Put trivariate PDF in terms of bivariate PDFs

Let there be the random variables $$X$$, $$Y$$, and $$Z$$. Let all the bivariate PDFs $$f_{X, Y}$$, $$f_{X, Z}$$, and $$f_{Y, Z}$$ be known.

Can we write the unknown trivariate PDF $$f_{X, Y, Z}$$ in terms of the known bivariate PDFs?

• No. Knowing the pairwise distributions between the variables is not sufficient to know the mutual distribution. – Graham Kemp May 22 at 3:19
• Ok, thank you for that. Might you please provide an informal proof by contradiction with some example? – Tyler Collins May 22 at 3:29

Here is an example, obtained by tweaking a 2D counterexample: Let $$(X,Y,Z)$$ be such that $$f_{X,Y,Z}(x,y,z)= \begin{cases} 2 \phi(x)\phi(y)\phi(z) & xyz>0\\ 0 & \text{otherwise} \end{cases}$$ where $$\phi$$ is the standard 1D Gaussian pdf. Then the bivariates $$f_{X,Y}, f_{Y,Z}, f_{Z,X}$$ are all standard 2D Gaussians, but of course $$(X,Y,Z)$$ is not Gaussian. Now both the standard 3D Gaussian and this $$f$$ give the same distribution of bivariates.
Let $$f(x,y,z)=1$$ when $$x,y,z\in [0,1]$$, and be zero otherwise. Then $$f_{X,Y}$$ is uniform on the unit square, and same for $$f_{X,Z}$$ and $$f_{Y,Z}$$. On the other hand, let $$g(x,y,z)=1+\sin(2\pi (x+y+z)) \hspace{1cm}\text{for }x,y,z\in [0,1]$$ Then $$g_{X,Y},g_{X,Z}$$ and $$g_{Y,Z}$$ are also uniform on the unit square.