difference of two independent exponentially distributed random variables











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Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.



I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.



Any help would be appreciated, thanks.






probability







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    up vote
    0
    down vote

    favorite












    Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.



    I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.



    Any help would be appreciated, thanks.






    probability







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.



      I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.



      Any help would be appreciated, thanks.






      probability







      share|cite|improve this question













      Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.



      I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.



      Any help would be appreciated, thanks.






      probability




      probability






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      asked Nov 22 at 17:15









      dxdydz

      1929




      1929






















          1 Answer
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          By symmetry,



          we have



          begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}



          Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$






          share|cite|improve this answer





















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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            By symmetry,



            we have



            begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}



            Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              By symmetry,



              we have



              begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}



              Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                By symmetry,



                we have



                begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}



                Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$






                share|cite|improve this answer












                By symmetry,



                we have



                begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}



                Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 17:32









                Siong Thye Goh

                97.8k1463116




                97.8k1463116






























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