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Suppose I have two distributions $X$ and $Y$, and I have a statistic $T(X, Y)$. If I have a random sample (iid) $X_1, ..., X_N$ taken from $X$ and a random sample (iid) $Y_1, ..., Y_M$ taken from $Y$, then are the statistics of the cross products (i.e., $\{T(X_i, Y_j) : i \in 1, ..., N, j \in 1, ..., M\}$ a random sample (iid) of the distribution of $T(X, Y)$?

Is this true in the case that $X$ and $Y$ are independent?

Here's what I'm thinking when we ignore the test statistic $T$, and just focus on the joint distribution of $X$ and $Y$: Suppose we want to determine whether $(X_i, X_i)$ and $(X_j, Y_j)$ are iid. We can show that by proving that $Pr(X_i, Y_i | X_j, Y_j) = Pr(X_j, Y_j)$, I believe.

On the right hand side, by independence, and iid of the $X_i$'s and $Y_i$'s, we have $Pr(X_j, Y_j) = Pr(X_j)Pr(Y_j) = Pr(X_i)Pr(Y_i)$.

On the left hand side, we can simplify $Pr(X_i, Y_i | X_j, Y_j) = Pr(X_i | X_j, Y_j) Pr (Y_i | X_j, Y_j) = Pr(X_i | X_j) Pr(Y_i | Y_j) = Pr(X_i) Pr(Y_i)$ where the first two equalities follow from the independence of $X$ and $Y$ and the last equality follows from the iid property of the $X_i$'s and $Y_i$'s.

So if the $(X_i, Y_j)$'s are iid from the joint distribution of $X$ and $Y$, then it seems that the $T(X_i, Y_j)$'s are iid from the distribution of $T(X, Y)$.

Does that proof seem correct?

Thanks in advance!

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    $\begingroup$ To get some insight into why the answers must be negative, suppose $T(X,Y)=X$ and $X$ is nondegenerate. Then the cross products consist of $M$ replications of each $X_i$, which when $M\gt 1$ obviously is not a random sample of anything. $\endgroup$
    – whuber
    Jun 18 '14 at 19:42
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    $\begingroup$ That's a good example. Thanks for your help @whuber! $\endgroup$ Jun 18 '14 at 19:55
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As @whuber pointed out, the error in your proof is that even though $(X_i, Y_i)$ is independent of $(X_j, Y_j)$, it is not independent of $(X_i, Y_j)$.

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