I have following exercise:

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This is what I did:

\begin{align} C(K)&= e^{-r\tau} \mathbb{E}^\mathbb{Q}[((S_T - K)^+] \\ &= e^{-r\tau}\mathbb{E}^\mathbb{Q}[((S_T - K)\mathbb{1}_{S_T>K}] \\ &=e^{-r\tau}\mathbb{E}^\mathbb{Q}[S_T \mathbb{1}_{S_T>K}]-Ke^{-r\tau}\mathbb{E}^\mathbb{Q}[\mathbb{1}_{S_T>K}] \end{align}

by$ \mathbb{1}$ I mean indicator function. Now I understand that $$ \mathbb{E}^\mathbb{Q}[\mathbb{1}_{S_T>K}] = \mathbb{Q}(S_T>K) $$

however I don't know how to deal with $$\mathbb{E}^\mathbb{Q}[S_T \mathbb{1}_{S_T>K}]$$

is this the right way how to solve this exercise?

  • $\begingroup$ This is simply an integration for S(T) > K. The density function is log-normal. You'll need to substitute the density function into the equation and do the integration. It's not very hard. $\endgroup$
    – SmallChess
    Commented Jul 28, 2016 at 12:32
  • 2
    $\begingroup$ Indeed. You can also do it the "hard way" through a change of numéraire (or Bayes' rule for conditional expectations). The latter option requires a good command of stochastic calculus though, while the approach proposed by @Student T only requires a good command of standard calculus. Also, this question is tackled in any good textbook (see the steps here: valonews.com/?p=612, although text is in french) and is a little bit basic for this forum. $\endgroup$
    – Quantuple
    Commented Jul 28, 2016 at 12:39
  • 2
    $\begingroup$ For such question, there are a lot of discussions on this site based on either a brute force approach, or a numeraire change approach. By the way, should $S_t\pmb{1}_{S_T>K}$ be $S_T\pmb{1}_{S_T>K}$? $\endgroup$
    – Gordon
    Commented Jul 28, 2016 at 13:10
  • 2
    $\begingroup$ For example, see this and this. You can search the tag black-scholes. $\endgroup$
    – Gordon
    Commented Jul 28, 2016 at 13:17
  • $\begingroup$ For the last term I think the question wants you to switch to the stock numeraire and then it equals S(t) times the expectation of the indicator under the stock measure. (Apologies I can't write symbols on my phone ) $\endgroup$
    – dm63
    Commented Jul 28, 2016 at 13:24


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