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?


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 $$MacDur=r\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t\cdot t = r \cdot\frac{r+1}{r^2}=\frac{r+1}{r}$$

  • $\begingroup$ Thanks, can you please show that your last result holds? $\endgroup$
    – emcor
    Dec 12 '15 at 14:33
  • $\begingroup$ @Nicholas The duration you calculate is MacDur (not ModDur). And how do you know that ModDur=1/r? ModDur assumes continuous compounding, but I think c/r is the bond value in discrete compounding. $\endgroup$
    – emcor
    Dec 12 '15 at 16:43
  • $\begingroup$ @emcor you are indeed correct. I had a typo in a last formula - should be MacDur instead of ModDur. $\endgroup$
    – Nicholas
    Dec 12 '15 at 19:54
  • 1
    $\begingroup$ @Nicholas This is close enough. Mod duration is simply $\text{MacDur} / (1 + r / f)$, where $f$ is the compounding frequency. $\endgroup$
    – Helin
    Dec 12 '15 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.