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I am working on showing the following.

There is a coupon bond redeemable at par with annual coupon rate $r$ per year. The yield to maturity is $i$. The total number of coupons is $n$. Show that the Macaulay duration for this coupon bond is

$$\frac{1+i}{i}-\frac{1+i+n(r-i)}{r[(1+i)^n-1]+i}$$

I understand the following.

To find the Macaulay Duration, we use the Present value of the bond, $P$ and the rate of change with respect to the yield rate $-\frac{d}{di}P$ and find the ratio, then multiply $1+i$.

So, letting $F$ be the face value and using some actuarial notation, I am thinking that

$$\begin{align} P &= Fr(v+v^2+ \cdots +v^n)+Fv^n \\ &=Fra_{\overline{n}\rceil i}+Fv^n\\ &=F(1+(r-i)a_{\overline{n}\rceil i})\\ \end{align}$$

I do not know which expression would be easier to use, but so far I have been trying to solve this using the last one.

Also, $-P'$ can be found as

$$\begin{align} -P'&=Fr(v+2v^2+ \cdots +nv^n)+Fnv^n\\ &=F(r(Ia)_{\overline{n}\rceil i}+nv^n)\\ \end{align}$$

The Macaulay Duration can be found from

$$D=\frac{F(r(Ia)_{\overline{n}\rceil i}+nv^n)}{F(1+(r-i)a_{\overline{n}\rceil i})}(1+i)$$

I do definitely see the bits and pieces in there, and as a matter of fact I was able to manipulate it so that the denominators are the same, but the numerator does not seem to match with my work.

For example, factoring out a $1 \over i$ on the given expression

$$\frac{1}{i} \left( (1+i)-\frac{1+i+n(r-i)}{ra_{\overline{n}\rceil i}+1}\right)$$

and combining the two terms into one,

$$\frac{1}{i} \left( \frac{r\ddot{a}{_{\overline{n}\rceil}}+n(r-i)}{ra_{\overline{n}\rceil}+1}\right)$$

Since

$$(Ia)_{\overline{n}\rceil i}= \frac{\ddot{a}_{\overline{n}\rceil i}-nv^n}{i}$$

I have a feeling that I am on the right track, but I haven't gotten the right expression. It's really driving me nuts!!

Can I have some help?

Thanks!

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  • $\begingroup$ I found my solution. The error I made was that the $D$ I described had an extra $(1+i)$ which threw off my calculation. Thank you for your interest. $\endgroup$
    – hyg17
    Dec 17 '14 at 23:43
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As you found $$D=\frac{F(r(Ia)_{\overline{n}\rceil i}+nv^n)}{Fra_{\overline{n}\rceil i}+Fv^n}=\frac{r(Ia)_{\overline{n}\rceil i}+nv^n}{ra_{\overline{n}\rceil i}+v^n}$$ Substituting $(Ia)_{\overline{n}\rceil i}= \frac{\ddot{a}_{\overline{n}\rceil i}-nv^n}{i}$ and $a_{\overline{n}\rceil i}=\frac{1-v^n}{i}$ and $\ddot{a}_{\overline{n}\rceil i}=(1+i)a_{\overline{n}\rceil i}$ we have $$ \begin{align} D&=\frac{r\frac{\ddot{a}_{\overline{n}\rceil i}-nv^n}{i}+nv^n}{r\frac{1-v^n}{i}+v^n}=\frac{r\ddot{a}_{\overline{n}\rceil i}-r\,nv^n+i\,nv^n}{r-rv^n+iv^n}=\frac{r\ddot{a}_{\overline{n}\rceil i}-r\,nv^n+i\,nv^n}{r-rv^n+iv^n}\\ &=\frac{r(1+i){a}_{\overline{n}\rceil i}+(i-r)\,nv^n}{r+(i-r)v^n}=\frac{r(1+i)\frac{1-v^n}{i}+(i-r)\,nv^n}{\color{red}{v^n}\{r[(1+i)^n-1]+i\}}\\ &=\frac{\color{red}{v^n}\left\{r\frac{1+i}{i}(v^{-n}-1)+(i-r)n\right\}}{\color{red}{v^n}\{r[(1+i)^n-1]+i\}} =\frac{r\frac{1+i}{i}[(1+i)^n-1]+(i-r)n}{r[(1+i)^n-1]+i}\\ &=\frac{r\frac{1+i}{i}[(1+i)^n-1]\color{red}{+\frac{1+i}{i}i-\frac{1+i}{i}i}+(i-r)n}{r[(1+i)^n-1]+i}\\ &=\frac{\color{red}{\frac{1+i}{i}}\{r[(1+i)^n-1]+\color{red}{i}\}-\color{red}{\frac{1+i}{i}i}+(i-r)n}{r[(1+i)^n-1]+i}\\ &=\color{red}{\frac{1+i}{i}}+\frac{\color{red}{-\frac{1+i}{i}i}+(i-r)n}{r[(1+i)^n-1]+i} \end{align} $$ that is $$\color{blue}{D=\frac{1+i}{i}-\frac{1+i+(r-i)n}{r[(1+i)^n-1]+i}}$$

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