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I have an expectation given as:

$\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$

where $K$ is just an arbitrary number (i.e. the strike price, but that's unimportant) and $S$ can be modelled by the equation $S_{t} = \exp((r-\frac{1}{2})t + \sigma W_{t})$. Also, the expectation is under the $\mathbb{P}$-measure, not the $\mathbb{Q}$-measure, so effectively this expectation is $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$

Now, when trying to evaluate $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right)$ under the $\mathbb{Q}$-measure, then solving this expectation is fairly easy, since you can integrate the SDE and take the $\log$ of $S$ to get $\log(S_{T}) = \log(S_{0}) + (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1)$ (since $W_{t}\approx \sqrt{T}N(0,1)$) and thus we have for $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = \mathbb{P}({S_{T}\geq K})$:

$\log(S_{T}) = (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1) > \log(K)$ and thus rearranging this equation gives

$N(0,1) > \frac{\log(K/S_{0}) - (r-\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$

and through a little bit more rearrangement (i.e. knowing that $N(0,1)<-x = 1 - N(x)$ we get finally $d_{2} = \frac{(r-\frac{1}{2}\sigma^{2})T + \log(S_{0}/K)}{\sigma\sqrt{T}}$ and thus $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = N(d_{2})$.

The problem I have now however is that I'm a bit unsure how to find $\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$, I honestly don't really know how to calculate it, so if someone could help me out I'd really appreciate it. Thanks in advance.

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    $\begingroup$ You also asked this question here. $\endgroup$ – Gordon Sep 12 '16 at 13:24
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You could take a look at the derivation of the Black-Scholes formula, in which a central part is computing the above expectation. Take a look here: Proof of the Black - Scholes pricing formula for European Call Option.

If you define $Z=\log S_T$ then $Z$ is a certain normally-distributed random variable (check yourself) with density $f(z)$, and you can then compute

$\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right) = \mathbb{E}\left(\exp(Z)\mathbb{1}_{Z\geq \log K} \right) = \int_{\log K}^\infty \, e^z \, f(z) \, \, \mathrm{d}z, $

which is done in a more general set-up in one of the answers in the above link. You should end up with (here $\Phi$ denotes the distribution function of a $\mathcal{N}(0,1)$-variable):

$\Phi\left(\frac{\log\frac{S_0}{K}+\left(r+\frac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}}\right)$.

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  • $\begingroup$ Thanks for that! I forgot to mention the important part which I was having difficulty - the thing about the expectation is that it's under the $\mathbb{P}$-measure, not the $\mathbb{Q}$-measure, so effectively the expectation is $\mathbb{E}^{\mathbb{P}}(S_{T}\mathbb{1}_{S_{T}\geq K})$, which is why I'm having so much difficulty with the problem (unfortunately I've never studied measure theory before). Will this change anything? I'll update my original post as well $\endgroup$ – ThePlowKing Sep 12 '16 at 0:39
  • $\begingroup$ The $Q$ measure would only change the mean of your $Z$ random variable and not the calculations. Just identify the parameters in the normal distribution of $Z $ and proceed as in the link. Contrary to what I wrote before, your result might differ sligtly due to different parameters. $\endgroup$ – Furrer Sep 12 '16 at 17:02

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