TITLE: Joint distribution of $\min(X_1,\ldots,X_n)$ and $\max(X_1,\ldots,X_n)$.
QUESTION [7 upvotes]: Let $X_1,\ldots,X_n$ be independent random variables with common cumulative distribution function $F$. I am trying to find the joint cumulative distribution function of
$$U=\min(X_1,\ldots,X_n)\quad\text{and}\quad V=\max(X_1,\ldots,X_n).$$
The result that I get is
$$F_{U,V}(u,v)=F(v)^n-(F(v)-F(u))^n,$$
but this seems wrong because letting $u\to\infty$ gives the marginal CDF
$$F_V(v)=F(v)^n-(F(v)-1)^n,$$
while it should be $F_V(v)=F(v)^n$. I obtained the result as follows
$$\begin{align} F_{U,V}(u,v) &= P(U\leq u,V\leq v) \\
&= P(V\leq v)-P(U>u,V\leq v) \\
&= P(X_1\leq v,\ldots,X_n\leq v)-P(u<X_1\leq v,\ldots,u<X_n\leq v) \\
&= F(v)^n-(F(v)-F(u))^n.
\end{align}
$$
But
$$F_V(v)=P(V\leq v)=P(X_1\leq v,\ldots,X_n\leq v)=P(X_1\leq v)\cdots P(X_n\leq v)=F(v)^n.$$
What is wrong with these computations?

REPLY [4 votes]: I see my mistake now:
The computation for $F_{U,V}$ was valid only under the assumption that $u<v$. If $u\geq v$ we get instead
$$F_{U,V}(u,v)=F(v)^n.$$
So everything is consistent and we are good.