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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

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

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

$endgroup$

add a comment | 

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

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

$endgroup$

add a comment | 

$begingroup$

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

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

$endgroup$

share|cite|improve this question

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

share|cite|improve this question

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

edited yesterday

StubbornAtom

6,14311339

edited yesterday

StubbornAtom

6,14311339

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

395113

asked yesterday

Mahamad A. KanoutéMahamad A. Kanouté

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

edited 22 hours ago

answered yesterday

StubbornAtomStubbornAtom

6,14311339

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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

edited 22 hours ago

answered yesterday

StubbornAtomStubbornAtom

6,14311339

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 22 hours ago

answered yesterday

StubbornAtomStubbornAtom

6,14311339

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 22 hours ago

answered yesterday

StubbornAtomStubbornAtom

6,14311339

$endgroup$

share|cite|improve this answer

edited 22 hours ago

answered yesterday

StubbornAtomStubbornAtom

6,14311339

answered yesterday

StubbornAtomStubbornAtom

6,14311339

answered yesterday

StubbornAtomStubbornAtom

6,14311339