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Suppose $X_1, X_2$ are independent $U(0, 1)$ random variables, and

$$Y = \min(X_1, X_2) $$ $$Z = \max(X_1, X_2) $$

By this question, they $Y$ and $Z$ should be independent:

Are functions of independent variables also independent?

But by this answer the covariance is not zero:

What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?

How do I reconcile these two things? The $\min$ and $\max$ are a function of independent random variables, yet they have covariance.

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    $\begingroup$ The min and the max cannot be independent. If you know one, you know something about the other. $\endgroup$ – Hans Engler Dec 17 '18 at 20:05
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    $\begingroup$ It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself. $\endgroup$ – Tki Deneb Dec 17 '18 at 20:10
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    $\begingroup$ In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$ $\endgroup$ – Sauhard Sharma Dec 17 '18 at 20:36
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If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.

In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.

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Think of a random vector $\bar{X} = [X_1 \; X_2]$. Now, both $Y$ and $Z$ are functions of $\bar{X}$. Whether the elements of $\bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $\bar{X}$.

Alternatively, if $X_1 < X_2$, the $\min$ ($=Y$) is $X_1$, which automatically implies that the $\max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.

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