# Delta derivation from the expectation

I'm trying to understand the following transformation leading to Delta

$\frac{dC}{dx} = e^{-r\tau} \mathbb{E}[ \frac{\partial}{\partial x}\text{max}(xY-K,0)] = e^{-r\tau} \mathbb{E}[Y \textbf{1}(xY>K)] = e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}[e^{-\sigma\sqrt{\tau}Z} \textbf{1}(Z>-d_2)] = \Phi(d_1)$

I get the first part, but I don't understand the last transformation.

$Y = e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, Z is Normal(0,1)

x - current stock price

• I have changed $\Phi(d_2)$ to $\Phi(d_1)$ in the last equality. – Gordon May 6 '15 at 18:10

Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}\big[e^{-\sigma\sqrt{\tau}Z}\mathbb{1}_{\{Z > -d_2\}} \big] \\ &= e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}\big[e^{\sigma\sqrt{\tau}Z}\mathbb{1}_{\{Z < d_2\}} \big]\\ &= \int_{-\infty}^{d_2} e^{-\frac{\sigma^2}{2}\tau + \sigma\sqrt{\tau} x} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_2} e^{-\frac{\sigma^2}{2}\tau + \sigma\sqrt{\tau} x -\frac{x^2}{2}}dx\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_2 + \sigma\sqrt{\tau}}e^{-\frac{x^2}{2}}dx\\ &= \Phi(d_1), \end{align*} where $d_1 = d_2 + \sigma\sqrt{\tau}$.
• @rollerboller, you can treat $xY$ as the stock price, however, $Z$ is standard normal. – Gordon May 7 '15 at 12:53