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Suppose $X$ and $Y$ are i.i.d. random variables. Also suppose they take the values from the set $\{1,2, \dots, n \}$. Then does this mean that $$P(X=1, Y= 1) = P(X=1) \cdot P(Y=1)$$ $$P(X=1, Y=2) = P(X=1) \cdot P(Y=2) \dots$$

So there are are $\binom{n}{2}$ cases for which $P(X \cap Y) =P(X) \cdot P(Y)$? If we didn't know that they were independent, we would have to check all these cases?

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    @Henry: I know that they are i.i.d. I would just like to confirm that they are by going through the process. So there would be $\binom{n}{2}$ calculations?2012-03-12
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    For i.i.d. discrete random variables $X$ and $Y$, $(X,Y) \in \{1,2,\ldots, n\}^2$ has $n^2$ different points where $P\{X = i, Y = j\} = P\{X = i\}P\{Y = j\}$ holds, $i, j \in \{1,2,\ldots, n\}$, not $\binom{n}{2}$ points. To _prove_ dependence, all you need to do is find _one_ $(i,j)$ such that $P\{X = i, Y = j\} \neq P\{X = i\}P\{Y = j\}$. To prove that a given joint pmf corresponds to i.i.d random variables, you need to work harder.2012-03-12

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