Distribution of integral wrt. to a Levy process Assume that a stochastic process is given by
$X_{t} = \int_0^t e^{-k(t-s)}dY_{s}$
where $Y_{s}$ is a Levy process. 
Is there any way I can use the knowledge about the Levy measure of $Y_{t}$ in order to determine the expectation and variance of $X_t$?
 A: For the sake of simplicity, we assume that $\mathbb{E}(Y_1^2)<\infty$. It is not difficult to see that
$$\tilde{Y}_t := Y_t-t \mathbb{E}Y_1$$
defines an $L^2$-martingale. In particular,
$$X_t = \mathbb{E}Y_1 \int_0^t e^{-k(t-s)} \, ds + \underbrace{\int_0^t e^{-k(t-s)} d\tilde{Y}_s}_{=:I_t}.$$
Now, as $(\tilde{Y}_t)_{t \geq 0}$ is a martingale we know that $(I_t)_{t \geq 0}$ is a martingale. Hence,
$$\mathbb{E}X_t = \mathbb{E}Y_1 \int_0^t e^{-k(t-s)} \, ds= \frac{\mathbb{E}Y_1}{k} (1-e^{-kt}).$$
This shows that $\text{var} X_t = \mathbb{E}(I_t^2)$. Using Itô's isometry, we see that
$$\mathbb{E}(I_t^2) = \mathbb{E} \left( \int_0^t e^{-2k(t-s)} \, d\langle \tilde{Y} \rangle_s \right)$$
where $\langle \tilde{Y} \rangle$ denotes the quadratic variation of the martingale $(\tilde{Y}_t)_{t \geq 0}$. Since $(Y_t)_{t  \geq 0}$ is a Lévy process, the quadratic variation can be expressed in terms of the Lévy triplet $(\gamma,\sigma^2,\nu)$; I leave this to you.
Remark The process $(X_t)_{t \geq 0}$ is a (particular case of) an Ornstein-Uhlenbeck process driven by the Lévy process $(Y_t)_{t \geq 0}$.
