Considering the following expression, with $x, y, z, c \in \mathbb{R}$, is such expression separable into $f(x, y, z) = g(x, z)h(y, z)$?
$c(x + y)^2 e^{yz}$
If not, why isn't it separable?
Note:
- $c$ is a constant, but I'm not sure if I can remove it, though.
Adding context: the original question concerns proving whether or not a given expression represents the function of probability distribution of two independent random variables $X$ and $Y$, conditioned to $Z$ -- $f(x, y \mid z)$.
This may also be written as $\mathbb{P} \models X \bot Y \mid Z$.
Here's what I've been able to do so far:
$c(x + y)^2 e^{yz} = $
$ = c[x^2 + 2xy + y^2]e^{yz} = $
$ = [x^2 + 2xy]ce^{yz} + cy^2 e^{yz}$
Using $t = xy$:
$c\left[x^2 + 2t\right] e^{\left(\frac{tz}{x}\right)} + c\left(\frac{t}{x}\right)^2 e^{\left(\frac{tz}{x}\right)}$
From here I really don't know what else to do.
Naturally we assume $c \ne 0$, otherwise the zero map is clearly seperable. Suppose $c(x+y)^2 e^{yz}=g(x,z)h(y,z).$ Define $k(y,z)=h(y,z)/[c e^{yz}].$ This is OK since $e^{yz}>0$ for all $y,z$. Then we have $$[1]\ \ (x+y)^2=g(x,z)k(y,z).$$ But this is not possible since the mixed partials $\partial^2/\partial x \partial y$ of the two sides are now $2$ for the left side and $0$ for the right side.
Note that [1] can also be shown impossible without partials: Let $z$ be constant and put $u(x)=g(x,z)$ and $v(y)=k(y,z).$ Then [1] becomes $$(x+y)^2=u(x)v(y).$$ The zero set of the left side is the line $y=-x$, whereas if the zero sets of $u,v$ are $A,B$ respectively, the zero set of the right side is $A \times B$, and no matter what the sets $A,B$ are one cannot express the line $y=-x$ as the cartesian product $A \times B.$