dstefan
06-06-2008, 12:03 AM
Chapter 2, Problem 5
Let K, T, \sigma and r be positive constants.
Define the function g : {\mathbb R} \to {\mathbb R} as
g(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~\int_{-\infty}^{b(x)} e^{-\frac{y^2}{2}} ~dy,
where
b(x) ~=~ \left( \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T \right) / \left(\sigma \sqrt{T}\right).
Compute g'(x).
Solution:
g'(x) = \frac{1}{\sqrt{2 \pi}} b'(x) e^{-\frac{(b(x))^2}{2}} = \frac{1}{x \sigma \sqrt{2 \pi T}} e^{-\frac{\left( \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T \right)^2}{2 \sigma^2 T}}
Let K, T, \sigma and r be positive constants.
Define the function g : {\mathbb R} \to {\mathbb R} as
g(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~\int_{-\infty}^{b(x)} e^{-\frac{y^2}{2}} ~dy,
where
b(x) ~=~ \left( \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T \right) / \left(\sigma \sqrt{T}\right).
Compute g'(x).
Solution:
g'(x) = \frac{1}{\sqrt{2 \pi}} b'(x) e^{-\frac{(b(x))^2}{2}} = \frac{1}{x \sigma \sqrt{2 \pi T}} e^{-\frac{\left( \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T \right)^2}{2 \sigma^2 T}}