Finding the probability density function (pdf) for a two variable 'function' Asking rather basic question.
Say we have two parameters:
$x,y$ with a pdf $X(x),Y(y)$.
I want to  calculate the pdf for a third parameter say: $z$ which is defined by:
$z=f(x,y)$
What would the pdf function $Z(z)$ would look like?
Sorry if that's embarrassing
Best regards anyone who spends time on this :)
 A: Unfortunately there is no general method that can give you $Z$ from $X,Y$.
If there is a variable transformation $f$ that is bijective, and such, invertable, then there does exist a formula. However, since $z = f(x,y)$ reduces the dimensionality of the input space from $2$ to $1$, $f$ is likely not invertable.
If you don't mind, I'd like to change your notation for the p.d.f. of $x$ to $f_x$, so we can use $F_x$ for its c.d.f.
So the method is the following:
If $f:\mathbb{R}^n \to \mathbb{R}^n$, $f(x_1,\dots,x_n)=(y_1,\dots,y_n)$, with $x := (x_1,\dots,x_n)$ and $y := (y_1,\dots,y_n)$, and $x$ having joint p.d.f. $f_x$ and joint c.d.f. $F_x$, then $\forall i \in \{1,\dots,n\}$:
$$\frac{\partial}{\partial a_i}\mathbb{P}(y_1 \le a_1,\dots,y_n \le a_n) = \frac{\partial}{\partial a_i}\mathbb{P}(f(x) \le (a_1,\dots,a_n)) = \\ = \frac{\partial}{\partial a_i}\mathbb{P}(x \le f^{-1}(a_1,\dots,a_n)) = \frac{\partial}{\partial a_i} F_x(f^{-1}(a_1,\dots,a_n)) = \\ = f_x(f^{-1}(a_1,\dots,a_n))\cdot \left(\frac{\partial f^{-1}}{\partial a_i}\right)(a_1,\dots,a_n)$$
where the derivative of the inverse of $f$ is often called the Hessian matrix with elements
$$H_{i,j}(a_1,\dots,a_n) = \left(\left(\frac{\partial f^{-1}}{\partial a_i}\right)(a_1,\dots,a_n)\right)_j$$
So the p.d.f. of $y$ in this case is
$$f_y(a) = f_x(f^{-1}(a))[H_1(a) \cdots H_n(a)]$$
where $H_k(a)$ is the $k$'th row of the Hessian matrix evaluated at $a$.
