I am working on computing the price of a standard European call option under a Black-Scholes market.
Using knowledge of the payoff, I can split the calculation into:
$ e^{-rT}(E[S_t] \mathbb{1}_{S_T > K} - KE[\mathbb{1}_{S_T > K}] $
for which the expected value of the indicator function is simply:
$P[S_T >K]$
Using the formula for $S_T$ as a Brownian motion I can derive a $Z = \frac{lnS_T - (lnS_0 +r - \frac{\sigma^2}{2})T)}{\sigma \sqrt{T}} $ ~ $ N(0, 1)$
and thus the calculation for the probability (second term) leads to $N(d_\_)$
However, I am struggling to compute the first term. Doing
$E[S_T \mathbb{1}_{S_T >K}] = E [S_o exp(r-\frac{\sigma^2}{2})T + \sigma W_T) \mathbb{1}_{S_T > K}]$, in my notes leads to:
$E[S_o exp(r-\frac{\sigma^2}{2})T + \sigma \sqrt{T}Z) \mathbb{1}_{Z> -d_+ + \sigma \sqrt{T}}]$, which I can not follow where it comes from?
It comes from simply rewriting the expression $S_T > K$. The following inequalities are all equivalent: \begin{align*} S_T &> K \\ S_0 \exp\left( \left(r-\frac{\sigma^2}{2}\right)T + \sigma \sqrt{T}Z\right) &> K \\ \left(r-\frac{\sigma^2}{2}\right)T + \sigma \sqrt{T}Z &> \ln\left(\frac{K}{S_0}\right) \\ Z &> \frac{\ln (K/S_0) - \left(r-\frac{\sigma^2}{2}\right)T}{\sigma \sqrt T} \\ Z &> -\left( \frac{\ln (S_0/K) + \left(r+\frac{\sigma^2}{2}\right)T}{\sigma \sqrt T}\right) + \sigma \sqrt T. \end{align*} Since $$d+ = \frac{\ln (S_0/K) + \left(r+\frac{\sigma^2}{2}\right)T}{\sigma \sqrt T},$$ we have that $1_{S_T > K} = 1_{Z > -d_+ + \sigma \sqrt T}$.