# Pricing a call option in the Black Scholes Market - calculation steps

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 $$1_{S_T > K} = 1_{Z > -d_+ + \sigma \sqrt T}$$.
• Going over this again, where does the $+ \sigma \sqrt{T}$ come from in the final line?
• The line before it has a term that is $+\frac{\sigma}{2} \sqrt T$ (if you split off the last term in the fraction), but $-d_+$ has a term with $-\frac{\sigma}{2} \sqrt T$, so we need to add back on $\sigma \sqrt T$ to make the lines the same. Commented Apr 23, 2022 at 17:56