$E(|X-Y|)$ for $X$, $Y$ i.i.d and uniform on $[0,1]$
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What is $E|X-Y|$ when $X$, $Y$ are i.i.d and uniform on $[0,1]$?
I want to do something like
$$
E(|X-Y|)=iint_{x>y}(x-y);dxdy+iint_{y>x}(y-x);dxdy=2iint_{x>y} (x-y);dxdy
$$
But am not sure how to solve that last double integral or if this setup is right.
probability
add a comment |
up vote
0
down vote
favorite
What is $E|X-Y|$ when $X$, $Y$ are i.i.d and uniform on $[0,1]$?
I want to do something like
$$
E(|X-Y|)=iint_{x>y}(x-y);dxdy+iint_{y>x}(y-x);dxdy=2iint_{x>y} (x-y);dxdy
$$
But am not sure how to solve that last double integral or if this setup is right.
probability
2
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
Do you think this setup is right?
– user616954
Nov 18 at 20:14
2
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is $E|X-Y|$ when $X$, $Y$ are i.i.d and uniform on $[0,1]$?
I want to do something like
$$
E(|X-Y|)=iint_{x>y}(x-y);dxdy+iint_{y>x}(y-x);dxdy=2iint_{x>y} (x-y);dxdy
$$
But am not sure how to solve that last double integral or if this setup is right.
probability
What is $E|X-Y|$ when $X$, $Y$ are i.i.d and uniform on $[0,1]$?
I want to do something like
$$
E(|X-Y|)=iint_{x>y}(x-y);dxdy+iint_{y>x}(y-x);dxdy=2iint_{x>y} (x-y);dxdy
$$
But am not sure how to solve that last double integral or if this setup is right.
probability
probability
edited Nov 18 at 20:14
asked Nov 18 at 19:46
user616954
2
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
Do you think this setup is right?
– user616954
Nov 18 at 20:14
2
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16
add a comment |
2
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
Do you think this setup is right?
– user616954
Nov 18 at 20:14
2
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16
2
2
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
Do you think this setup is right?
– user616954
Nov 18 at 20:14
Do you think this setup is right?
– user616954
Nov 18 at 20:14
2
2
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16
add a comment |
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2
The last double integral is just $2int_0^1 int_y^1 (x-y),dx,dy=2int_0^1left(int_y^1 x,dx-yint_y^1,dxright),dy$.
– StubbornAtom
Nov 18 at 20:08
Do you think this setup is right?
– user616954
Nov 18 at 20:14
2
Yes the setup is correct because of symmetry.
– StubbornAtom
Nov 18 at 20:16