Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $U,V$

share|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Johanna, 1999, anomaly, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $f_{U,V}(u,v) = \frac{\partial^2}{\partial u \partial v} F_{U,V}(u,v)$.

share|improve this answer

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 $f_{U,V}$.

share|improve this answer

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