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QUESTION

A regular stream of payments (for a given number of years) is
called an annuity.

\begin{description}

\item[(a)]
If the amount to be paid at the end of each year is $d$, and the
guaranteed rate of return is $r$, write down the formula which
gives the value of the annuity after $T$ years.

\item[(b)]
Hence deduce the size of the annuity payments necessary to
accumulate a given sum in the future.

\item[(c)]
Repeat the calculation for $m$ compoundings a year.

\item[(d)]
Show that the amount accumulated under continuous compounding is
$\frac{d(e^{rT}-1)}{(re^r-1)}$.

\end{description}


ANSWER

\begin{description}

\item[(a)]
Pay in $d$ at the end of each year, annual compounding for $T$
year.

-Consider payment for year 1. It earns interest for $T-1$ years.
Thus it grows to $d(1+r)^{T-1}$.

-Consider payment in year 2. It earns interest for $T-2$ years.
Thus it grows to $d(1+r)^{T-2}$

\hspace{3cm}\vdots

-Consider payment in year $t(&lt;T)$. It earns interest for $T-t$
years. Thus it grows to $d(1+r)^{T-t}$.

\hspace{3cm}\vdots

-Consider payment in year $T$. It doesn't earn interest. Thus it
stays at $d$.

Add all these up: $F_T=$ value at year $T$

\begin{eqnarray*}
F_T&amp;=&amp;d(1+r)^{T-1}+d(1+r)^{T-2}+\ldots+d(1+r)^{T-t}+\ldots+d\\
&amp;=&amp;\sum_{t=1}^Td(1+r)^{T-t}
\end{eqnarray*}

This is a geometric progression with common factor $(1+r)^{_1}$
(and overall prefactor of $d(1+r)^T$

$$F_T=d\left[\frac{(1+r)^T-1}{r}\right]$$ by standard sum of a GP.

\item[(b)]
If you require a given sum in the future of $F_T$ with an annual
compounding at $r$ for $T$ years, the annual payment must be $d$
where

$$d=\left[\frac{F_Tr}{(1+r)^T-1}\right]$$

\item[(c)]
Similar arguments for $m$ compoundings a year give

$$F_T=d\sum_{t=1}^T\left(1+\frac{r}{m}\right)^{m(T-t)}$$

Again this is a GP common factor $\left(1+\frac{r}{m}\right)^{-m}$

$$F_T=d\left\{\frac{\left(1+\frac{r}{m}\right)^{mT}-1}{\left(1+\frac{r}{m}\right)^m-1}\right\}$$

so

$$d=F_T\left\{\frac{\left(1+\frac{r}{m}\right)^m-1}{\left(1+\frac{r}{m}\right)^{mt}-1}\right\}$$

\item[(d)]
To find the same result for continuous compounding take the limit
of (c) as $m\to\infty$.

$$\lim_{m\to\infty}\left(i+\frac{r}{m}\right)^{mt}=e^{rT}$$

from question 2 above.

$$\lim_{m\to\infty}\left(1+\frac{r}{m}\right)^m=e^r\ (t=1)$$

Thus

$$F_T=d\frac{(e^{rt}-1)}{(e^r-1)}$$

and

$$d=\frac{F_T(e^r-1)}{(e^{rT}-1)}$$

\end{description}



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