I have this process:
$dx_t = -\frac{k}{2}x_t \, dt + \frac{\beta}{2} \, dz_t$
and must prove it's normally distributed with first two moments:
$\mu = e^{-\frac{1}{2}kt}x_0$
$\sigma^2 = \frac{\beta^2}{4k}(1-e^{-kt})$
I tried to multiply $x_t$ by $e^{kt}$ and apply Ito's Lemma to this 'product process' in order to eventually recover back $x_t$ by taking exponentials.
The normality is straightfoward; the variance is ok but the mean isn't since I'm left with an integral whose integrand includes $x_t$ and I'm stuck.
I don't know whether I made some mistakes or adopted the wrong approach since the beginning.
Since the stochastic integral has zero expectation, we have
$$\mathbb{E}(X_t)-\mathbb{E}(X_0) = -\frac{k}{2} \mathbb{E} \left( \int_0^t X_s \right) \, ds.$$
Applying Fubini's theorem, we find that $\varphi(t) := \mathbb{E}(X_t)$ satisfies
$$\varphi(t)-\varphi(0) = -\frac{k}{2} \int_0^t \varphi(s) \, ds,$$
i.e. it solves the ODE
$$\varphi'(t) = - \frac{k}{2} \varphi(t), \qquad \varphi(0) = \mathbb{E}(X_0).$$
It is well-known that the unique solution of this ODE is given by
$$\varphi(t) = \mathbb{E}(X_0) \cdot e^{-\frac{k}{2} t}.$$