# Proof of the Black - Scholes pricing formula for European Call Option

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 !!

• 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? – Chinny84 Aug 17 '14 at 11:26
• 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, – Teodor Fredriksson Aug 17 '14 at 12:17
• Tried the substitution, still nothing... I will try to find that book you mentioned. – Teodor Fredriksson Aug 17 '14 at 12:31
• 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! – Chinny84 Aug 17 '14 at 12:42
• It seems i figured it out :) – Teodor Fredriksson Aug 17 '14 at 16:39

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 !