Let us consider a random variable $X^{(n)}$ which depends on a deterministic quantity $n$. Then $P(X^{(n)}=x)=f(x,n).$
Now let us set the $n$ to random variable $N$. Can one say generally that $P(X^{(N)}=x|N=n)=f(n)$? Which conditions have to be fulfilled for that?
I saw it a few times that some authors write $P(X^{(N)}=x)=f(x,N)$. That does not make sense imo as probability is deterministic, not random. So I thought maybe they mean $P(X^{(N)}=x|N)=f(N)$.
It feels like what you makes sense. $P(X^{(N)}=x)=f(N)$ is valid, and in this statement $N$ is random, so $f(N)$ is some random density (but I think it would be better still to write $f(x,N)$). However if you condition on a particular setting of $N$ it becomes a deterministic density function, so that you now arrive at, $P(X^{(N)}=x|N=n)=f(n)$ (and I would still prefer to write $f(x,n))$. The conditioning on a particular setting of $N$ is important.
This is not a very clear statement. The evaluation of a probability mass function at a particular point is not random, but a probability measure is meant to help encode a sense of randomness via the random variable you define.
This does not make sense to me. As I said before you need to condition on an actual value. You need to be given a value, hence you should condition at $N= n $. A event must have occurred.
I think it could be insightful to understand the difference between the parameterisation of a density, and its density dummy variable used to define the density. What is a common example is something like: $\mathcal{N}(x;\mu,\sigma^2)$, where $x$ is density variable, and the parameterisation are non-random quantities. Under your notation this could be expressed as, $P(X^{(\mu,\sigma^2)}\leq x) = \int_{-\infty}^x \mathcal{N}(x;\mu,\sigma^2) dx$, where $X\sim\mathcal{N}(x;\mu,\sigma^2)$, since remember the probability for a continuous R.V. requires an integration over the pdf.