Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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 ?

share|improve this question
3  
No, B depends on C too –  pbr142 Mar 16 at 2:01
add comment

1 Answer 1

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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