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I have 2 random hypergeometric variables $X \sim \mathcal{H}(N, K, n)$ and $Y(X) \sim \mathcal{H}(N-n, L-n+X, n)$ with $N = K + L$. How can I write the joint pdf of $X$ and $Y$?

I guess that $P(X=a \wedge Y = b) = P(X=a)P(Y=b)$ but I'm not really sure that $X$ and $Y$ are independent.


I know that $P(X=a)$ follows the first hypergeometric distribution and $P(Y=b \mid X=a)$ follows the second. Then, I can state the following:

$$P(Y=b \mid X=a) = \frac{P(Y=b \wedge X=a)}{p(X=a)} \Rightarrow P(Y=b \wedge X=a) = P(Y=b \mid X=a)P(X=a) $$

Is this the right way to evaluate the joint pdf?

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The Bayes rule that you wrote is correct. For independency one needs to check if $P(Y=b|X=a)=P(Y=b)$ is correct or not. – Seyhmus Güngören Feb 3 '13 at 15:05
Instead of $Y(X) \sim \mathcal{H}(N-n, L-n+X, n)$, I'd normally write $Y\mid X \sim \mathcal{H}(N-n, L-n+X, n)$. That way it's clear that it's not $Y$ that is determined by $X$, but rather it's the conditional probability distribution of $Y$ that is determined by $X$. But don't make the mistake of construing "$Y\mid X$" as a noun. There's no object called $Y\mid X$. – Michael Hardy Feb 3 '13 at 16:51
I've retagged: please do not use an ambiguous abbreviation ("pdf") when the expanded tag is already available ("probability-distributions"). – Willie Wong Feb 4 '13 at 8:58

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