How to compute $ \mathop{\mathbb{E^{}}}\left\lbrace 1_{S_T > K} \; S_T \right\rbrace $ ?


$ dS_t = S_t r dt + \sigma dW_t $


$ 1_{S_T > K} $ is the indicator function being one when the condition is satisfied.

I would try

$ \mathop{\mathbb{E^{}}}\left\lbrace 1_{S_T > K} \; S_T \right\rbrace = \int_{S_T > K}^{\infty} n(\varepsilon) S_T d\varepsilon $


$ S_T = S_t e^{r(T-t)} + \sigma e^{rT} \int_t^T e^{-rs} dW_s $

but I cannot handle the `expectation-integral' because $S_T$ is a sum --- and of course a lack of knowledge in general. My experience is only with making a GBM into a standard normal .

  • $\begingroup$ See quant.stackexchange.com/questions/11577/…. $\endgroup$ – Gordon Dec 8 '15 at 0:27
  • $\begingroup$ @Gordon In that case the SDE is a GBM, so I do not see how it applies here? $\endgroup$ – GuestNo3829297 Dec 8 '15 at 0:50
  • $\begingroup$ oops. I miss read your question. will come back later. $\endgroup$ – Gordon Dec 8 '15 at 3:13
  • $\begingroup$ Did your question there generate any content? Please ask a mod there to handle it as I can find it. Please also make an account, that makes it easier to manage for us mods and also earns you privileges. $\endgroup$ – Bob Jansen Dec 8 '15 at 10:59

Since \begin{align*} S_T = S_0 e^{rT} + \sigma e^{rT} \int_0^T e^{-rs} dW_s, \end{align*} $S_T$ is normal with mean \begin{align*} a &=S_0 e^{rT}, \end{align*} and variance \begin{align*} b^2 &= \sigma^2 e^{2rT} \int_0^T e^{-2rs} ds\\ &=\frac{\sigma^2}{2r} \left(e^{2rT} - 1 \right). \end{align*} That is, $S_T = a + b\, \xi$, where $\xi$ is a standard normal random variable. Consequently, \begin{align*} \mathbb{E}\left(1_{S_T > K} S_T \right) &= \int_{-\infty}^{\infty} 1_{a + b\, x> K} (a + b \, x) \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx\\ &=\int_{\frac{K-a}{b}}^{\infty}(a + b \, x) \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx\\ &= a N\left(\frac{a-K}{b}\right) + b \int_{\frac{K-a}{b}}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx\\ &=a N\left(\frac{a-K}{b}\right) + \frac{b}{\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{K-a}{b} \right)^2}, \end{align*} where $N$ is the cumulative distribution function of a standard normal random variable.


To solve this, you need to use the property of Radon-Nikodym derivative $(L)$, which states:
$E[L.X] = E[X]$ under the new measure (where X can be your indicator function).

Next, to convert S(t) to RN derivative, do:
$S(t) = S(0) * exp(rt) * E\left(\frac{S(T)}{exp(rt)S(0)}\right)$
$S(t) = S(0) * exp(rt) * L$
as $E\left(\frac{S(T)}{exp(rt)S(0)}\right)$ can be used as a RN derivative.

RN derivative helps you move into the new measure, with expectation of just the indicator function (which is nothing but the probability of S(T) > K). To get this probability you need to find the process of S(t) under this new measure, which can be obtained by replacing $dW$ with $\sigma dt + dW$ (this comes from Girsanov theorem). This will give you the answer.

Watch this video for a more complete explanation (https://www.youtube.com/watch?v=W8YG5O1GGjE from 30min mark)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.