Generating probability distribution function from continuous data We have a continuous function $F(x,y)$ defined on a bounded domain $(x, y) \in [0, L_x] \times [0, L_y]$. Suppose the function $F$ (the explicit form is irrelevant here) is defined such that $F(x,y)$ always lies between, say, 0 and 1.
By varying $x$ and $y$ continuously over the finite region, we obtain a set of corresponding (continuous) values for $F$. Now, I would like to find the PDF of the values assumed by $F$ itself. In other words, what I am looking for is the probability $P\, [F = \zeta;\,\, 0 \le \zeta \le 1]$ as a function of $\zeta$ .
This answer suggests a numerical prescription for such a situation as follows:

Way of generate a PDF from discrete / continuous data:

*

*Find a continuous equation that models the collected data, let say normal distribution equation


*Calculate the parameters required in the equation from the collected data.For example, parameters for normal distribution equation are mean and standard deviation. Calculate them from collected data


*Based on the parameters, plot the equation with continuous x-value --> that is called PDF

However, I was wondering if there is a method to obtain the PDF analytically, given that we know the exact functional form of $F$.
I would greatly appreciate any help in this regard. Many thanks!
 A: $F(x,y) = \zeta \Longrightarrow P(X<x, Y<y) = \zeta.$
If you want the joint pdf, you take
$$\frac{\partial^2}{\partial x\,\partial y}F(x,y) = f(x,y),$$
if I understand your question correctly.

Allow me to try again. 
After some thought, I realized that it might be the case that what you call $F(x,y)$ is something I would call $h(x,y)$. $F(x,y)$ is generally understood to mean the joint CDF. I think then that to get the joint pdf, you want $$f_{X,Y}(x,y) = \frac{P(X\in dx, Y\in dy)}{dx\,dy},$$ where $P(X\in dx, Y\in dy)$ is to be determined using $h(x,y)$.
A: General method to find pdf: 1) find cdf, 2) take derivative...
e.g. we can do this for the standard normal.  Let:
$N$ = pdf for standard normal
$N^{-1}$ = the inverse of the standard normal on $(0,\infty)$
Then:
$P(N(x) < y) = P(x<-N^{-1}(y)) + P(x>N^{-1}(y))$
This second probability is easy to compute, if we know the distribution for $x$.  E.g. if we choose an $x$ uniform on [0,1] and ask about the distribution of N(x):
$P(x>N^{-1}(y)) = 0$ if $y < N(1)$ or $y > N(0)$
$P(x>N^{-1}(y)) = N^{-1}(y)$ else
so $pdf(N) = \frac{d}{dy} N^{-1}(y)$
there's no simple formula, really... More examples here: https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables
