Inflation Adjustment

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 economics, inflation 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.

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