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Computing $mathbbPX_(1)+X_(2) > X_(3)$ where $X_1,X_2,X_3$ are i.i.d $U(0,1)$


Find $P(x_2/x_3 leq a)$ where $x_1,x_2,dots,x_n $ are iid $operatornameUniform(0,1)$For exponential random variables $X_i$, how to find $P(t-X_1<X_2mid t-X_1<X_3)$?Simply question about i.i.d uniform random variables on (0,1)Probability and expectation of three ordered random variablesGiving an example of a t-distribution with independent $X_1, X_2, X_3$Finding $P(X_1+X_2 > 1.9X_3)$ where $X_1$, $X_2$, and $X_3$ are independent, normal distributed random variablesHow can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?Finding $P(X_1+X_2+X_3+X_4ge3)$ for independent $X_isim U(0,1)$Find $P(X_1 > max(X_2, X_3)| X_1, X_2, X_3 geq d)$Compute $P(X_(2) ≤ 3X_(1))$ by using the integration technique













1












$begingroup$



Suppose that $X_1 , X_2 , X_3$ are independent $U (0, 1)$-distributed random variables and let $(X_(1) , X_(2) , X _(3) )$ be the corresponding order statistic. Compute $mathbbPX_(1)+X_(2) > X_(3)$.




I have the joint of the distribution which is $3!$



Also, I have:



$$mathbbPX_(1)+X_(2) > X_(3)=1-mathbbPX_(1)+X_(2) le X_(3)$$ with $mathbb 0<x_1<minx_2,x_3-x_2, quad 0<x_2<x_3, quad0<x_3<1$



But when I do:



$$1-6left(int_0^1int_0^x_3int_0^x_2 ,dx_1,dx_2,dx_3 +int_0^1int_0^x_3int_0^x_3-x_2 ,dx_1,dx_2,dx_3right)=-3$$



I do not really know what I'm doing wrong.






probability probability-theory probability-distributions uniform-distribution order-statistics







share|cite|improve this question











$endgroup$
















    1












    $begingroup$



    Suppose that $X_1 , X_2 , X_3$ are independent $U (0, 1)$-distributed random variables and let $(X_(1) , X_(2) , X _(3) )$ be the corresponding order statistic. Compute $mathbbPX_(1)+X_(2) > X_(3)$.




    I have the joint of the distribution which is $3!$



    Also, I have:



    $$mathbbPX_(1)+X_(2) > X_(3)=1-mathbbPX_(1)+X_(2) le X_(3)$$ with $mathbb 0<x_1<minx_2,x_3-x_2, quad 0<x_2<x_3, quad0<x_3<1$



    But when I do:



    $$1-6left(int_0^1int_0^x_3int_0^x_2 ,dx_1,dx_2,dx_3 +int_0^1int_0^x_3int_0^x_3-x_2 ,dx_1,dx_2,dx_3right)=-3$$



    I do not really know what I'm doing wrong.






    probability probability-theory probability-distributions uniform-distribution order-statistics







    share|cite|improve this question











    $endgroup$














      1












      1








      1


      0



      $begingroup$



      Suppose that $X_1 , X_2 , X_3$ are independent $U (0, 1)$-distributed random variables and let $(X_(1) , X_(2) , X _(3) )$ be the corresponding order statistic. Compute $mathbbPX_(1)+X_(2) > X_(3)$.




      I have the joint of the distribution which is $3!$



      Also, I have:



      $$mathbbPX_(1)+X_(2) > X_(3)=1-mathbbPX_(1)+X_(2) le X_(3)$$ with $mathbb 0<x_1<minx_2,x_3-x_2, quad 0<x_2<x_3, quad0<x_3<1$



      But when I do:



      $$1-6left(int_0^1int_0^x_3int_0^x_2 ,dx_1,dx_2,dx_3 +int_0^1int_0^x_3int_0^x_3-x_2 ,dx_1,dx_2,dx_3right)=-3$$



      I do not really know what I'm doing wrong.






      probability probability-theory probability-distributions uniform-distribution order-statistics







      share|cite|improve this question











      $endgroup$





      Suppose that $X_1 , X_2 , X_3$ are independent $U (0, 1)$-distributed random variables and let $(X_(1) , X_(2) , X _(3) )$ be the corresponding order statistic. Compute $mathbbPX_(1)+X_(2) > X_(3)$.




