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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 !!

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  • $\begingroup$ change of variables to $z\rightarrow z-1$ but I have solved for the European by not transforming to Z first. I went through Baxter and Rennie (last night in fact) and one of the problems was this exact problem. Do you have baxter and rennie? $\endgroup$ – Chinny84 Aug 17 '14 at 11:26
  • $\begingroup$ Actually i tried that subsituiton, although come to think of it I might have forgotten to change z = ln K/s to z = ln K/s + 1, $\endgroup$ – Teodor Fredriksson Aug 17 '14 at 12:17
  • $\begingroup$ Tried the substitution, still nothing... I will try to find that book you mentioned. $\endgroup$ – Teodor Fredriksson Aug 17 '14 at 12:31
  • $\begingroup$ I was going to go through the derivation, but I think it is such a wonderful (if not laborious) process that I think if you dig out any decent financial mathematics text book it will enlighten you at the steps you went wrong. Good luck! $\endgroup$ – Chinny84 Aug 17 '14 at 12:42
  • $\begingroup$ It seems i figured it out :) $\endgroup$ – Teodor Fredriksson Aug 17 '14 at 16:39
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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 !

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  • $\begingroup$ Nicely done! Now onto the American ;). Once you gone through this process once the put, and other pay offs are a variation on a theme right (I hope). So what is this in aid of anyway? Course or interview? $\endgroup$ – Chinny84 Aug 19 '14 at 20:01
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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. :)

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