Macaulay and Modified Duration of a Perpetual Bond

# Duration of perpetual bond


# Duration of perpetual bond












I am trying to derive the duration of a perpetual bond with coupon $c$ in two ways:

$$D=-\frac{\frac{\partial P}{\partial r}}{P},$$ $$P=\frac{c}{r}$$ $$\Rightarrow D = -\frac{-\frac{c}{r^2}}{\frac{c}{r}}=\frac{1}{r}$$

In the second approach, I want to derive the duration using the Macauley Duration (average PV-weighted time to maturity):

$$D=\sum_{t=1}^T \frac{c_t}{(1+r)^tP}\cdot t$$ $$\Rightarrow D=\sum_{t=1}^\infty\frac{ c\cdot t}{(1+r)^t\frac{c}{r}}=\sum_{t=1}^\infty\frac{ r\cdot t}{(1+r)^t}=r\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t\cdot t$$ However, I am unable to show the convergence of this sum to $1/y$.

I came as far as to rewriting the sum as: $$S_m=\sum_{k=1}^mkx^k=\sum_{k=0}^{m-1}(k+1)x^{k+1}=x+x\sum_{k=1}^{m-1}kx^k+x\sum_{k=1}^{m-1}x^k.$$ $$\Rightarrow (1-x)S_m=x\frac {1-x^m}{1-x}$$

For $y>0$ we have $x=\dfrac1{1+r}<1$ and so the sum converges to $$\Rightarrow S_m=\frac {x}{(1-x)^2}=\frac {\dfrac1{1+r}}{(1-\dfrac1{1+r})^2}$$

$$\Rightarrow D=\frac {\dfrac{r}{1+r}}{(1-\dfrac1{1+r})^2}$$ However, I was unable to show the desired result $D=\frac{1}{r}$.

Can someone show the correct solution?

## Answer by Nicholas (score 3, accepted)

https://quant.stackexchange.com/a/22289

You were on a right track. In the first approach you've shown Modified Duration of perpetuity is $ModDur=\frac{1}{r}$. In your second approach keep in mind that $ModDur=\frac{MacDur}{(1+y_k/k)}$ so for annual compounding your second approach should converge to $MacDur=ModDur \cdot (1+r) = \frac{1+r}{r}$, which should be the case.

$$S_m=\sum_{k=1}^mkx^k=x+2x^2+3x^3+4x^4+...$$ now $$xS_m=x\sum_{k=1}^mkx^k=x^2+2x^3+3x^4+4x^5+...$$ subtracting $S_m-xS_m$ we get $$S_m-xS_m=x+x^2+x^3+x^4+...+:=A$$ now we note that $A-xA=x$ which yields $A=\frac{x}{1-x}$ and from $S_m-xS_m=\frac{x}{1-x}$ we find $S_m$ which is $S_m=\frac{x}{(1-x)^2}=$ and in your notions $x=\frac{1}{1+r}$ so $S_m=\frac{1}{1+r}\cdot(1-\frac{1}{1+r})^{-2}=\frac{r+1}{r^2}$

now we substitute the result into your formula $$MacDur=r\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t\cdot t = r \cdot\frac{r+1}{r^2}=\frac{r+1}{r}$$