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Let $S_t$ be ageometric brownian motion with parameters $\sigma$ and $r$ and fix $T,K\in (0,\infty)$.

How can I show that: \begin{align} \mathbb{E}[e^{-rT}max\{(S_T-K),0\}] & = x\Phi(d_+(T-t,x))-e^{-r(T-t)}\Phi(d_+(T-t,x))\\ \mbox{where:} \, d_{\pm}(\tau,x)& :=\frac{ln(\frac{x}{K})+(r+\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}? \end{align}

I expect it to be a direct calculation but I am not seeing it; I don't want an exact step by step answer, just an outlien of how to do it (I want to figure it out myself).
Thanks :)

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  • Compute the integral $\int_{\mathbb{R}} (S_T-K,0)\phi(x)dx$ where $\phi$ is the standard normal density
  • Use the fact that we know the solution to $S_T$ (the discounted GBM solution)
  • Notice that you're integrating only over a positive domain (This will give tow integral with opposing signs)
  • Make the substitution $u:=d_{\pm}$ to convert each integral to $\int \phi(u)du = \Phi(u) = \Phi(d_{\pm})$
  • Add the two integral together and voila!

As you requested I'll leave the details to you; goodluck! $\overset{..}{\smile}$

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  • $\begingroup$ Perfect, thanks for the hint-type answer I was looking for :) $\endgroup$ – Mr Library Guy Feb 27 '16 at 4:51
  • $\begingroup$ My pleasure Mr Library Guy & goodluck $\endgroup$ – AIM_BLB Feb 27 '16 at 4:52

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