# Support of joint pdf being the Cartesian product of supports of the one-dimensional conditionals

Note: This is a homework problem so please DO NOT PROVIDE A COMPLETE SOLUTION. Some gentle hints would be good.

Suppose $$X_1,\, \ldots ,\, X_p$$ have a $$p$$-variate joint pdf whose support is the Cartesian product of the supports of the one-dimensional conditionals, show that the joint pdf is uniquely determined from the one-dimensional conditionals.

My Attempt

Without loss of generality, let $$p=2$$ and call the two random elements $$X$$ and $$Y$$. Let $$\text{supp}\, f_Y (y) = \{ y_0 \leq y \leq y_2\}$$ and $$\text{supp}\, f_{X\vert Y} (x) = \{ x_0 \leq x \leq x_2\}$$ ($$x_0,\, x_2$$ may be dependent on $$Y$$). Then by assumption, $$\text{supp}\, f_{X,\, Y} (x, y) = \{ (x,\,y): x_0 \leq x \leq x_2,\,y_0 \leq y \leq y_2 \}$$.

For any $$(x_1,\, y_1)\in \text{supp}\, f_{X,\, Y}(x,\, y)$$, \begin{align} F_{X,\, Y} (x_1,\, y_1) &= \int_{x_0}^{x_1} \int_{y_0}^{y_1} f_{X,\,Y}(x,\, y) dy\, dx\\ &= \int_{y_0}^{y_1} \int_{x_0}^{x_1} f_{X\vert Y}(x)dx f_Y(y) dy. \end{align} Now, $$f_{X,\,Y}(x,\, y)$$ is obtained by differentiating $$F_{X,\,Y}(x,\, y)$$ but from the above equality, we know that for any $$(x_1,\, y_1)$$ in the support of $$f_{X,\,Y}(x,\, y)$$, $$F_{X,\,Y}(x_1,\, y_1)$$ is uniquely determined by $$f_{X\vert Y}(x)$$ and $$f_Y(y)$$. Therefore, $$f_{X,\,Y}(x,\, y)$$ is uniquely determined by the one-dimensional conditionals.

My Question

I don't feel like the above argument is enough because it seems too trivial to me. So I must have missed something? Could anyone tell me what that is please? Many thanks in advance!

• Simply, for $f_{X,Y}(x,y)$ to be non-zero, both multiplicands in its expansion must be non-zero, i.e. $f_Y(y)$ and $f_{X|Y}(x)$. So, why did you prefer using joint CDF? – gunes Mar 24 '19 at 17:08
• @gunes I was trying to bring in the supports of the pdfs involved. But I guess you're right. – msd15213 Mar 24 '19 at 17:13
• @gunes $f_{X,\, Y}(x,\, y) = f_{X\vert Y}(x) f_Y(y)$ alone does not guarantee that the joint is uniquely determined by the conditionals, right? – msd15213 Mar 24 '19 at 17:14
• Having $f_X(x)$, $f_Y(y)$ (the marginals) doesn't determine the joint, if I understand you correctly. – gunes Mar 24 '19 at 17:19
• In the title, by "one-dimensional conditionals" you must mean the $p$ marginals, right? Otherwise you're asking about an uncountable Cartesian product, which makes little sense in this context. By the way, the whole interest in this question lies in the case $p\gt 2:$ when $p=2$ it is indeed trivial. You should therefore rethink the justification for your "wlg" assertion at the outset. – whuber Mar 24 '19 at 18:20