Prophet
Jeremy's
notes

Navigate: ⇐

Rent or buy, that is the question§

I have just make some algebra to understand how the amortization works and which is the best way to pay a Loan. A few definitions are necessary to understand the following calculus.

  • \(N\) = Notional: Initial amount borrowed by the lender.

  • \(O_k\) = Outstanding: Remaining or unpaid loan amount.

  • \(A_k\) = Amortization: Portion of the nominal amount that is paid in each payment.

  • \(r_k\) = Interest Rate: Annualized interest rate applied to the Loan. It could be fixed or variable.

  • \(\Delta t_k\) = Year fraction: Fraction of the year since last payment.

  • \(n\) = Total number of payments: The number of payments in which the loan is expected to be settled.

  • \(I_k\) = Interest Payment. Interest to be paid on each payment.

    \[I_k = O_{k-1} \cdot r_k \cdot \Delta t_k\]

3 are the most common ways to amortise a loan:

  1. Bullet: During the life of the loan only interest is paid and at the end the nominal is settled.

  2. Constant: The same nominal amount is amortised in each payment.

  3. French: A minimally sophisticated formula is applied in which several of the defined variables intervene to calculate the fee to be paid, which includes interest:

    \[P_k = A_k + I_k = O_{k-1} \cdot \frac{r_k \cdot \Delta t_k}{1-(1 + r_k \cdot \Delta t_k)^{n-k+1}}\]

In this post I have focus on the methods that have payments during the loan’s life. So I won’t concern about bullet method, which amortization is pretty straightforward.

Due to the amortizations selected, each amortization depends on the remaining amount after each payment, which we can write as:

\[\begin{split}A_k = \alpha_k \cdot O_{k-1}\\ with\ O_0 = N\end{split}\]

For the successive payments:

\[\begin{split}\begin{aligned} A_1 &= N \cdot \alpha_1\\ A_2 &= O_1 \cdot \alpha_2 = N(1 - \alpha_1) \cdot \alpha_2 \\ A_3 &= O_2 \cdot \alpha_3 = N(1 - \alpha_1)(1 - \alpha_2) \cdot \alpha_3 \\ \ldots \end{aligned}\end{split}\]

Which reveals a clear pattern, the Outstanding can be written as:

\[O_k = N \cdot \prod_{i=1}^{k}(1 - \alpha_i)\]

So now we can just focus on \(\alpha\), that depends on the amortization type.

Constant (warmup)§

The constant amortization is not a big deal, so here I am showing how to work with it as a way to get used to the method, but the french amortization is the one that motivate this post. First of all the \(\alpha\), which comes from the amortization definition:

\[A_k = O_{k-1} \cdot \frac{1}{n-k+1} \rightarrow \alpha_k = \frac{1}{k-n-1}\]

So, as it has been shown:

\[\begin{split}\begin{aligned} O_k &= N \cdot \prod_{i=1}^{k} \left(1 - \frac{1}{n-i+1}\right)\\ O_k &= N \cdot \prod_{i=1}^k \frac{n-i}{n-i+1}\\ \end{aligned}\end{split}\]

Here the telescopic property for the product: \(\prod_{i=1}^{k} \frac{a_i}{a_{i-1}} = \frac{a_k}{a_0}\) is used to simplify the expression with \(a_i = n - i\):

\[O_k = N \cdot \left(1 - \frac{k}{n}\right)\]

Now it becomes obvious that at \(k=n\) the Outstanding is 0, as it should. We can evaluate the amortization making finite differences:

\[A_k = \frac{N}{n}\]

So the Amortization does not depend on \(k\) (that’s probably why it is called constant), if there are not prepayments, the amortised amount is always the same. Finally we should evaluate how this combines with the interest payment and the total payment. For simplicity I am going to use \(\beta_k = r_k \cdot \Delta t_k\).

\[\begin{split}\begin{aligned} I_k = O_{k-1} \cdot \beta_k = N \cdot \beta_k \cdot \frac{n-k+1}{n}\\ P_k = \frac{N}{n} \cdot \left(1 + \beta_k \cdot (n-k+1) \right) \end{aligned}\end{split}\]

In view of the formula it is clear that the instalment to be paid declines linearly with the number of payments due only to the payment of interest. If we suppose a fixed annual Interest Rate (\(r\)) of 3%, a monthly payment frequency (it is \(\beta = 0.0025\)) and 20 years as time to maturity of the loan (\(n=240\)); we can evaluate what percentage of each installment will represent the interest:

\[IR_k = \frac{I_k}{P_k} = \frac{\beta \cdot (n-k+1)}{1 + \beta \cdot (n-k+1)}\]

But to make the reasoning cleaner, we will calculate the percentage of the amount amortised, in this way the sum of these percentages will give the percentage of interest paid on the nominal, since the sum of the amounts amortised is the nominal:

\[TIR = \sum_{i=1}^n\frac{I_i}{N} = \frac{1}{n}\sum_{i=1}^n \beta_i \cdot (n-i+1)\]

Assuming that \(\beta\) is constant (as in our example):

\[TIR = \frac{\beta \cdot (n+1)}{2}\]

Which, with the figures given above, totals 30.125%. So you maybe want to think twice before taking a loan with this amortization type.

