2

$i^{(p)}$ is the nominal interest converted p-thly i.e the total interest per unit of time paid on a loan of amount 1 at time 0 where interest paid in p equal installments at the end of each p-th sub-interval.

I can't understand the statement given as follows :

Since $i^{(p)}$ is the total interest paid and each interest payment is of amount $\dfrac{i^{(p)}}{p}$ then the accumulated value at time 1 of the interest payments is :

$\dfrac{i^{(p)}}{p}(1+i)^{(p-1)/p} + \dfrac{i^{(p)}}{p}(1+i)^{(p-2)/p} +$.. ..... $+ \dfrac{i^{(p)}}{p} = i$

where $i$ is the effective rate of interest.

Can anyone explain the above statement and thus the equation ? Thanks in advance !

I tried and tried , finally I interpreted it in the following way , please correct me if wrong ,

We are adding the interest applicable per interval i.e in ( 1/p , 2/p , .... , 1) and equating it to the annual effective rate of interest $i$.

So the interest applicable per interval can be thought of as : $ \dfrac{i^{(p)}}{p} \times$ (Value accumulated just before that interval ) , so , interest applicable in the (1/p)th interval would be $\dfrac{i^{(p)}}{p} (1 + i) ^{(1 - \frac{1}{p})}$ , So ,

$\dfrac{i^{(p)}}{p}(1+i)^{(p-1)/p} + \dfrac{i^{(p)}}{p}(1+i)^{(p-2)/p} +$.. ..... $+ \dfrac{i^{(p)}}{p} = i$

Is this correct ??

User9523
  • 2,094

2 Answers2

1

If the effective rate is $i$, then the rate $r$ for each of the $p$ periods is given by $1+r = (1+i)^{1 \over p}$.

There are $p$ interest payments made at the end of each interval. Each payment is ${i^{(p)} \over p}$, hence the first payment accumulates interest over $p-1$ periods, the second over $p-2$ periods, ... , and the last over $0$ periods.

Hence the value of the first payment at the time 1 is ${i^{(p)} \over p} (1+r)^{p-1} = {i^{(p)} \over p} (1+i)^{p-1 \over p}$, the value of the second payment at the end of time 1 is ${i^{(p)} \over p} (1+r)^{p-2} = {i^{(p)} \over p} (1+i)^{p-2 \over p}$, etc, etc, and the value of the last payment (made at time 1) at time 1 is ${i^{(p)} \over p}$.

Hence the total is \begin{eqnarray} i &=& \sum_{k=0}^{p-1} {i^{(p)} \over p} (1+r)^{p-k} \\ &=& {i^{(p)} \over p} { (1+r)^p -1\over r} \\ &=& {i^{(p)} \over p} { (1+i) -1\over r} \\ &=& {i^{(p)} \over p} { i \over r} \\ \end{eqnarray} It follows that ${i^{(p)} \over p} = r $ and so $1+{i^{(p)} \over p} = (1+i)^{1 \over p}$, or equivalently $1+i = (1+{i^{(p)} \over p})^p$.

copper.hat
  • 172,524
  • Actually we have to establish the relation you used above , i.e $1+i = (1+{ i^{(p)} \over p})^p$ , so can we proceed without using this relation ? – User9523 Dec 23 '15 at 13:37
  • I misunderstood. If I understand correctly, you are using the last equation to define the nominal rate. Let me update accordingly. – copper.hat Dec 23 '15 at 16:29
  • Would you please go through the interpretation I came up with.. I've edited the question above.. @copper.hat – User9523 Jan 13 '16 at 16:58
  • Let $r$ be as I have above. There are $p$ equal payments of ${i^{(p)} \over p}$. The first payment is made at the end of the first period, so the value at the end is ${i^{(p)} \over p}(1+r)^{p-1}$. The value of the second payment at the end is ${i^{(p)} \over p}(1+r)^{p-2}$, etc, etc. When you add them all up they must equal $i$. – copper.hat Jan 14 '16 at 05:42
1

Let's look at this with an example. Suppose $t = 1$ represents one year of time from $t = 0$. Let $p = 12$, so that $i^{(12)}$ represents a nominal interest rate convertible monthly (12 times per year). As this is a rate, it is also equivalent to an amount of interest paid per monetary unit. The meaning of "convertible monthly" is that the interest accrues (i.e., is compounded) at the end of each month, but at an effective monthly rate of $i^{(12)}/12$.

We are interested in the equivalent effective annual rate $i$ corresponding to the nominal monthly rate $i^{(12)}$; that is to say, if interest were to be accrued at the end of one year, what would the required amount be in order to be equal to the total interest accrued if compounded monthly?

The idea, then, is to note that payments of $i^{(12)}/12$ are made at time points $t = 1/12, 2/12, 3/12, \ldots, 12/12$, and under the effective annual rate $i$, these payments have had time $1-t = 11/12, 10/12, \ldots, 0$ to accrue interest, respectively. Therefore, the future value of these payments at time $t = 1$ are $$\frac{i^{(12)}}{12} (1 + i)^{11/12} + \frac{i^{(12)}}{12} (1 + i)^{10/12} + \cdots + \frac{i^{(12)}}{12} (1 + i)^0,$$ and this must equal $i$ itself, since the total interest accrued under both situations are equivalent. Now, using the formula for a geometric series with common ratio $r = (1+i)^{1/12}$, we find that the LHS is $$\frac{(1+i)^{12/12} - 1}{(1+i)^{1/12} - 1} \frac{i^{(12)}}{12} = \frac{i^{(12)} i}{12((1+i)^{1/12} - 1)},$$ and equating this with $i$ and solving gives $$i = \left(1 + \frac{i^{(12)}}{12}\right)^{\!12} - 1,$$ and in general, it is easy to see that for a nominal rate of $i^{(p)}$ convertible $p$ times from $t = 0$ to $t = 1$, we have $$i = \left(1 + \frac{i^{(p)}}{p}\right)^{\!p} - 1.$$

heropup
  • 135,869