Conditional probability implication

Question

I'm having a hard time figuring out whether $$\mathbb{P}\left(X\cap Y\mid Z\right)=\mathbb{P}\left(X\mid Z\right)\mathbb{P}\left(Y\mid Z\right)$$ and $\mathbb{P}\left(Z\right)<1$ implies $$\mathbb{P}\left(X\cap Y\mid\overline{Z}\right)=\mathbb{P}\left(X\mid\overline{Z}\right)\mathbb{P}\left(Y\mid\overline{Z}\right)$$ I fail to see how to go past $$\mathbb{P}\left(X\cap Y\cap Z\right)=\frac{\mathbb{P}\left(X\cap Z\right)\mathbb{P}\left(Y\cap Z\right)}{\mathbb{P}\left(Z\right)}$$ so it makes me guess the implication isn't even true, but it was true for every example I could think of.

Answer 1

For a counterexample, consider a sample space of cardinality $7$, with all sample points equally likely.

Let events $X,Y,Z$ be chosen to match the following Venn diagram

where each of the $7$ pictured regions has cardinality $1$.

Then we have $$ P(X|Z)=P(Y|Z)=\frac{1}{2}\;\;\text{and}\;\;P(X\cap Y|Z)=\frac{1}{4} $$ so $P(X\cap Y|Z)=P(X|Z)P(Y|Z)$, but $$ P(X|\overline{Z})=P(Y|\overline{Z})=\frac{2}{3}\;\;\text{and}\;\;P(X\cap Y|\overline{Z})=\frac{1}{3} $$ so $P(X\cap Y|\overline{Z})\ne P(X|\overline{Z})P(Y|\overline{Z})$.

Answer 2

It is not true.

A (somewhat degenerate) counter-example: $$(\Omega, \mathcal A, \mathsf P)=\big(\{0,1,2\}, \mathfrak P(\{1,2,3\}), \text{counting measure}\big)$$ and $$X=\{0\}, Y=\{1\}, Z=\{2\}.$$