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suppose $X\sim\ E(\alpha)$ and $Y\sim\ E(\beta)$ be two independent random variable. if $U=\min(X,Y)$ , $V=\max(X,Y)$ how can find joint density function $U,V$

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closed as off-topic by Did, Johanna, 1999, anomaly, Daniel W. Farlow Mar 21 at 4:11

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Answer now undeleted. I must admit being curious about whether the indications in Robert's answer are enough for the OP to reach a full solution... –  Did Apr 29 '12 at 9:38

2 Answers 2

up vote 1 down vote accepted

The joint CDF is $F_{U,V}(u,v) = P(U\le u, V \le v)$. For $u \ge v$ this is $P(V \le v) = P(X \le v) P(Y \le v)$, for $u < v$ it is $P(X\le u) P(Y \le v) + P(X \le v) P(Y \le u) - P(X\le u) P(Y \le u)$. The joint density is $f_{U,V}(u,v) = \frac{\partial^2}{\partial u \partial v} F_{U,V}(u,v)$.

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Almost surely, $0\leqslant U\leqslant V$. For every $u\leqslant v$, $[u\leqslant U,V\leqslant v]=[u\leqslant X\leqslant v]\cap[u\leqslant Y\leqslant v]$. Since $\mathrm P(u\leqslant X\leqslant v)=\mathrm e^{-\alpha u}-\mathrm e^{-\alpha v}$, $\mathrm P(u\leqslant Y\leqslant v)=\mathrm e^{-\beta u}-\mathrm e^{-\beta v}$, and the random variables $X$ and $Y$ are independent, one gets $$ \mathrm P(u\leqslant U,V\leqslant v)=(\mathrm e^{-\alpha u}-\mathrm e^{-\alpha v})\cdot(\mathrm e^{-\beta u}-\mathrm e^{-\beta v}) $$ The density $f_{U,V}$ is defined, on every $0\leqslant u\leqslant v$, by $$ f_{U,V}(u,v)=-\frac{\partial^2}{\partial u\partial v}\mathrm P(u\leqslant U,V\leqslant v), $$ that is, $$ \color{red}{f_{U,V}(u,v)=\alpha\beta\cdot(\mathrm e^{-\alpha u-\beta v}+\mathrm e^{-\alpha v-\alpha u})}. $$ The infinitesimal justification of this formula is direct: for $u\lt v$, $$ [U\in(u,u+\mathrm du),V\in(v,v+\mathrm dv)]=A_1\cup A_2 $$ with $$ A_1=[X\in(u,u+\mathrm du),Y\in(v,v+\mathrm dv)],\quad A_2=[Y\in(u,u+\mathrm du),X\in(v,v+\mathrm dv)]. $$ The events $A_1$ and $A_2$ are disjoint. Since $X$ and $Y$ are independent, $$ \mathrm P(A_1)=\mathrm P(X\in(u,u+\mathrm du))\cdot\mathrm P(Y\in(v,v+\mathrm dv))=\alpha\mathrm e^{-\alpha u}\mathrm du\cdot\beta\mathrm e^{-\beta v}\mathrm dv. $$ Likewise, $\mathrm P(A_2)=\alpha\mathrm e^{-\alpha v}\mathrm dv\cdot\beta\mathrm e^{-\beta u}\mathrm du$. Considering $\mathrm P(A_1)+\mathrm P(A_2)=f_{U,V}(u,v)\mathrm du\mathrm dv$, one gets the formula above for $f_{U,V}$.

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