I’m studying the probability theory for finite sets (in my field we call them Random Finite Sets, but they are known as (simple) Point Processes). In this context, a random finite set is a random variable whose values are finite subsets of $\mathbb{R}^d$. Let’s say that a random finite set is composed by two random points $x_1$, $x_2$ of $\mathbb{R}^d$. Since sets are not ordered, the joint PDF $f(x_1,x_2)$ of $x_1$, $x_2$ must be symmetric, in the sense that $$f(x_1,x_2)=f(x_2,x_1)$$ must hold. Let’s suppose that $\tilde{f}(x_1,x_2)$ is a non-symmetric PDF, then we can define its symmetrization as $$f(x_1,x_2)\triangleq \frac{\tilde{f}(x_1,x_2)+\tilde{f}(x_2,x_1)}{2}.$$ This is the proposed solution by the theory.
Moreover, the theory explicitly forbids sets containing repeated elements, in the sense that it must satisfy $f(x_1,x_1)=0$ and $f(x_2,x_2)=0$.
Question
(1) In my opinion we can achieve symmetry even by considering the transformation $$f(x_1,x_2) = \tilde{f}(x_1,x_2)\tilde{f}(x_2,x_1).$$ Why don’t we do it that way? What are the disadvantages of this strategy?
From comments:
In general, your suggested alternative would usually not have the property $\iint f(x_1,x_2) =1$ needed to be a probability density. The proposed solution has this property provided that $\tilde{f}(x_1,x_2)$ does.
You could try $f(x_1,x_2) = \dfrac{\tilde{f}(x_1,x_2)\tilde{f}(x_2,x_1)}{\iint \tilde{f}(x_1,x_2)\tilde{f}(x_2,x_1)}$ so long as the denominator is finite and positive,
which is similar to the proposed solution being $ \frac{\tilde{f}(x_1,x_2)+\tilde{f}(x_2,x_1)}{\iint (\tilde{f}(x_1,x_2)+\tilde{f}(x_2,x_1))}$ where the denominator is clearly $2$.