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Given that $X$ is independent of $Y$ and of $Z$, and that $Y$ and $Z$ are dependent on each other.

Can $P(X\cap Y|Z)$ be shown as P(X)P(Y|Z)?

This reason I believe this is because if X is independent of Y then $P(X\cap Y|Z)$ is equal to P(P(X)P(Y)|Z) and if X is also independent of Z then it would be P(X)P(Y|Z).

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What on earth is the probability of a probability? – Rahul Feb 28 '13 at 1:47
There is a notational problem here. Please rephrase. – ncmathsadist Feb 28 '13 at 2:10
Are you asking if $P(X\cap Y|Z)$ is equal to $P(X)P(Y|Z)$? – Rahul Feb 28 '13 at 7:52

The answer depends on what exactly you mean by "X is independent of Y and Z".

Suppose a hat contains the numbers 111, 100, 010, and 001, and you pick a number uniformly at random from the hat. Let X be "the hundreds digit is 1," let Y be "the tens digit is 1," let Z be "the units digit is 1."

X is independent of Y; $P(X\cap Y)=1/4=(1/2)(1/2)=P(X)P(Y)$.

Similarly, X is independent of Z.

$P(X\cap Y\mid Z)=1/2$; $P(X)P(Y\mid Z)=(1/2)(1/2)=1/4$. So $$P(X\cap Y\mid Z)\ne P(X)P(Y\mid Z)$$

What's happening here is X is independent of Y and (of) Z, but X is not independent of $Y\cap Z$.

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Wow, i just noticed that mistake, when i wrote Y and Z I did mean Y and of Z – Probs with Stats Mar 2 '13 at 17:01
OK. In that case, the answer to your question is negative, and my answer gives a counterexample. – Gerry Myerson Mar 2 '13 at 22:16

First, let $B = X \cap Y$, so then you'll have $P(B \mid Z)$.

So then you'll have: $P( B \mid Z) = \frac{P(B \cap Z)}{P(Z)}$ by the definition of conditional probability

Substituting back in for $B$, you will have $\frac{P((X \cap Y) \cap Z)}{P(Z)}$.

By the distributive law in probability (from theorem 1.1.4 in $\textit{Statistical Inference}$ By Casella and Berger), it equals $\frac{P((X \cap Z) \cap (Y \cap Z))}{P(Z)}$, which you can then manipulate to try to get what you're trying to show.

Hope this helps! Good Luck!

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Isnt P(XintersectY)intersectZ) equal to P(XintersectYintersectZ) – Probs with Stats Feb 28 '13 at 23:21

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