Quiero probar los siguientes
El precio de un europeo llaman opción con strike price $K$ y tiempo de madurez $T$ está dada por la fórmula $\Pi(t) = F(t,S(t))$, donde $$F(t,s) = sN[d_1(t,s)]-e^{-r(T-t)}KN[d_2(t,s)]$ $ $$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} $ $ y $$N(x) = \int^{\infty}_{-\infty} e^{-\frac{x^2}{2}}\,dx,\,\,X\sim N(0,1)$ $ he conseguido este acuerdo... $$S(T) = s\exp\left[\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right)\right]$ $ y definir $$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right) $ $ con\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*} así \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*} entonces\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*} wi ¿to el caso de una opción europea ponemos $\Phi(x) = \max\left[x-K,0\right]$ y luego\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*} fueron voy de aquí? ¡Yo no puedo resolver la integral restante!
