# If two random variables are independent, why isn't their min and max?

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.

• The min and the max cannot be independent. If you know one, you know something about the other. – Hans Engler Dec 17 '18 at 20:05
• 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. – Tki Deneb Dec 17 '18 at 20:10
• 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 $f(X_1,X_2)$ – 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 $$g(X_1, X_2)$$.
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.