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

where

$ dS_t = S_t r dt + \sigma dW_t $

and

$ 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 $

with

$ 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 .

GuestNo3829297
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  • See http://quant.stackexchange.com/questions/11577/how-to-use-the-stock-as-a-numeraire-to-price-a-derivative-with-payoff-of-the-for/21872#21872. – Gordon Dec 08 '15 at 00:27
  • oops. I miss read your question. will come back later. – Gordon Dec 08 '15 at 03:13
  • @Gordon In that case the SDE is a GBM, so I do not see how it applies here? – GuestNo3829297 Dec 08 '15 at 00:50
  • 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. – Bob Jansen Dec 08 '15 at 10:59

2 Answers2

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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)

Uditg_ucla
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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.

Gordon
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