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I recently asked a question about pairwise versus mutual independence (also related to this and this q).

However,

(1) I inadvertently used incorrect terminology:

three events, A, B, C are mutually independent when:

P[A,B]=P[A]P[B], P[B,C]=P[B]P[C], P[A,C]=P[A]P[C], P[A,B,C]=P[A]P[B]P[C]

Did and others pointed out that

"Mutual independence means the four identities you copied, pairwise independence means the first three of these identities." -- Did

Note that the term mutual has varying definitions across math. For example, mutual information is a pairwise relation.

(2) Going back to probability, GC Rota said the theory can be approached two ways: focusing on random variables (event algebra) or focusing on distributions. Here I am interested in distributions, where independence can be interpreted as factorization of the probability distribution function. The conditions are the same as above, where P is interpreted as the PDF function.

The following graphic based on a standard example from Counterexamples in Probability and Statistics of a 3-dimensional binomial PDF that factorizes pairwise (ie, each of the 3 pairs of random variables are independent and the 2-dim joint distributions can all be written as the product of the respective marginals) but not 3-way independent (the joint distribution cannot be written as the product of the individual marginal distributions)

enter image description here

My question is whether the opposite can happen, ie if the 3-dim (or perhaps higher) joint distribution factorizes into the 1-dim marginals, does that imply the pairwise factorization of all 2-dim joint distributions into the marginals?

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alancalvitti
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  • Could you perhaps phrase the question in terms of PDFs? At the moment this sounds a lot like asking: if three variables $X_1,X_2,X_3$ are mutually independent, then are $X_1$ and $X_2$ independent? – Colin McQuillan Feb 03 '13 at 01:14
  • @ColinMcQuillan, It's not just about the independence of $X_1$ and $X_2$ but all pairs formed from the set ${X_1,X_2,X_3}$. Or do you have an issue with the R.V. versus distributional definition? Also, as I state above, mutual statistical independence includes independence of the pairs not just the joint, so I am avoiding that definition to focus on the 3-way to 2-way implication, which is unambiguous. – alancalvitti Feb 03 '13 at 02:23
  • Got something from an answer below? – Did Feb 09 '13 at 10:25
  • Yes, thank you Did - both your and Dilip's answers are good, and although his displays the marginalization directly, I accepted yours because it is more general of the two. – alancalvitti Feb 09 '13 at 21:00

2 Answers2

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Indeed, assume that $\mu$ is the product of the probability measures $\mu_1$, $\mu_2$ and $\mu_3$, hence, for every $(A_1,A_2,A_3)$, $\mu(A_1\times A_2\times A_3)=\mu_1(A_1)\cdot\mu_2(A_2)\cdot\mu_3(A_3)$.

Then, for example, the 2-marginal distribution $\nu$ corresponding to the two first coordinates is such that $\nu(A)=\mu(A\times\mathbb R)$ for every 2-dimensional $A$.

One sees that, for every $(A_1,A_2)$, $\nu(A_1\times A_2)=\mu(A_1\times A_2\times\mathbb R)=\mu_1(A_1)\cdot\mu_2(A_2)$, that is, that $\nu$ is indeed the product of $\mu_1$ and $\mu_2$.

Did
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If the joint density of $X$,$Y$, and $Z$ is $f_{X,Y,Z}(x,y,z)$, then for any pair, say $Y$ and $Z$, $$f_{Y,Z}(y,z) = \int_{-\infty}^{\infty}f_{X,Y,Z}(x,y,z)\,\mathrm dx.$$ Now suppose that $f_{X,Y,Z}(x,y,z) = f_X(x)f_Y(y)f_Z(z)$ for all real numbers $x,y$ and $z$. Substitution in the integral gives $$f_{Y,Z}(y,z) = \int_{-\infty}^{\infty}f_X(x)f_Y(y)f_Z(z) \mathrm dx = f_Y(y)f_Z(z)\int_{-\infty}^{\infty}f_X(x)\mathrm dx = f_Y(y)f_Z(z).$$ I will leave it to you to verify that a similar result holds for the other two pairs of random variables.

Dilip Sarwate
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