probability conditioned on a function Is it correct to write that a probability model is conditioned on a function?
For example if $X \sim \mathcal{N}(\mu,\sigma^2)$ then $p(x | \mu,\sigma^2) \propto \frac{1}{\sigma} \exp(-\frac{1}{2\sigma^2}(x-\mu)^2)$. if $\mu$ is generated from a function $f(\cdot)$ and a constant $c$, is it correct to write the probability in the form of $p(x|f(\cdot),c,\sigma^2)$
 A: Let me point out two things:
First, you are using the bayesian definition of probability, in which $\mu$ and $\sigma$ (or rather $M=\mu$ and $S=\sigma$, see below) are random variables. In frequentist setting you would write $p(x;\mu,\sigma)$ where $\mu$ and $\sigma$ are function parameters and not random variables.
Second, $p(x)$ is a shorthand for $p(X=x)$. It is helpful to write $p(x|\mu,\sigma^2)$ in full detail as $p(X=x|M=\mu,S=\sigma^2)$, where $M$ and $S$ are random variables with domain $M \in \mathbb{R}$ and $S \in \mathbb{R}^+$.
You ask whether $p(x|f(\cdot),c,\sigma^2)$ is a valid expression in bayesian probability. It is difficult to tell, because we are provided only the shorthand form without sufficient detail to reconstruct the full expression. 
Roughly, any expression of the form $p(E_1,\dots,E_n|F_1,\dots, F_m)$, where each $E_1,\dots,E_n$ and $F_1,\dots, F_m$ can be written in the form $X \bullet x$, where $X$ is a random variable, $\bullet$ is $=$, $\gt$,$\lt$, $\le$ or $\ge$ and $x$ is a constant; any such expression is valid.
$p(x|f(\cdot),c,\sigma^2)$ can be written as $p(X=x|M=f(\cdot),C=c,S=\sigma^2)$, which is valid as long as $f(\cdot)$ is not a random function. If  $f(\cdot)$ is a random function we may write $p(X=x|M-f(\cdot)=0,C=c,S=\sigma^2)$, which is valid. 
I interpret, 

$\mu$ is generated from a function $f(\cdot)$ and a constant $c$

as saying that $M=g(f(\cdot),c)$
Then $p(X=x|M-g(f(\cdot),c)=0, S=\sigma^2 )$ is valid.
