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?

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    $\begingroup$ See p.8 of this. $\endgroup$ Dec 6, 2014 at 20:08

1 Answer 1


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.


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