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The answer didn't consider the case when x = -1. This would be 0/0.

Yet, it's always true that the column where y = 1 is twice the column where y = -1 (since 0 = 2 * 0).

I'm a bit confused and not sure whether the answer is correct or not.


You can say there is conditional independence. Conditioned on $Y\not=0$ the probability that $X=-1$ is $\mathbb P(X=-1 \mid Y \not=0)=0$ so does not affect the argument. The table becomes

y=1  ||  4k      0     2k     8k 
y=-1 ||  2k      0      k     4k 
       ====== ====== ====== ====== 
        x=-2   x=-1    x=0    x=1

You then have $$\mathbb P(Y=y, X=x \mid Y \not=0)=\mathbb P(Y=y \mid Y \not=0)\,\mathbb P(X=x \mid Y \not=0)$$ for all $x$ and $y$, even for $x=-1$, and this means there is conditional independence.


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