The payoff function of a call is $f(S_T, K) = (S_T - K)^+$, so the expected payoff should allow me to value the price of this call.

$$ \mathbb{E}[f(S_T, K)] = \mathbb{E}[(S_T - K)^+] = \mathbb{E}[(S_T - K) \cdot \mathbb{1}(S_T - K > 0)] $$ $$ = e^{-rT} (S_T - K) \mathbb{E}[\mathbb{1}(S_T - K > 0)] $$ $$ = e^{-rT} (S_T - K) \mathbb{P}[(S_T - K > 0)] $$

Now the question is simplified to calculating the probability that $S_T$ would be greater than $K$. $$ \mathbb{P}[(S_T - K > 0)] = \mathbb{P}[(S_T > K)] $$ $$ = \int_K^{\infty} dx $$ where $dx = dS_T = \mu S_0 dt + \sigma S_0 dW$ $$ = \int_K^{\infty} \mu S_0 dt + \sigma S_0 dW $$

I do not think this is the correct way to go, and I would appreciate any input on this matter. Thanks.

  • $\begingroup$ The derivation of the BS formula is covered in any good textbook. I'm voting to close this question as it is too basic. $\endgroup$
    – Quantuple
    Commented Oct 24, 2016 at 9:46
  • $\begingroup$ @Quantuple Sorry about that. I've actually did my research online and shuffled through Hull and Shreve, but they did not have this way of proving B-S. I'm in a position where I can't really ask my colleagues for such rudimentary concepts. I'll delete the post after a several days once I have some answers to my other pending questions, if that is okay with you. $\endgroup$
    – Astaboom
    Commented Oct 24, 2016 at 18:01
  • $\begingroup$ Hmmm pretty weird. Did you see this SE question: quant.stackexchange.com/questions/19038/…. Seems exactly like what you are looking for to me. $\endgroup$
    – Quantuple
    Commented Oct 24, 2016 at 18:06

1 Answer 1


You know the solution to the SDE

\begin{equation} \mathrm{d}S_t = \mu S_t \mathrm{d}t + \sigma S_t \mathrm{d}W_t \end{equation}


\begin{equation} S_T = S_0 \exp \left\{ \left( \mu - \frac{1}{2} \sigma^2 \right) T + \sigma W_T \right\} \end{equation}

Now, $W_T \sim \mathcal{N}(0, T)$, so

\begin{equation} S_T > K \qquad \Leftrightarrow \qquad S_0 \exp \left\{ \left( \mu - \frac{1}{2} \sigma^2 \right) T + \sigma \sqrt{T} X \right\} > K, \end{equation}

where $X \sim \mathcal{N}(0, 1)$. Rearranging yields

\begin{equation} X > \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{K}{S_0} \right) - \left( \mu - \frac{1}{2} \sigma^2 \right) T \right) := \alpha \end{equation}

Now, $\mathbb{P} \left\{ X > \alpha \right\} = \mathbb{P} \left\{ X < -\alpha \right\}$ and thus $\mathbb{P} \left\{ S_T > K \right\} = \mathbb{P} \left\{ X < -\alpha \right\} = \mathcal{N}(-\alpha)$.

  • $\begingroup$ Thanks, this is exactly what I wanted. Though, I've worked through the steps myself now, but I am stuck with another problem. You see, the $d_1$ formula and the $d_2$ formula or $d_{\pm}$, is different by $\sigma \sqrt{T}$. Since we would just multiply $\mathbb{P}{[X > \alpha]}$ with $e^{-rT}(S_T - K)$, I cannot see where the difference originates. $\endgroup$
    – Astaboom
    Commented Oct 24, 2016 at 17:53
  • $\begingroup$ You have $\mathbb{P} \left\{ S_T > K \right\} = \mathcal{N} \left( d_- \right)$ and $\mathbb{E} \left[ S_T \mathrm{1} \left\{ S_T > K \right\} \right] = S_0 \mathcal{N} \left( d_+ \right)$. Try explicitly computing the expectation in the second expression and you'll get there. Also - search a bit, this is probably the most commonly asked Black/Scholes related question. $\endgroup$ Commented Oct 24, 2016 at 18:40

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