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.

  • 7
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.