Differentiating Modified Duration with Respect to Coupon Rate

Summary

This question considers how a three-year annual-coupon bond’s modified duration changes when its coupon rate changes, with bond price expressed as the discounted value of its cash flows. It starts from the standard yield sensitivity definition of modified duration and attempts to differentiate that quantity with respect to the coupon. The response explains that duration is a ratio: the negative price derivative with respect to yield divided by price.

Consequently, the coupon sensitivity requires the quotient rule. The numerator depends on coupon through the yield-weighted cash flows, while the denominator also changes with coupon through the bond price. Differentiating only the numerator and dividing by price, as the question suggests, omits the denominator’s contribution. The document gives a symbolic expression for the corrected derivative but does not simplify or evaluate it for particular yields or coupons. Its result applies to the stated bond cash-flow setup and assumes yield is held fixed when taking the coupon derivative.

Key ideas

• Modified duration is the negative yield derivative of price divided by price.
• Coupon changes affect both the discounted cash-flow sensitivity and the bond price.
• The coupon derivative of duration must apply the quotient rule to the full ratio.
• The displayed result is symbolic and is not evaluated for a particular bond yield or coupon.

Tags

# Calculating the sensitivity of the modified bond duration to changes in the coupon rate


# Calculating the sensitivity of the modified bond duration to changes in the coupon rate












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

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

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

- Then find $\frac{dD}{dC}$

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

Am I correct ?

## Answer by Probilitator (score 1)

https://quant.stackexchange.com/a/10617

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

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.