Pdf of $|X-Y|$ when $X,Y$ are independent Uniform $[0,a]$ variables
$begingroup$
Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
As it is symmetrical I have taken one part of triangle.
Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
Not getting expected answer.
Could any help.
Answer:
$f_{z}(z)=frac{2}{a}(1-frac{z}{a})$
probability probability-distributions random-variables uniform-distribution
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
$endgroup$
add a comment |
$begingroup$
Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
As it is symmetrical I have taken one part of triangle.
Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
Not getting expected answer.
Could any help.
Answer:
$f_{z}(z)=frac{2}{a}(1-frac{z}{a})$
probability probability-distributions random-variables uniform-distribution
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
$endgroup$
add a comment |
$begingroup$
Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
As it is symmetrical I have taken one part of triangle.
Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
Not getting expected answer.
Could any help.
Answer:
$f_{z}(z)=frac{2}{a}(1-frac{z}{a})$
probability probability-distributions random-variables uniform-distribution
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
$endgroup$
Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
As it is symmetrical I have taken one part of triangle.
Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
Not getting expected answer.
Could any help.
Answer:
$f_{z}(z)=frac{2}{a}(1-frac{z}{a})$
probability probability-distributions random-variables uniform-distribution
probability probability-distributions random-variables uniform-distribution
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
edited Dec 22 '18 at 19:58
StubbornAtom
6,06811239
6,06811239
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
asked Nov 22 '18 at 12:35
Pramod_acharPramod_achar
34
34
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$
So for $0< z<a$,
begin{align}
P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
\&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
\&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
\&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
\&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
\&=frac{2}{a^2}left(az-frac{z^2}{2}right)
end{align}
Hence the pdf of $Z=|X-Y|$ is
$$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$
Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.
Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
$endgroup$
add a comment |
$begingroup$
|X-Y| can be written as below ,
$ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $
The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.
$ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $
After differentiating w.r.t z ,
$ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $
$ =frac{2}{a^{2}}(a-z) $
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$
So for $0< z<a$,
begin{align}
P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
\&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
\&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
\&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
\&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
\&=frac{2}{a^2}left(az-frac{z^2}{2}right)
end{align}
Hence the pdf of $Z=|X-Y|$ is
$$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$
Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.
Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
$endgroup$
add a comment |
$begingroup$
Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$
So for $0< z<a$,
begin{align}
P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
\&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
\&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
\&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
\&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
\&=frac{2}{a^2}left(az-frac{z^2}{2}right)
end{align}
Hence the pdf of $Z=|X-Y|$ is
$$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$
Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.
Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
$endgroup$
add a comment |
$begingroup$
Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$
So for $0< z<a$,
begin{align}
P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
\&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
\&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
\&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
\&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
\&=frac{2}{a^2}left(az-frac{z^2}{2}right)
end{align}
Hence the pdf of $Z=|X-Y|$ is
$$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$
Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.
Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
$endgroup$
Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$
So for $0< z<a$,
begin{align}
P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
\&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
\&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
\&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
\&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
\&=frac{2}{a^2}left(az-frac{z^2}{2}right)
end{align}
Hence the pdf of $Z=|X-Y|$ is
$$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$
Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.
Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
edited Dec 22 '18 at 19:54
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
answered Dec 22 '18 at 19:19
StubbornAtomStubbornAtom
6,06811239
6,06811239
add a comment |
add a comment |
$begingroup$
|X-Y| can be written as below ,
$ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $
The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.
$ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $
After differentiating w.r.t z ,
$ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $
$ =frac{2}{a^{2}}(a-z) $
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
$endgroup$
add a comment |
$begingroup$
|X-Y| can be written as below ,
$ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $
The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.
$ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $
After differentiating w.r.t z ,
$ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $
$ =frac{2}{a^{2}}(a-z) $
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
$endgroup$
add a comment |
$begingroup$
|X-Y| can be written as below ,
$ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $
The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.
$ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $
After differentiating w.r.t z ,
$ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $
$ =frac{2}{a^{2}}(a-z) $
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
$endgroup$
|X-Y| can be written as below ,
$ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $
The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.
$ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $
After differentiating w.r.t z ,
$ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $
$ =frac{2}{a^{2}}(a-z) $
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
answered Dec 22 '18 at 17:32
Pramod_acharPramod_achar
34
34
add a comment |
add a comment |
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