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
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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
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited yesterday
StubbornAtom
6,14311339
6,14311339
asked yesterday
Mahamad A. KanoutéMahamad A. Kanouté
395113
395113
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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
$endgroup$
$begingroup$
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 Answer
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active
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votes
$begingroup$
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
$endgroup$
$begingroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
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