There's this exercise that's been giving me trouble over the last couple of days :
Given $X$, $Y$ and $Z$ discrete r.v verifying ($X$ and $Y$ independant) and ($X$ and $Y$ independant conditionnally to $Z$), proof or disproof the following :
(X and Z are independant) or (Y and Z are).
I tried first the case where $Z$ is binary and after some calculations, we obtain a quadratic polynomial in $P(Z=0)$ and the solution follows nicely.
To generalize for the case where $Z$ takes $n$ different values, the problem can be rewritten as follows:
Given that $$ \sum_{i=1}^{n} \frac{a_kb_k}{c_k} = 1$$ and $$ \sum a_k > = \sum b_k = \sum c_k =1$$ Show that either $(\forall k ~~ a_k = c_k)$ or $(\forall k ~~ b_k = c_k)$
I tried to do that by induction with no success and I can't see how it can be solved. Seeing how much I struggled, I've been thinking that maybe the result doesn't stand for n>2 but then again, I can't find a counterexample for that.
Any ideas ?
The assertion is false in general. Here is a counterexample when $X,Y,Z$ take four possible values.
Let $X$ and $Y$ be independent, each one taking values $-2, -1, 1, 2$ with equal probability. Define $Z$ to be the quadrant in the $(x,y)$-plane to which the point $(X,Y)$ belongs. So $Z=1$ when $(X,Y)$ take the values $(1,1)$, $(1,2)$, $(2,1)$, or $(2,2)$; and $Z=2$ when $(X,Y)$ take the values $(-1,1)$, $(-1,2)$, $(-2,1)$, or $(-2,2)$, and so on. It is clear that $X$ and $Y$ are independent conditional on $Z$, since the conditional distribution of $(X,Y)$ given $Z$ is uniform over a square. However, $X$ and $Z$ are not independent, and neither are $Y$ and $Z$, since knowledge of the value of $Z$ will restrict the quadrant where $X$ and $Y$ are found.
As an even simpler version of the above, let $(X,Y,Z)$ be uniform over the four points $(1,1,1)$, $(1,2,2)$, $(2,1,3)$, $(2,2,4)$. Here again $Z$ serves as a label of the region where $X$ and $Y$ can be found. Unconditionally, $X$ and $Y$ are independently uniform over $\{1,2\}$. Conditional on $Z$, the distribution of $(X,Y)$ is constant, so $X$ and $Y$ are independent given $Z$.