Given that $B=Ce^{-y} + Ce^{-2y}+ (100+C)e^{-3y}$ where B is the bond price, C is the coupon. and It is a 3 years annual coupon bond.

I want to find $\frac{dD}{dC}$ where $D$ is the modified duration.

My steps:

1.modified duration = D

  1. $D = \frac{-1}{B} * \frac{dB}{dy}$

  2. $D = \frac{-1}{B} (-Ce^{-y} - 2Ce^{-2y} - 3Ce^{-3*y} - 300e^{-3y})$

  3. Then find $\frac{dD}{dC}$

  4. $\frac{dD}{dC} = \frac{1}{B} (e^{-y} + 2e^{-2y} + 3e^{-3y}) $

Am I correct ?

  • 3
    $\begingroup$ No, B depends on C too $\endgroup$ – pbr142 Mar 16 '14 at 2:01

For the sake of completeness:

Taking pbr142's comment into account and working in the setting you described.

Set $f(C,y)=Ce^{-y} + 2Ce^{-2y} + 3Ce^{-3*y} + 300e^{-3y}$. Write $B(C,y)$ instead of $B$. Applying the quotient rule to $\frac{\partial D(C,y)}{\partial C}$ with $D(C,y)=\frac{f(C,y)}{B(C,y)}$. This leads to the following expression

$\frac{\partial D(C,y)}{\partial C}=\frac{(e^{-y} + 2e^{-2y} + 3e^{-3y})B(C,y)-(e^{-y} + e^{-2y} + e^{-3y})f(C,y)}{B(C,y)}$

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