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Suppose we have independence of two random variables, $X$ and $Y$. This of course then implies that $p(x,y)=p(x)p(y)$ where $p(x,y)$ is the joint pmf of X and Y (ie. $p(x,y) = P(X=x, Y=y) = P(X=x)P(Y=y) = p(x)p(y)$.

This may be silly, but that does this imply that $P(X=x, Y=y | Z=z) = P(X=x | Z=z)*P(Y=y | Z=z)$, where Z is an arbitrary random variable?

My first thought would be yes, because given Z=z, X and Y should still be independent. However, when trying to prove this, I found: $P(X=x , Y=y | Z=z) = \frac{P(X=x,Y=y,Z=z)}{P(Z=z)} = \frac{P(X=x|Y=y,Z=z)P(Y=y,Z=z)}{P(Z=z)}=P(X=x|Y=y,Z=z)P(Y=y|Z=z)$.

So where I'm having trouble is the last step, can we remove the conditioning of Y=y on the left side of the product given X is independent of Y?

Note: This is not a homework problem or anything, but rather I tried to make the assumption that this was correct in another proof I was doing and was concerned about the validity.

StackExchangeDisplayName
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2 Answers2

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No. The relation $P(X,Y\mid Z) = P(X \mid Z) P(Y \mid Z)$ means that $X$ and $Y$ are conditionally independent given $Z$, and that is not implied by independence of $X$ and $Y$, see for example here.

G. Gare
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It does not work.

E.g. let $Z$ be prescribed by $\omega\mapsto1$ if $X(\omega)=Y(\omega)$ and $\omega\mapsto 0$ otherwise.

If $x\neq y$ then: $$P(X=x,Y=y\mid Z=1)=0$$ but it is quite well possible that: $$P(X=x\mid Z=1)P(Y=y\mid Z=1)>0$$

drhab
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