理解這個積分的解
以下積分錶示幾何布朗運動的期望值 $ S_T>K $ （即 Black-Scholes 看漲期權價格的一部分）：
$$ \int_{z^} (S_te^{\mu\tau-\frac{1}{2}\sigma^2\tau+\sigma\sqrt{\tau}z})\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =e^{\mu\tau}S_tN(d_1^) $$ 和 $ z^=\frac{\ln\frac{K}{S_t}-(\mu-\sigma^2/2)\tau}{\sigma\sqrt{\tau}} $ , $ d_1^=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}} $ ， 和 $ N(\cdot) $ 累積標準正態分佈。 你能解釋一下這個平等是如何獲得的嗎？
讓 $ \tau = T-t $ . 然後
$$ \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, Z}, \end{align*} $$ 在哪裡 $ Z $ 是標準正態隨機變數，獨立於 $ \mathcal{F}t $ . 而且， $$ \begin{align*} E\left(S_T 1{{S_T >K}}\mid \mathcal{F}t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, Z}, 1{\left{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, Z} >K\right}}\mid \mathcal{F}t \right)\ &=\int{-\infty}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, z}, 1_{\left{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, z} >K\right}} \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\ &=\int_{z^}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, z}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\ &=\int_{z^}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}, z - \frac{z^2}{2}}dz\ &=\int_{z^}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau -\frac{1}{2}(z-\sigma \sqrt{\tau})^2 +\frac{1}{2}\sigma^2 \tau }dz\ &=S_t e^{u\tau} \int_{z^ - \sigma \sqrt{\tau}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\ &=S_t e^{u\tau}N(\sigma \sqrt{\tau}-z^)\ &=S_t e^{u\tau}N(d_1^), \end{align*} $$ 在哪裡 $$ \begin{align*} d_1^* &= \sigma \sqrt{\tau}-z^\ &=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}. \end{align} $$