dstefan
03-22-2008, 01:30 PM
Chapter 8, Problem 4:
Recall that finding the implied volatility from the given price of a call option is equivalent to solving the nonlinear problem f(x) = 0, where
f(x) ~=~ S e^{-q T} N(d_1(x)) - K e^{-r T} N(d_2(x)) ~-~ C
and
d_1(x) ~=~ \frac{\ln\left(\frac{S}{K}\right) ~+~ \left(r-q+\frac{x^2}{2}\right)T}{x \sqrt{T}} , d_2(x) ~=~ \frac{\ln\left(\frac{S}{K}\right) ~+~ \left(r-q-\frac{x^2}{2}\right)T}{x \sqrt{T}} .
(i) Show that \lim_{x \to \infty} d_1(x) = \infty and \lim_{x \to \infty} d_2(x) = -\infty , and conclude that \lim_{x \to \infty} f(x) ~=~ S e^{-q T} ~-~ C .
(ii) Show that
\lim_{x \searrow 0} d_1(x) ~=~ \lim_{x \searrow 0} d_2(x)~=~ \left\{ \begin{array}{cl}-\infty, & \mbox{if}~~ S e^{(r-q)T} ~<~ K; \\ 0, & \mbox{if}~~ S e^{(r-q)T} ~=~ K; \\ \infty, & \mbox{if}~~ S e^{(r-q)T} ~>~ K. \end{array} \right.
(Recall that F = S e^{(r-q) T} is the forward price.)
Conclude that
\lim_{x \searrow 0} f(x) ~=~ \left\{ \begin{array}{cl}-C, & \mbox{if}~~ S e^{(r-q)T} ~\leq~ K; \\ S e^{-q T} - K e^{-r T} ~-~ C, & \mbox{if}~~ S e^{(r-q)T} ~> ~ K \\ \end{array} \right.
(iii) Show that f(x) is a strictly increasing function and
\begin{array}{rccclcl}-C &<& f(x) &<& S e^{-q T} - C, & \mbox{if} & S e^{(r-q)T} ~\leq~ K; \\ S e^{-q T} - K e^{-r T} - C &<& f(x) &<& S e^{-q T} - C, & \mbox{if} & S e^{(r-q)T} ~>~ K. \end{array}
(iv) For what range of call option values does the problem f(x) = 0 have a positive solution?
Solution:
a
Recall that finding the implied volatility from the given price of a call option is equivalent to solving the nonlinear problem f(x) = 0, where
f(x) ~=~ S e^{-q T} N(d_1(x)) - K e^{-r T} N(d_2(x)) ~-~ C
and
d_1(x) ~=~ \frac{\ln\left(\frac{S}{K}\right) ~+~ \left(r-q+\frac{x^2}{2}\right)T}{x \sqrt{T}} , d_2(x) ~=~ \frac{\ln\left(\frac{S}{K}\right) ~+~ \left(r-q-\frac{x^2}{2}\right)T}{x \sqrt{T}} .
(i) Show that \lim_{x \to \infty} d_1(x) = \infty and \lim_{x \to \infty} d_2(x) = -\infty , and conclude that \lim_{x \to \infty} f(x) ~=~ S e^{-q T} ~-~ C .
(ii) Show that
\lim_{x \searrow 0} d_1(x) ~=~ \lim_{x \searrow 0} d_2(x)~=~ \left\{ \begin{array}{cl}-\infty, & \mbox{if}~~ S e^{(r-q)T} ~<~ K; \\ 0, & \mbox{if}~~ S e^{(r-q)T} ~=~ K; \\ \infty, & \mbox{if}~~ S e^{(r-q)T} ~>~ K. \end{array} \right.
(Recall that F = S e^{(r-q) T} is the forward price.)
Conclude that
\lim_{x \searrow 0} f(x) ~=~ \left\{ \begin{array}{cl}-C, & \mbox{if}~~ S e^{(r-q)T} ~\leq~ K; \\ S e^{-q T} - K e^{-r T} ~-~ C, & \mbox{if}~~ S e^{(r-q)T} ~> ~ K \\ \end{array} \right.
(iii) Show that f(x) is a strictly increasing function and
\begin{array}{rccclcl}-C &<& f(x) &<& S e^{-q T} - C, & \mbox{if} & S e^{(r-q)T} ~\leq~ K; \\ S e^{-q T} - K e^{-r T} - C &<& f(x) &<& S e^{-q T} - C, & \mbox{if} & S e^{(r-q)T} ~>~ K. \end{array}
(iv) For what range of call option values does the problem f(x) = 0 have a positive solution?
Solution:
a