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For N variables $X_1,X_2,\ldots,X_N$, the PDFs of all $X_i$s are given. Can we prove a joint distribution always exists?

What if some consistent PDFs of combinations of these variables are given, i.e., one PDF does not contradict with another? By "consistency" and "contradiction", I meant you cannot produce different margins from any two given PDFs. For example, one can calculate the marginal distribution of $X_1$ from the marginal distribution of $X_1,X_2$. If the marginal distribution of $X_1,X_2$ on $X_1$ is different from the given PDF of $X_1$, then there is an inconsistency (contradiction).

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  • $\begingroup$ Hint: can you answer the question with the added assumption that $X_1,\ldots, X_n$ are independent? $\endgroup$ – Jonathan Y. Jan 7 '16 at 8:10
  • $\begingroup$ @JonathanY. Thanks for pointing it out. See my update. $\endgroup$ – xuhdev Jan 7 '16 at 8:20
  • $\begingroup$ I'm sorry, I don't follow the clarification. What do we mean by one PDF doesn't contradict with another? $\endgroup$ – Jonathan Y. Jan 7 '16 at 8:22
  • $\begingroup$ @JonathanY. Please see my update. $\endgroup$ – xuhdev Jan 7 '16 at 8:32
  • $\begingroup$ Let me rephrase, and tell me if I've got your question correctly: Suppose that for some $J_1,\ldots,J_k\subseteq \{X_1,\ldots,X_n\}$ we're given joint PDF's $F_{J_i}$, $i=1,\ldots, k$, such that whenever $J\subset J_i\cap J_j$, the marginal PDF $F_J$ derived from $F_{J_i}$ and $F_{J_j}$ is the same. Then can we make a consistent choice for $F_{X_1,\ldots,X_n}$? $\endgroup$ – Jonathan Y. Jan 7 '16 at 8:36
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Given $\mathcal{X}=\{X_1,\ldots,X_n\}$ and some collection $J_1,\ldots,J_k\in \mathcal{P}(\mathcal{X})$, with PDF's $F_{J_i}$ for all $i=1,\ldots,k$, such that whenever $J\subset J_i\cap J_j$ we have that $F_{J_i}$ and $F_{J_j}$ induce the same marginal PDF $F_J$, we don't necessarily have $F_{\mathcal{X}}$ inducing $F_{J_i}$ for all $i=1,\ldots,k$.

Consider, e.g., $X,Y,Z$ where $$F_{X,Y}(x,y) = \begin{cases}0 & \min(x,y)<0\\ \frac{1}{2} & 0\leq\min(x,y)<1\\ 1 & \min(x,y)\geq 1\end{cases}$$ (which amounts to having $X\stackrel{a.s.}{=}Y\sim B(0.5)$), $$F_{Y,Z}(y,z) = \begin{cases}0 & \min(y,z)<0\\ \frac{1}{2} & 0\leq\min(y,z)<1\\ 1 & \min(y,z)\geq 1\end{cases}$$ ($Y\stackrel{a.s.}{=}Z\sim B(0.5)$) and $$F_{X,Z}(x,z) = \begin{cases}0 & \min(x,z)<0 \vee \max(x,z)<1\\ \frac{1}{2} & 0\leq\min(x,z)<1\leq\max(x,z)\\ 1 & \min(x,z)\geq 1\end{cases}$$ ($(1-X)\stackrel{a.s.}{=}Z\sim B(0.5)$).

Then from any pair of PDF's we consistently get $$F_X(t) = F_Y(t) = F_Z(t) = \begin{cases}0 & t<0\\ \frac{1}{2} & 0\leq t<1\\ 1 & t\geq 1\end{cases}$$ but no joint PDF $F_{X,Y,Z}$ generates all three marginal distribution functions (since that would imply $X\stackrel{a.s.}{=}(1-X)$).


However, if the maximal elements of $J_1,\ldots,J_k$ (w.r.t. inclusion) are all pairwise-disjoint, then we can construct $F_{\mathcal{X}}$ by assuming independence.

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Yes. One can produce such a joint distribution by imposing the additional condition of independence.

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  • $\begingroup$ Sorry I really messed up my question... I've updated the question, which is the one I'm interested in. $\endgroup$ – xuhdev Jan 7 '16 at 8:20

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