Drift of stochastic process from initial known value from known spectral density Consider a process $\phi$ with know spectral density $S_\phi(\omega)$.
Suppose at time $t=0$ the process takes on a known value $\phi(0)$.
Ho does one compute the time dependent variance referenced to this intial value?
In other words, how does one compute $\langle (\phi(0) - \phi(\tau))^2 \rangle$ for $\tau > 0$?
If the answer depends on more than just the spectral density, please just state whatever additional assumptions are needed to make the problem well defined.
This question is motivated by trying to understand how much the phase of a microwave oscillator will drift over a given amount of time, so I think we should assume that the process is not stationary, i.e. the phase can drift arbitrarily far away from an initial value as time goes on.
The real underlying question here is how one goes from Figure 10 in this datasheet, which gives the phase noise spectral density, to the time dependent drift of the oscillator's phase.
 A: Note on conventions: In this answer, the symbol $S(\omega)$ refers to a single-sided spectral density.
In other words, $\int_0^\infty S(\omega) d\omega/(2\pi)$ is the total power in the process.
It's easier to work in terms of the spectral density of $\dot{\phi}$, which we denote $S_\dot{\phi}(\omega)$.
Note that $S_\dot{\phi}(\omega) = \omega^2 S_\phi(\omega)$.
We can write a particular realization of the process $\phi$ as
$$\phi(\tau) = \int_0^\tau \dot{\phi}(t) \, dt \, ,$$
so then
$$\langle \phi(\tau)^2 \rangle = \int_0^\tau \int_0^\tau \left \langle \dot{\phi}(t') \dot{\phi}(t'') \right \rangle \, dt' dt'' \, .$$
Using the Wiener-Khinchin theorem we can replace
$$ \left \langle \dot{\phi}(t') \dot{\phi}(t'') \right \rangle = \int_0^\infty S_\dot{\phi}(\omega) \cos \left(\omega \left( t' - t'' \right) \right) \frac{d\omega}{2\pi} \, ,$$
giving
\begin{align}
\left \langle \phi(\tau)^2 \right \rangle
&=\int_0^\infty S_\dot{\phi}(\omega) \frac{d\omega}{2\pi} \int_0^\tau \int_0^\tau \cos \left( \omega \left( t' - t'' \right) \right) \, dt' dt'' \\
&= \tau^2 \int_0^\infty S_\dot{\phi}(\omega) \left( \frac{\sin \left( \omega \tau / 2 \right)}{\left( \omega \tau / 2 \right)} \right)^2 \frac{d \omega}{2\pi}
\end{align}
which is what we wanted to find if we assume $\phi(0)=0$.
