Black–Scholes but probably basic stats

Question

Hello friends! I'm rusty (bad) with my statistics and this problem is bugging me, so any help would be greatly appreciated!

Just really bad at figuring out how the 1-N() gets transformed into the 2nd line.

EDIT: sorry I got too excited and didn't ask about part (d). Any explanation on where the first 2 equations (St) come from?

Answer 1

eltigrechino's answer is correct after the correction by Ian. That is, if your distribution is normally distributed ~N($\mu, \sigma$) then it is a property of normal distributions that 1 - N(x) = N(-x). In particular, using your expression, we have: x = $\frac{lnK - lnS_0 - (\mu - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$ Note that $lnK - lnS_0 = ln(\frac{K}{S_0})$ then -x = $\frac{-lnK + lnS_0 + (\mu - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$ = $\frac{ln(\frac{S_0}{K}) + (\mu - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$. Since this quantity is $d_2$ in the Black-Scholes model, we see that P($S_T$ > K) = N($d_2$).

To get the first two equations in (d) note that we may write $ln(S_T) = ln(S_0) + (\mu - \frac{\sigma^2}{2})T + \sigma\sqrt{T}Z$ where Z ~ N(0, 1) since this will be normally distributed with the apprpriate mean and variance. Exponentiating both sides gives you the result.

Answer 2

The normal distribution is symmetric about its mean. Therefore you can flip the CDF calculation as follows

$1 - \Phi(x) = \Phi(-x) $ where $\Phi$ represents the CDF of a $N(\mu,\sigma)$ distribution.

Edit: whoops forgot to flip the sign, thanks