dstefan
06-10-2008, 04:31 PM
Chapter 3, Problem 12
Show that the price of a plain vanilla European call option is a convex function of the strike of the option, i.e., show that \frac{\partial^2 C}{\partial K^2} ~\geq~ 0
Solution:
Recall that
S e^{-q(T-t)} ~N'(d_1) ~=~ K e^{-r(T-t)} ~N'(d_2)
By differentiating the Black-Scholes formula C = S e^{-q(T-t)} N(d_1) - K e^{-r(T-t)} N(d_2) with respect to K, we obtain that
\frac{\partial C}{\partial K} = S e^{-q(T-t)} N'(d_1) \frac{\partial d_1}{\partial K} ~-~ K e^{-r(T-t)} N'(d_2) \frac{\partial d_2}{\partial K} ~-~ e^{-r(T-t)} N(d_2)
= S e^{-q(T-t)} N'(d_1) ~\left( \frac{\partial d_1}{\partial K} - \frac{\partial d_2}{\partial K} ~\right) ~-~ e^{-r(T-t)} N(d_2) = - e^{-r(T-t)} N(d_2)
since d_1 = d_2 + \sigma \sqrt{T-t} and therefore \frac{\partial d_1}{\partial K} ~=~ \frac{\partial d_2}{\partial K}
By differentiating once more with respect to K, we find that
\frac{\partial^2 C}{\partial K^2} = - e^{-r(T-t)} ~N'(d_2) ~\frac{\partial d_2}{\partial K} = - e^{-r(T-t)} ~\frac{1}{\sqrt{2 \pi}} ~e^{\frac{-d_2^2}{2}} \left( - \frac{1}{\sigma K \sqrt{T-t}} \right)
= \frac{1}{\sigma K \sqrt{2 \pi (T-t)}} e^{-r(T-t)} ~e^{\frac{-d_2^2}{2}} ~\geq~ 0;
recall that
d_2 = \frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}} = - \frac{\ln(K)}{\sigma \sqrt{T-t}} ~+~ \frac{\ln(S) + \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}}
Show that the price of a plain vanilla European call option is a convex function of the strike of the option, i.e., show that \frac{\partial^2 C}{\partial K^2} ~\geq~ 0
Solution:
Recall that
S e^{-q(T-t)} ~N'(d_1) ~=~ K e^{-r(T-t)} ~N'(d_2)
By differentiating the Black-Scholes formula C = S e^{-q(T-t)} N(d_1) - K e^{-r(T-t)} N(d_2) with respect to K, we obtain that
\frac{\partial C}{\partial K} = S e^{-q(T-t)} N'(d_1) \frac{\partial d_1}{\partial K} ~-~ K e^{-r(T-t)} N'(d_2) \frac{\partial d_2}{\partial K} ~-~ e^{-r(T-t)} N(d_2)
= S e^{-q(T-t)} N'(d_1) ~\left( \frac{\partial d_1}{\partial K} - \frac{\partial d_2}{\partial K} ~\right) ~-~ e^{-r(T-t)} N(d_2) = - e^{-r(T-t)} N(d_2)
since d_1 = d_2 + \sigma \sqrt{T-t} and therefore \frac{\partial d_1}{\partial K} ~=~ \frac{\partial d_2}{\partial K}
By differentiating once more with respect to K, we find that
\frac{\partial^2 C}{\partial K^2} = - e^{-r(T-t)} ~N'(d_2) ~\frac{\partial d_2}{\partial K} = - e^{-r(T-t)} ~\frac{1}{\sqrt{2 \pi}} ~e^{\frac{-d_2^2}{2}} \left( - \frac{1}{\sigma K \sqrt{T-t}} \right)
= \frac{1}{\sigma K \sqrt{2 \pi (T-t)}} e^{-r(T-t)} ~e^{\frac{-d_2^2}{2}} ~\geq~ 0;
recall that
d_2 = \frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}} = - \frac{\ln(K)}{\sigma \sqrt{T-t}} ~+~ \frac{\ln(S) + \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}}