French§

This is the amortization type I get most insterested in, because it is the most common in mortgage loans accessed by individuals. As far as I know this method was developed to keep the payment constant (almost), which is what people is looking for when they are asking for money.

To follow the previous path first we have to remove the interests from the payment formula, using the previously given \(\beta_k\) definition:

\[\begin{split}\begin{aligned} A_k &= O_{k-1} \cdot \left(\frac{\beta_k \cdot (1 + \beta_k)^{n-k+1}}{(1 + \beta_k)^{n-k+1} - 1} - \beta_k\right) = O_{k-1} \cdot \left(\frac{\beta_k}{(1 + \beta_k)^{n-k+1} - 1}\right) \\ \alpha_k &= \frac{\beta_k}{(1 + \beta_k)^{n-k+1} - 1} \end{aligned}\end{split}\]

Now I can apply the cumulative product formula that has been developed previously:

\[\begin{split}\begin{aligned} O_k &= N \cdot \prod_{i=1}^k \left(1 - \frac{\beta_i}{(1 + \beta_i)^{n-i+1} - 1}\right) \\ &= N \cdot \prod_{i=1}^k \left(\frac{(1 + \beta_i)\left((1 + \beta_i)^{n-i} - 1\right)}{(1 + \beta_i)^{n-i+1} - 1}\right) \\ &= N \cdot \left[ \prod_{i=1}^k (1+\beta_i) \cdot \frac{(1 + \beta_i)^{n-i} - 1}{(1 + \beta_i)^{n-i+1} - 1} \right] \end{aligned}\end{split}\]

This expression doesn’t allow further simplification because \(\beta_k\) is different for each payment, but if we use a constant \(\beta\), which could be think about as an approximation that is true when the interest is fixed (as in our example). Then we have:

\[O_k = N \cdot (1+\beta)^k \cdot \prod_{i=1}^k \frac{(1+\beta)^{n-i} - 1}{(1+\beta)^{n-i+1} - 1}\]

And again, using the telescopic property with \(a_i = (1+\beta)^{n-i} - 1\) then we have:

\[\begin{split}\begin{aligned} O_k &= N \cdot (1+\beta)^k \cdot \frac{(1+\beta)^{n-k} - 1}{(1+\beta)^n - 1}\\ &= N \cdot \frac{(1+\beta)^{n} - (1+\beta)^k}{(1+\beta)^n - 1} \end{aligned}\end{split}\]

As a check we can see that for \(k=n\) this amount (the outstanding) is zero, as it must be. And the amortization can be evaluated using the difference between 2 outstanding amounts:

\[A_k = O_{k-1} - O_k = N \cdot \frac{(1+\beta)^{k-1} \cdot \beta}{(1+\beta)^n - 1}\]

As the time the amortization depends on the payment (\(k\)) we need to compute the percentage of the notional finally paid as interest by hand doing (remember \(I_k = \beta_k \cdot O_k\)):

\[TIR = \sum_{k=1}^n \frac{I_k}{N} = \beta \cdot \sum_{k=1}^n \frac{(1+\beta)^n - (1+\beta)^k}{(1 + \beta)^n - 1}\]

Which, for the figures given in our example, is: 32,853%. Almost 273 bps higher than the constant method.

Real world data§

Now I feel trained enough to give my opinion in the debate: buying or renting, how is it less likely to get ruined?. In my city, Madrid, I have found the following prices for the purchase and rental of a home:

  • Average price per square meter at purchase (last 12 months): 3,718 €/m2.

  • Average price per square meter in the rent (last 12 months): 16’22 €/m2.

So lets suppose you are getting a house of 85 m2, if you buy it asking for its entire price, you are going to pay, considering that it is a second-hand property, around (this is an approximation) 15% of the house price as taxes, and looking for fixed rate mortgage I have found the average value of last year MLRI (Mortgage Loan Reference Index) to be 1’768% (this is a negative interest rate scenario), with this data, considering a 20 years maturity loan, you are going to pay for buying the house:

\[\begin{split}\text{house price} &= 3718 \cdot 85 = 316,030 \text{ €} \\ \text{taxes and interests} &= \text{house price} \cdot (0'15 + 0'1865) = 106,344'10 \text{ €}\end{split}\]

With that money, you could rent the same house during:

\[\frac{106,344'10}{16'22 \cdot 85} = 76'10 \text{ months}\]

So you can actually rent the same house that you are buying for more than 6 years. You would have paid on average a mortgage fee of 1562’53€ and the rent would be instead 1378’70€. Both this situations are out of scope for most of us (people around 30 years) but, if you are thinking about buying, you should know that you are reducing your savings / investment capacity for the next 20 years (or more), and paying to the Bank and the Government more than 33% of the house price for letting you buy it. I consider these results a proper approximation to the current (2020) scenario.

©2021, Prophet Jeremy. Powered by: Sphinx 7.2.6  & PJnotes 1.0.3