dstefan
03-22-2008, 01:11 PM
Chapter 1, Problem 6:
Let K, T, \sigma and r be positive constants, and define the function g : {\mathbb R} \to {\mathbb R} as
g(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~\int_{0}^{b(x)} e^{-\frac{y^2}{2}} ~dy , where b(x) ~=~ \frac{ \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T}{\sigma \sqrt{T}} .
Compute g'(x) .
Solution:
g'(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~e^{-\frac{(b(x))^2}{2}} \cdot b'(x) ~=~ \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, and define the function g : {\mathbb R} \to {\mathbb R} as
g(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~\int_{0}^{b(x)} e^{-\frac{y^2}{2}} ~dy , where b(x) ~=~ \frac{ \ln\left(\frac{x}{K}\right) ~+~ \left( r + \frac{\sigma^2}{2} \right) T}{\sigma \sqrt{T}} .
Compute g'(x) .
Solution:
g'(x) ~=~ \frac{1}{\sqrt{2 \pi}} ~e^{-\frac{(b(x))^2}{2}} \cdot b'(x) ~=~ \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}}