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!

  • $\begingroup$ 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? $\endgroup$ – gunes Mar 24 '19 at 17:08
  • $\begingroup$ @gunes I was trying to bring in the supports of the pdfs involved. But I guess you're right. $\endgroup$ – msd15213 Mar 24 '19 at 17:13
  • $\begingroup$ @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? $\endgroup$ – msd15213 Mar 24 '19 at 17:14
  • $\begingroup$ Having $f_X(x)$, $f_Y(y)$ (the marginals) doesn't determine the joint, if I understand you correctly. $\endgroup$ – gunes Mar 24 '19 at 17:19
  • $\begingroup$ 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. $\endgroup$ – whuber Mar 24 '19 at 18:20

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