# Result linked to Black-Scholes evaluation

Why does this

$$Se^{-D(T-t)}e^{-d_1^2/2} - Ee^{-r(T-t)}e^{-d_2^2/2}$$

equal to $0$? (Where $E$ is a strike)

Note that this question is similar to Verifying an identity of an equation for Black Scholes formula.

You need to use the fact that \begin{align*} d_1 &= \frac{\ln (S/E) + (r-D)(T-t) + \frac{\sigma^2}{2}(T-t)}{\sigma \sqrt{T-t}}, \\ d_2 &= d_1 - \sigma \sqrt{T-t}. \end{align*} Then, \begin{align*} E e^{-r(T-t)} e^{-d_2^2/2} &=E e^{-r(T-t)} e^{-d_1^2/2 - \frac{\sigma^2}{2}(T-t)+d_1 \sigma \sqrt{T-t}}\\ &=E e^{-r(T-t)} e^{-d_1^2/2 - \frac{\sigma^2}{2}(T-t)+\ln (S/E) + (r-D)(T-t) + \frac{\sigma^2}{2}(T-t)}\\ &=Se^{-D(T-t)} e^{-d_1^2/2}. \end{align*}