      I have the joint of the distribution which is $3!$



      Also, I have:



      $$mathbbPX_(1)+X_(2) > X_(3)=1-mathbbPX_(1)+X_(2) le X_(3)$$ with $mathbb 0<x_1<minx_2,x_3-x_2, quad 0<x_2<x_3, quad0<x_3<1$



      But when I do:



      $$1-6left(int_0^1int_0^x_3int_0^x_2 ,dx_1,dx_2,dx_3 +int_0^1int_0^x_3int_0^x_3-x_2 ,dx_1,dx_2,dx_3right)=-3$$



      I do not really know what I'm doing wrong.






      probability probability-theory probability-distributions uniform-distribution order-statistics




      probability probability-theory probability-distributions uniform-distribution order-statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      StubbornAtom

      6,14311339




      6,14311339










      asked yesterday









      Mahamad A. KanoutéMahamad A. Kanouté

      395113




      395113




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          There could possibly be a simpler argument that does not require doing the integration explicitly.



          For the integration, I felt it convenient to use a change of variables.



          You are looking for



          $$P(X_(1)+X_(2)<X_(3))=6iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz$$



          Change variables $(x,y,z)to(u,v,w)$ with $$u=x+y,,,v=y,,,w=z$$



          Then, $$x+y<z,,,0<x<y<z<1implies u<w,,,0<u-v<v<w<1$$



          Therefore,



          beginalign
          iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz&=int_0^1int_0^wint_v^min(2v,w),mathrmdu,mathrmdv,mathrmdw
          \&=int_0^1int_0^w/2int_v^2v,mathrmdu,mathrmdv,mathrmdw+int_0^1int_w/2^wint_v^w,mathrmdu,mathrmdv,mathrmdw
          \&=frac112
          endalign






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            why does v goes from 0 to w/2 ?
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            I get -1 as answer
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            @MahamadA.Kanouté See the breaking of the integral.
            $endgroup$
            – StubbornAtom
            22 hours ago










          Your Answer





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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          There could possibly be a simpler argument that does not require doing the integration explicitly.



          For the integration, I felt it convenient to use a change of variables.



          You are looking for



          $$P(X_(1)+X_(2)<X_(3))=6iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz$$



          Change variables $(x,y,z)to(u,v,w)$ with $$u=x+y,,,v=y,,,w=z$$



          Then, $$x+y<z,,,0<x<y<z<1implies u<w,,,0<u-v<v<w<1$$



          Therefore,



          beginalign
          iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz&=int_0^1int_0^wint_v^min(2v,w),mathrmdu,mathrmdv,mathrmdw
          \&=int_0^1int_0^w/2int_v^2v,mathrmdu,mathrmdv,mathrmdw+int_0^1int_w/2^wint_v^w,mathrmdu,mathrmdv,mathrmdw
          \&=frac112
          endalign






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            why does v goes from 0 to w/2 ?
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            I get -1 as answer
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            @MahamadA.Kanouté See the breaking of the integral.
            $endgroup$
            – StubbornAtom
            22 hours ago















          1












          $begingroup$

          There could possibly be a simpler argument that does not require doing the integration explicitly.



          For the integration, I felt it convenient to use a change of variables.



