$\begingroup$

I want to prove the following

The price of a European call option with strike price $K$ and time of maturity $T$ is given by the formula $\Pi(t) = F(t,S(t))$, where $$F(t,s) = sN[d_1(t,s)]-e^{-r(T-t)}KN[d_2(t,s)]$$ $$d_1(t,s) = \frac{1}{\sigma\sqrt{T-t}}\left[\ln \frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)\right]$$ $$d_2(t,s) = d_1(t,s) - \sigma\sqrt{T-t} $$ and $$N(x) = \int^{\infty}_{-\infty} e^{-\frac{x^2}{2}}\,dx,\,\,X\sim N(0,1)$$ I've gotten this far... $$S(T) = s\exp\left[\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right)\right]$$ and we define $$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right) $$ with \begin{align*} \mathbb{E}\left(Z\right) & = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) \\ \mbox{Var}\left(Z\right) & = \sigma^2\left(T-t\right) \end{align*} and so \begin{align*} Z\sim N\left(\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right),\sigma^2\left(T-t\right)\right) \end{align*} then \begin{align*} F(t,s) & = e^{-r(T-t)}\mathbb{E}^Q\left(\Phi\left(S(T)\right)\right) \\ \,\, & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(S(T)\right)f(z)\,dz \\ \,\; & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(se^z\right)f(z)\,dz \end{align*} with the case of an European option we put $\Phi(x) = \max\left[x-K,0\right]$ and then \begin{align*} F(t,s) & = e^{-r(T-t)}\int^\infty_{-\infty} \max\left[se^z-K,0\right]f(z)\,dz \\ \,\, & = e^{-r(T-t)}\left(\int^{\ln \frac{K}{s}}_{-\infty} 0\cdot f(z)\,dz + \int^{\infty}_{\ln\frac{K}{s}} \left(se^z-K\,f(z)\right)\,dz\right) \\ \,\, & = e^{-r(T-t)}\int^{\infty}_{\ln\frac{K}{s}}\left(se^z-K\right)\,f(z)\,dz \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^ze^{-\frac{z^2}{2}}\,dz -K\int^{\infty}_{\ln\frac{K}{s}}e^{-\frac{z^2}{2}}\,dz \right) \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^ze^{-\frac{z^2}{2}}\,dz -KN\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{z-\frac{z^2}{2}}\,dz -KN\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(z-1\right)^2+\frac{1}{2}}\,dz-KN\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\ \,\, & = e^{-r(T-t)} \left(se^{\frac{1}{2}}\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(z-1\right)^2}\,dz-KN\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \end{align*} Were do I go from here ?? I can't solve the remaining integral !!

$\endgroup$ 6

2 Answers

$\begingroup$

Now lets solve the above $SDE$. This is just a GBM with solution $$S(T) = s\exp\left[\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right)\right]$$ and we define $$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right) $$ $$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}Y,\,\,Y\sim N(0,1)$$ with \begin{align*} \mathbb{E}\left(Z\right) & = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) \\ \mbox{Var}\left(Z\right) & = \sigma^2\left(T-t\right) \end{align*} and so \begin{align*} Z\sim N\left(\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right),\sigma^2\left(T-t\right)\right) \end{align*} then \begin{align*} F(t,s) & = e^{-r(T-t)}\mathbb{E}^Q\left(\Phi\left(S(T)\right)\right) \\ \,\, & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(S(T)\right)f(z)\,dz \\ \,\; & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(se^z\right)f(z)\,dz \end{align*} with the case of an European option we get have $\Phi(x) = \max\left[x-K,0\right]$ hen \begin{align*} F(t,s) & = e^{-r(T-t)}\int^\infty_{-\infty} \max\left[se^z-K,0\right]f(z)\,dz \\ \,\, & = e^{-r(T-t)}\left(\int^{\ln \frac{K}{s}}_{-\infty} 0\cdot f(z)\,dz + \int^{\infty}_{\ln\frac{K}{s}} \left(se^z-K\right)\,f(z)\,dz\right) \\ \,\, & = e^{-r(T-t)}\int^{\infty}_{\ln\frac{K}{s}}\left(se^z-K\right)\,f(z)\,dz \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\ \,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\ \,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}y}e^{-\frac{y^2}{2}}\,dy -K\int^{\infty}_{\ln\frac{K}{s}}e^{-\frac{z^2}{2}}\,dz \right) \\ \,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}y-\frac{y^2}{2}}\,dy\right) -Ke^{-r(T-t)}\Phi\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}} e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}\left(s\int^{\infty}_{\ln\frac{K}{s}} e^{\sigma\sqrt{T-t}y-\frac{y^2}{2}}\,dy\right) -Ke^{-r(T-t)}\Phi\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y^2-2\sigma\sqrt{T-t} y+\sigma^2\left(T-t\right)\right)}e^{\frac{1}{2}\sigma^2\left(T-t\right)}\,dy\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2+\frac{1}{2}\sigma^2\left(T-t\right)}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}e^{\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\ & = \frac{1}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = s\Phi\left(-\frac{\ln\frac{K}{s}-\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}+\sigma\sqrt{T-t}\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\ \,\, & = s\Phi\left(\frac{\ln\frac{s}{K}+\left(r+\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \end{align*} and thats it !

$\endgroup$ 1 $\begingroup$

I think the American case is out of scope for the book I'm using. One has to know how to solve optimization problems to derive the pricing formula for American options, This i will be able to do after i've read a course on in next semester. :)

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy