Convexity of a Bond

[mroberts]

Hey Guys, here's my question....

What is the convexity of a zero-coupon bond?
Hint: The convexity of a bond that pays annual coupons for T years is:

\(\displaystyle C\, =\, \dfrac{1}{B\, (1\, +\, y)^2}\, \sum_{t=1}^T \left(\dfrac{C_t}{(1\, +\, y)^t}\, (t^2\, +\, t)\right)\)

My Workings:

Zero-coupon bonds have the highest convexity.
Let C_t stand for the Face Value, as the only cash flow occurs in the terminal year.

\(\displaystyle C\, =\, \dfrac{1}{675.5642\, (1\, +\, 0.04)^2}\, *\, \left(\dfrac{1000}{(1\, +\, 0.04)^{10}}\, (10^2\, +\, 10)\right)\)

B = 0*(1-(1.04)^-10)/(0.04)+1000/(1.04)^10
B = 675.5642
R=4%

C = 74,312.05857 / (675.5642(1.0816))
C = 101.701

Does this seem right??

Thanks

 

[DexterOnline]

Price of Zero Coupon Bond

\(\displaystyle V = C_t \times (1+i)^{-t}\)

Convexity of Zero Coupon Bond

\(\displaystyle C = V^{-1} \times \frac{d^{2}V}{di^{2}}\)

\(\displaystyle \frac{dV}{di} = -t \times C_t \times (1+i)^{-(t+1)}\)

\(\displaystyle \frac{d^{2}V}{di^{2}} = t \times (t+1) \times C_t \times (1+i)^{-(t+2)}\)

Assuming annual compounding of interest

\(\displaystyle C_t=1,000\)
\(\displaystyle i=4\%\)
\(\displaystyle n=10\)

\(\displaystyle V = 1,000 \times (1+0.04)^{-10}\)
\(\displaystyle V = 1,000 \times (1.04)^{-10}\)
\(\displaystyle V = 1,000 \times 0.67556\)
\(\displaystyle V = 675.56\)

\(\displaystyle \frac{dV}{di} = -10 \times 1,000 \times (1+0.04)^{-(10+1)}\)
\(\displaystyle \frac{dV}{di} = -10,000 \times (1.04)^{-11}\)
\(\displaystyle \frac{dV}{di} = -10,000 \times 0.649581\)
\(\displaystyle \frac{dV}{di} = -6,495.81\)

\(\displaystyle \frac{d^{2}V}{di^{2}} = 10 \times (10+1) \times 1,000 \times (1+0.04)^{-(10+2)}\)
\(\displaystyle \frac{d^{2}V}{di^{2}} = 10 \times 11 \times 1,000 \times (1+0.04)^{-12}\)
\(\displaystyle \frac{d^{2}V}{di^{2}} = 110,000 \times 0.62459705\)
\(\displaystyle \frac{d^{2}V}{di^{2}} = 68705.68\)

\(\displaystyle C = V^{-1} \times \frac{d^{2}V}{di^{2}}\)
\(\displaystyle C = 675.56^{-1} \times 68705.68\)
\(\displaystyle C = 0.00148 \times 68705.68\)
\(\displaystyle C = 101.701 years^{2}\)