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The partial expectation $\mathbb{E}(X;_{X>K})$ for an alpha-stable distributed random variable:

By playing with convolutions of Characteristic Functions of alpha-Stable distributions $S(\alpha, \beta, \mu, \sigma)$ and a payoff $K$, assuming $\mu=0$ and deriving under the integral sign, found the partial expectation $\mathbb{E}(X;_{X>K})$ , i.e., where F(x) is the distribution function of X, $\int_K^\infty x\, \mathrm{d}F(x)$ (which is not to be confused with the conditional expectation).

I am ending up with a difficult integral (easy to evaluate numerically but hard to get explicitly). With $1<\alpha\leq 2$:

$$\psi (\alpha, \beta, \sigma, K) = \frac{1}{2 \pi }\int_{-\infty }^{\infty } \alpha \sigma ^{\alpha } \left| u\right| ^{\alpha -2} \left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right) \text{sgn}(u)\right) \exp \left(\left| u \sigma \right| ^{\alpha } \left(-1-i \beta \tan \left(\frac{\pi \alpha }{2}\right) \text{sgn}(u)\right)+i K u\right) du$$

The solutions is easy for $K=0$, so $$\psi(\alpha,\beta,\sigma,0)=-\sigma\frac{\Gamma \left(-\frac{1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }\right)}{\pi \alpha } .$$ Also, there is a well known solution for symmetric cases in Zolotarev's book. But it is the $K \ne 0$ that is critical.

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    $\begingroup$ One never knows. I was able to explicit form for the PDF when $\beta=\pm 1$ and we can get explicit expectations for $\mu=0$. $\endgroup$ – Nero Dec 25 '14 at 11:52
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    $\begingroup$ Alas, $K=0$ so far. But I can play by shifting $\mu$ and $K$. $\endgroup$ – Nero Dec 25 '14 at 11:58
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    $\begingroup$ Sorry but my first comment seems to fully apply to your question as it is stated. If you have a restricted setting in mind (such as K=0, or some other), please mention it. $\endgroup$ – Did Dec 25 '14 at 12:00
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    $\begingroup$ Figured out an explicit solution for the general case where $\mu=0$ and $K=0$, cases for which we do not have explicit PDF and added to question above. $\endgroup$ – Nero Dec 25 '14 at 12:58
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    $\begingroup$ Hello, Nero. Comments can only be edited for 5 minutes, then they are locked. Of course, you can delete comments anytime you wish. Next, based on this, may we infer that you found a solution? If so, please consider posting it as an answer. Such things are allowed, actually encouraged! It is always nice to see you here. $\endgroup$ – Ellie Kesselman Dec 30 '14 at 1:00
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Finally, the semi-explicit (but quite useable) answer for $K \neq 0$, which I jot down quickly before coming back with clean derivations and proofs.

$$\psi[\alpha, \beta,\sigma,K]=\sigma \frac{\Gamma \left(\frac{\alpha -1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }\right)}{2 \pi }+\sum _{k=1}^\infty \frac{i^k K^k \Gamma \left(\frac{k+\alpha -1}{\alpha }\right) \left(\beta ^2 \tan ^2\left(\frac{\pi \alpha }{2}\right)+1\right)^{\frac{1-k}{\alpha }} \left((-1)^k \left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{\frac{k-1}{\alpha }}+\left(1-i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{\frac{k-1}{\alpha }}\right)}{2 \pi \sigma^{k-1} k!}$$

Summing to $p$, we get an approximation, with $p=3$ that seems to beat numerical integration for $K$ values "at-the-money", i.e., not far from the center).

The idea is to get rid of $\exp(i K u)$ which caused trouble because of oscillations around the center.

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