          You are looking for



          $$P(X_(1)+X_(2)<X_(3))=6iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz$$



          Change variables $(x,y,z)to(u,v,w)$ with $$u=x+y,,,v=y,,,w=z$$



          Then, $$x+y<z,,,0<x<y<z<1implies u<w,,,0<u-v<v<w<1$$



          Therefore,



          beginalign
          iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz&=int_0^1int_0^wint_v^min(2v,w),mathrmdu,mathrmdv,mathrmdw
          \&=int_0^1int_0^w/2int_v^2v,mathrmdu,mathrmdv,mathrmdw+int_0^1int_w/2^wint_v^w,mathrmdu,mathrmdv,mathrmdw
          \&=frac112
          endalign






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            why does v goes from 0 to w/2 ?
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            I get -1 as answer
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            @MahamadA.Kanouté See the breaking of the integral.
            $endgroup$
            – StubbornAtom
            22 hours ago













          1












          1








          1





          $begingroup$

          There could possibly be a simpler argument that does not require doing the integration explicitly.



          For the integration, I felt it convenient to use a change of variables.



          You are looking for



          $$P(X_(1)+X_(2)<X_(3))=6iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz$$



          Change variables $(x,y,z)to(u,v,w)$ with $$u=x+y,,,v=y,,,w=z$$



          Then, $$x+y<z,,,0<x<y<z<1implies u<w,,,0<u-v<v<w<1$$



          Therefore,



          beginalign
          iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz&=int_0^1int_0^wint_v^min(2v,w),mathrmdu,mathrmdv,mathrmdw
          \&=int_0^1int_0^w/2int_v^2v,mathrmdu,mathrmdv,mathrmdw+int_0^1int_w/2^wint_v^w,mathrmdu,mathrmdv,mathrmdw
          \&=frac112
          endalign






          share|cite|improve this answer











          $endgroup$



          There could possibly be a simpler argument that does not require doing the integration explicitly.



          For the integration, I felt it convenient to use a change of variables.



          You are looking for



          $$P(X_(1)+X_(2)<X_(3))=6iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz$$



          Change variables $(x,y,z)to(u,v,w)$ with $$u=x+y,,,v=y,,,w=z$$



          Then, $$x+y<z,,,0<x<y<z<1implies u<w,,,0<u-v<v<w<1$$



          Therefore,



          beginalign
          iiintmathbf1_x+y<z,,,0<x<y<z<1,mathrmdx,mathrmdy,mathrmdz&=int_0^1int_0^wint_v^min(2v,w),mathrmdu,mathrmdv,mathrmdw
          \&=int_0^1int_0^w/2int_v^2v,mathrmdu,mathrmdv,mathrmdw+int_0^1int_w/2^wint_v^w,mathrmdu,mathrmdv,mathrmdw
          \&=frac112
          endalign







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 22 hours ago

























          answered yesterday









          StubbornAtomStubbornAtom

          6,14311339




          6,14311339











          • $begingroup$
            why does v goes from 0 to w/2 ?
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            I get -1 as answer
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            @MahamadA.Kanouté See the breaking of the integral.
            $endgroup$
            – StubbornAtom
            22 hours ago
















          • $begingroup$
            why does v goes from 0 to w/2 ?
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            I get -1 as answer
            $endgroup$
            – Mahamad A. Kanouté
            yesterday










          • $begingroup$
            @MahamadA.Kanouté See the breaking of the integral.
            $endgroup$
            – StubbornAtom
            22 hours ago















          $begingroup$
          why does v goes from 0 to w/2 ?
          $endgroup$
          – Mahamad A. Kanouté
          yesterday




          $begingroup$
          why does v goes from 0 to w/2 ?
          $endgroup$
          – Mahamad A. Kanouté
          yesterday












          $begingroup$
          I get -1 as answer
          $endgroup$
          – Mahamad A. Kanouté
          yesterday




          $begingroup$
          I get -1 as answer
          $endgroup$
          – Mahamad A. Kanouté
          yesterday












          $begingroup$
          @MahamadA.Kanouté See the breaking of the integral.
          $endgroup$
          – StubbornAtom
          22 hours ago




          $begingroup$
          @MahamadA.Kanouté See the breaking of the integral.
          $endgroup$
          – StubbornAtom
          22 hours ago

















          draft saved

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