How can i prove that the duration of a bond with a zero coupon is equal to its maturity for discrete case compounded?

Question

The duration of a bond is given by : $$ \frac{dP}{dy}\frac{1}{P} = - \frac{D}{1+y} $$ solving for D we have : $$D = -\frac{1}{P}\frac{dP}{dy}.$$ Now we know that the bond price formula is : $$P = \frac{c}{(1+y)^1} + \frac{c}{(1+y)^2}+\cdots + \frac{c+P_p}{(1+y)^T}$$ how can i prove that for zero coupon $c=0$ the duration of a bond is equal the time to maturity for the discrete compounding or it isn't and why? Secondly how the modified duration and convexity adjustment formulas changes when the coupon is zero? Any help would be appreciated since i am new to bond pricing.

Answer 1

When solving for D you forgot to multiply for $1+y$, so that the duration is

$$ D = - \frac{1+y}{P} \frac{dP}{dy} $$

The price of a ZCB is just $P = P_p/(1+y)^T$. The derivative returns $-T P_p/(1+y)^{T+1}$. If you multiply for the other terms almost everything cancels out and you are left with $T$.

To compute the convexity use the price I gave you to compute the second derivative.

Mild consideration: if you have a ZCB, the price is simply given by what you wrote but considering the coupon $0$ ($c = 0$).

Edit [computations]:

The derivative is the derivative wrt the yield (it is the term $dP/dy$). It turns out that:

$$ \frac{dP}{dy} = \frac{-T P_p}{(1+y)^{T+1} }$$ If we multiply this by $-(1+y)/P$ we get:

$$ -\frac{-T P_p}{(1+y)^{T+1} } \frac{1+y}{P} = \frac{T P_p}{(1+y)^{T+1} }\frac{(1+y)(1+y)^T}{P_p} = T$$ as we wanted.

The convexity is instead defined as

$$ C = \frac{1}{P} \frac{d^2P}{dy^2} $$ so we need to compute the second derivative and divide by the price like before