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I'm struggling to find the example asked above.

  • Provide an example of a distribution P(X1,X2,X3) where for each i ≠ j, we have that (Xi ⊥ Xj) ∈ I(P), but we also have that (X1,X2 ⊥ X3) ∉ I(P). Where I(P) is a set of independencies that hold in P.

What actually does the state (X1,X2 ⊥ X3) mean? I guess it is something like "The joint distribution of X1 and X2 is independent of X3" is that correct? Can you guys give me an example of a joint P(X1,X2,X3) that satisfies those statements above? (Note that (X1,X2 ⊥ X3) DOESN'T belong to I(P))

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Assuming that "(X1,X2 ⊥ X3)" (which makes no sense to me) is a typo for (X1,X2 | X3), and translating the rest of your jargon from an unspecified source as asking

"Find an example of three random variables $X,Y,Z$ that are pairwise independent but for which $X$ and $Y$ are NOT conditionally independent given $Z$",

see this answer of mine for an example of three Bernoulli random variables that have this property as well as an example of three normal random variables that that have this property.

A different example can be found in this answer on math.SE where $(X,Y,Z)$ is uniformly distributed on four of the eight identical subcubes into which we can partition the cube $[0,1]^3$. The subcubes are

  • $[0,\frac 12]^3$ and $[\frac 12,1]^2\times[0,\frac 12]$ which sit on the $z=0$ plane.

  • $[0,\frac 12]\times [\frac 12,1]^2$ and $[\frac 12,1]\times[0,\frac 12]\times[\frac 12,1]$ which sit on the $z=\frac 12$ plane.

In the answer on math.SE, it is shown that $X, Y, Z$ are pairwise independent $\mathcal U([0,1]$ random variables, that is, $$f_{X,Y}(x,y) = f_{X}(x)f_{Y}(y) = \begin{cases}1 & 0 \leq x, y \leq 1,\\0, & \text{otherwise}\end{cases}.$$ and similarly for the other two pairs. In particular, note that the pairwise independence of $X$ and $Z$ means that $f_{X\mid Z}(x\mid z) = f_X(x)$ and similarly, we have that $f_{Y\mid Z}(y\mid z) = f_Y(y)$. But, given that $Z = z \in [0,\frac 12]$, $(X,Y)$ is uniformly distributed on the set $$\left(\left[0,\frac 12\right]\times\left[0,\frac 12\right]\right) \bigcup\left(\left[\frac 12,1\right]\times\left[\frac 12,1\right]\right).$$ That is, $$f_{X,Y\mid Z}(x,y\mid z) = \begin{cases}2,& 0 \leq x,y \leq \frac 12,\\ 2,& \frac 12 \leq x,y \leq 1,\\ 0, &\text{otherwise},\end{cases} \quad \neq f_{X\mid Z}(x\mid z)f_{Y\mid Z}(y\mid z)$$

showing that $X$ and $Y$ are NOT conditionally independent for any given value of $Z$ in $[0,\frac 12)$, and a similar calculation holds when the value of $Z$ is in $(\frac 12, 1]$.

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  • $\begingroup$ Your explanation seems Ok, but I'm a bit confused with your translation, I'm not sure if the statement "give Z", translates well. Correct me if I'm wrong but for us to have a Conditional Independence we should have the " | " character in the statement isn't that right? $\endgroup$ – Diego Tsutsumi Dec 1 '16 at 16:09
  • $\begingroup$ I have put in the statements you want. $\endgroup$ – Dilip Sarwate Dec 1 '16 at 20:04

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