I'll try to clear one possible source of confusion. If picking each sample value from a single distribution feels "not random enough", then let's try to make things "more random" by adding another layer of randomness. This will be found to be futile.
Imagine that for each sample the noise is random in the sense that it comes from a distribution that is randomly selected for that sample from a list of possible distributions, each with their own probability of occurrence and a list of probabilities for the possible sample values. Keeping it simple with just three distributions and four possible sample values:
$$\begin{array}{l|llll}&\rlap{\text{Sample value and its prob-}}\\
\text{Probability}&\rlap{\text{ability in the distribution}}\\
\text{of distribution}&-2&-1&0&1\\
\hline
\color{blue}{0.3}&0.4&0.2&0.3&0.1\\
\color{blue}{0.2}&0.5&0.1&0.2&0.2\\
\color{blue}{0.5}&0.1&0.4&0.4&0.1\end{array}$$
Here we have actually a distribution of distributions. But there is a single distribution that says everything about the probabilities of the values for that sample:
$$\begin{array}{llll}\rlap{\text{Sample value and}}\\
\rlap{\text{its total probability}}\\
-2&-1&0&1\\
\hline
0.27&0.28&0.33&0.12
\end{array}$$
The total probabilities were obtained as sums of conditional probabilities of the sample values over the possible distributions:
$$0.4\times\color{blue}{0.3} + 0.5\times\color{blue}{0.2} + 0.1\times\color{blue}{0.5} = 0.27\\
0.2\times\color{blue}{0.3} + 0.1\times\color{blue}{0.2} + 0.4\times\color{blue}{0.5} = 0.28\\
0.3\times\color{blue}{0.3} + 0.2\times\color{blue}{0.2} + 0.4\times\color{blue}{0.5} = 0.33\\
0.1\times\color{blue}{0.3} + 0.2\times\color{blue}{0.2} + 0.1\times\color{blue}{0.5} = 0.12$$
The laws of probability that were applied:
$$P(A_i\cap B_j) = P(A_i|B_j)\color{blue}{P(B_j)}\quad\text{conditional probability}$$
$$P(A_i) = \sum_jP(A_i\cap B_j)\quad\text{total probability}$$
where $A_i$ are the events of the $i\text{th}$ sample value occurring, and $B_j$ are mutually exclusive and exhaustive events of choosing the $j\text{th}$ distribution.
With continuous distributions, similar things would take place, because those can be modeled as discrete distributions in the limit that the number of possible events approaches infinity.