Why is transient behaviour of a stochastic system not captured in the frequency domain? Consider a stochastic differential equation:
$$
dx(t) = a x(t)dt + b y(t)dt \quad (1)
$$
where $y(t)$ is a stochastic process satisfying $\langle y(t)y(t')\rangle = \delta(t-t')$. We will assume that the first statistical moments are zero and that $a<0$. I am interested in the variance $\langle x(t)^2 \rangle$. When I solve this equation in the frequency domain, I obtain the value $\langle x(t)^2 \rangle = \frac{-b^2}{2a}$. This is the steady state variance, given by $\lim_{t\rightarrow \infty} \langle x(t)^2 \rangle$ but does not give any information about the transient variance $\langle x(t)^2 \rangle$ at $t\neq \infty$.
My question is: why does solving this equation in the frequency domain not capture the transient behavior of the variance?
More mathematical details are below:
My question is: why does solving in the frequency domain give the steady state variance, and not a function which depends on $t$?
We can be solve in the time domain to give:
$$
\langle x(t)^2 \rangle = e^{2at}\langle x(0)^2 \rangle - \frac{b^2}{2a}(1-e^{2at})
$$
The steady state can then be found by taking $t \rightarrow \infty$.
We can also solve in the frequency domain. Fourier transforming $(1)$ gives:
$$
x(\omega) = \frac{b}{-a-i\omega} y(\omega) \quad (2)
$$
where $y(\omega)$ is the Fourier transform of $y(t)$ and now obeys $\langle y(\omega)y(\omega')\rangle = \delta(\omega + \omega')$. We then invert the Fourier transform to find the variance:
$$
\langle x(t)^2 \rangle = \frac{1}{2 \pi} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}\langle x(\omega)x(\omega')\rangle^{-it(\omega + \omega')} d\omega d\omega' \quad (3)
$$
Plugging $(2)$ into $(3)$ and integrating over the delta function gives a value of $\frac{-b^2}{2a}$, which is the steady state variance obtained earlier, but does not contain any information about the transient behavior. Why is this?
 A: If we define for some $-\infty<t_1<t_2<\infty$
$$\hat X_{\omega}^{(t_1,t_2)} = \int_{t_1}^{t_2}X_te^{i\omega t}dt,$$
we can use Itô's lemma to write
$$ d(X_te^{i\omega t})=i\omega X_te^{i\omega t}dt+e^{i\omega t}dX_t$$
and hence
$$i\omega \hat X_{\omega}^{(t_1,t_2)}=\left[e^{i\omega t} X_t\right]_{t_1}^{t_2}-\int_{t_1}^{t_2}e^{i\omega t}dX_t.$$
So far we have not used the SDE. If we now substitute $dX_t=aX_tdt+bdW_t$, we obtain
$$i\omega \hat X_{\omega}^{(t_1,t_2)}=\left[e^{i\omega t} X_t\right]_{t_1}^{t_2}-a\hat X_{\omega}-b\hat W_{\omega}^{(t_1,t_2)},$$
where $\hat W_{\omega}^{(t_1,t_2)}=\int_{t_1}^{t_2}e^{i\omega t}dW_t.$
Therefore,
$$\hat X_{\omega}^{(t_1,t_2)}=\frac{\left[e^{i\omega t} X_t\right]_{t_1}^{t_2}-b\hat W_{\omega}^{(t_1,t_2)}}{a+i\omega},$$
I'm not sure the limits $t_{1,2}\to\mp\infty$ make sense but if you set $t_1=0$ and compute your double integral for the variance it is possible that the $X_{t_2}$ cancels out (if you use the SDE to express it in terms of $X_{t_1}$) and you get the transient. I did not do the full calculation so let me know what comes out.
