-1
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ No, B depends on C too $\endgroup$ – pbr142 Mar 16 '14 at 2:01
1
$\begingroup$

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

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

| improve this answer | |
$\endgroup$

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.