Suppose that we want to invest € $P$ over $t$ years with an interest rate of $i$ (if the interest rate is 4%, then $i=0.04$).

Initially we have € $P$ – called the principal. After 1 year however, we will have the principal plus the interest: which is $i$ of $P$. Hence the value after one year will be: $P+Pi=P(1+i)$.

So to get the value of the investment you multiply the previous years value by $1+i$. So after three years, for example, the investment has value: $P(1+i)(1+i)(1+i)=P(1+i)^3$,

and we can generalise to any number of years.

The final value of an investment of a principal $P$ after $t$ years at an interest rate of $i$, $F$, is given by: $F=P(1+i)^t$.

p. 30 of the tables – Compound Interest

But what about inflation?

In economicsinflation is a rise in the general level of prices of goods and services in an economy over a period of time. When the general price level rises, each unit of currency buys fewer goods and services. Consequently, inflation also reflects an erosion in the purchasing power of money. A chief measure of price inflation is the inflation rate, the annualized average percentage change in a price of products.

Toy Example: Does it make sense to invest €100 at 4% in March 2011 for one year, when inflation is running at 5%?

Well after one year, in March 2012, you will have your €100 – plus your interest of 4% gives €104.

Suppose, for example, that in March 2011, that we wanted to buy a new pair of shoes for €100. Well in March 2012, with inflation running at 5%, the shoes are now going to cost €105. So although we have more euros at the end of the year (104 vs 100), in real terms we have less money – because our spending power has diminished.

What can €104 buy us after one year? What pair of shoes could we buy for €104 in March 2012? Well suppose $A$ was enough to buy a particular pair of shoes in March 2011 that would cost €104 in March 2012. So we want: $A+$ inflation $=104$ $\Rightarrow A+(0.05)A=A(1+0.05)=104$

So $A=\frac{104}{1+0.05}\approx 99.05$    (*)

So €104 in March 2012 was only enough money to buy slightly inferior (!) shoes.

To generalise (*), the real value of a final value $F$, after one year of inflation at a rate of $j$ is given by (again, for example, 2% translates to $j=0.02$): $A=\frac{F}{1+j}$.

Over, for example, three years: $A=\frac{F}{(1+j)(1+j)(1+j)}=\frac{F}{(1+j)^3}$

Hence to generalise, and include $F$ in terms of the principal:

The inflation adjusted final value of an investment of a principal $P$ after $t$ years at an interest rate of $i$, subject to an inflation rate of $j$ $F$, is given by: $F=\frac{P(1+i)^t}{(1+j)^t}$.

This formula is actually in the tables, it is called Present Value: $P=\frac{F}{(1+j)^t}=\frac{P(1+i)^t}{(1+j)^t}$

p. 30 of the tables – Present Value

# Mortgage Amortisation

Suppose that we take out a loan/ mortgage of $P$ over, for example, $t$ months. The bank is going to charge us interest monthly at a rate of $i$ per month but want us to pay back the same amount, $A$, every month over the lifetime of the mortage.

Let $p(n)$ denote the amount of money we owe the bank after $n$ months. The following proof is going to show us where the formula on p.31 (in my copy at least – it is the Amortisation formula) of the tables comes from.

What is $p(0)$? Well after 0 months, we owe the full amount: $p(0)=P$

What is $p(1)$? Well we still owe the principal $P$, they’ve thrown on some interest, $Pi$, but thankfully we have paid off $A$. So after one month we owe: $p(1)=P+Pi-A=\underbrace{P(1+i)}_{:=Pr}-A$

To simplify our calculations, let $r:=1+i$.

What about $p(2)$? Well we still owe what we owed last month, $p(1)$, and they’re going to put interest on this, but again we are going to reduce the loan by $A$ (remember $p(1)r=p(1)+p(1)i$ – what we owed plus interest): $p(2)=p(1)r-A=Pr^2-Ar-A$.

Similarly after three months we owe: $p(3)=p(2)r-A=Pr^3-Ar^2-Ar-A$

We can prove by induction that this pattern persists so that: $p(n)=Pr^n-A(r^{n-1}+r^{n-2}+\cdots r^2+r+1)$ $\Rightarrow p(n)=Pr^n-A\sum_{k=0}^{n-1}r^k$    (***)

Now looking at $\sum_{k=0}^{t-1}$: $1+r+r^2+\cdots+r^{n-1}$,

this is a geometric series with $n$ terms, first term $1$, and common ratio $r$. By p.22 of the tables, the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ is given by: $S_n=\frac{a(1-r^n)}{1-r}$ $\Rightarrow \sum_{k=0}^{n-1}r^k=\frac{r^n-1}{r-1}$.

Putting this back into (***): $p(n)=Pr^n-A\frac{r^n-1}{r-1}$

For $t$ payment periods, we expect the principal amount will be completely paid off at the last payment period, or $p(t)=0$ $\Rightarrow Pr^t-A\frac{r^t-1}{r-1}=0$.

Solving for $A$, we get $A=\frac{Pr^t(r-1)}{r^t-1}$

However note that: $r=1+i$ and hence $r-1=i$ $\Rightarrow A=\frac{Pi(1+i)^n}{(1+i)^n-1}$

p.31 of the tables – Amortisation – mortgages and loans

If we know how to use these formula, and what each letter stands for, we can get full marks in this question by using the tables properly.