexchange.com/q/31866/8387  

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## 1 answers

### Answer by username_0 on 2019-09-16. Score: 0 | Accepted answer

Note most of the argumentation adapts off the standard derivation for [constant nominal payments from Wikipedia](https://en.wikipedia.org/wiki/Mortgage_calculator#Monthly_payment_formula).

Let us assume $c$ is a yearly payment for the sake of simplicity. Consider how much is left to be repaid after each year. The principal remaining after the first year is $P\_1=(1+r)P-c$. That is, the initial amount plus interest less the payment. After the second year $P\_2=(1+r) P\_1 - c$ is left, so $P\_2=(1+r)((1+r)P-c)-c(1+\pi)$. Note we need to pay $c(1+\pi)$ instead of $c$ so that we are still paying $c$ in real terms for the second year. If the whole loan was repaid after two years, $P\_2=0$, so $P = \frac{c}{1+r}+\frac{c(1+\pi)}{(1+r)^2}$.

This equation generalizes for a term of $n$ years, $$P = \frac{c}{1+r} \sum\_{k=0}^{n-1} \frac{(1+\pi)^k}{(1+r)^k}$$. This is a [geometric series](https://en.wikipedia.org/wiki/Geometric_series) which has the sum
$P=\frac{c}{1+r}\left(\frac{1-\frac{1+\pi}{1+r}^n}{1-\frac{1+\pi}{1+r}}\right)$
which can be rearranged to give $$c=\frac{P(1+r)}{\left(\frac{1-\frac{1+\pi}{1+r}^n}{1-\frac{1+\pi}{1+r}}\right)}$$. The nominal payment for year $n$ will be $c(1+\pi)^{(n-1)}$.


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