# Is this intuitive understanding of “averaging over all realizations of a noise” correct?

Consider a continuous random variable $$\eta(t)$$ such that $$\langle \eta(t)\rangle=0, \forall t$$ where the average $$\langle ...\rangle$$ is taken over all possible realizations of the noise at any time $$t$$. This means, at a given time $$t=t_0$$, the average of the values $$\eta(t_0)$$ that are intersections of the vertical line $$t=t_0$$ and different realizations of the noise in the $$\eta(t)$$-$$t$$ plane is zero.

Now, for a Gaussian random process, can we think that the intersection points to be Gaussian distributed at $$t=t_0$$ i.e. $$p(\eta(t_0))=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\eta^2(t_0)/2\sigma^2}?$$ If so, when we say that $$\eta(t)$$ is a Gaussian random process with zero mean $$\langle \eta(t)\rangle=0, \forall t$$, can we think of $$\langle\eta(t)\rangle$$ as $$\langle \eta(t)\rangle=\int \eta(t)p(\eta(t))=\int \eta(t)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\eta^2(t)/2\sigma^2}=0?$$

Yes. For a given $$t$$, $$\eta(t)$$ is drawn from a Gaussian distribution with zero mean and variance $$\sigma^2$$. You can compute mean values in the usual way, so indeed $$\langle \eta(t) \rangle = \int d \eta(t) p(\eta(t)) = 0$$, where $$p(\eta(t))$$ is a zero mean Gaussian with variance $$\sigma^2$$.
You certainly can have more interesting Gaussian processes where different points in time are correlated, or the variance changes, but I am assuming that the distribution you wrote down in the question is correct for your use case. The words for what you wrote would be "a stationary, white Gaussian process with variance $$\sigma^2$$".