Independence of ratios of independent variates
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$begingroup$
If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
$x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
respectively, are $X$ & $Y$ independent?
I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?
distributions chi-squared independence gamma-distribution
edited Jan 13 at 15:28
Xi'an
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
$endgroup$
add a comment |
$begingroup$
If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
$x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
respectively, are $X$ & $Y$ independent?
I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?
distributions chi-squared independence gamma-distribution
edited Jan 13 at 15:28
Xi'an
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
$endgroup$
add a comment |
$begingroup$
If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
$x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
respectively, are $X$ & $Y$ independent?
I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?
distributions chi-squared independence gamma-distribution
edited Jan 13 at 15:28
Xi'an
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
$endgroup$
If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
$x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
respectively, are $X$ & $Y$ independent?
I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?
distributions chi-squared independence gamma-distribution
distributions chi-squared independence gamma-distribution
edited Jan 13 at 15:28
Xi'an
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
edited Jan 13 at 15:28
Xi'an
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
edited Jan 13 at 15:28
Xi'an
59.6k897368
edited Jan 13 at 15:28
Xi'an
59.6k897368
edited Jan 13 at 15:28
Xi'an
59.6k897368
59.6k897368
asked Jan 13 at 14:28
Jor_ElJor_El
917
asked Jan 13 at 14:28
Jor_ElJor_El
917
asked Jan 13 at 14:28
Jor_ElJor_El
917
917
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
begin{align*}
x_1&={varrho sin(theta)}^2\
x_1&={varrho cos(theta)}^2\
end{align*}
we get that
$$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
and
$$Y=frac{varrho^2}{varrho^2+x_3}$$
are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
$endgroup$
add a comment |
$begingroup$
A geometrical interpretation/intuition
You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.
The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.
See here a similar (but much simpler) calculation.
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
begin{align*}
x_1&={varrho sin(theta)}^2\
x_1&={varrho cos(theta)}^2\
end{align*}
we get that
$$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
and
$$Y=frac{varrho^2}{varrho^2+x_3}$$
are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
$endgroup$
add a comment |
$begingroup$
It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
begin{align*}
x_1&={varrho sin(theta)}^2\
x_1&={varrho cos(theta)}^2\
end{align*}
we get that
$$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
and
$$Y=frac{varrho^2}{varrho^2+x_3}$$
are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
$endgroup$
add a comment |
$begingroup$
It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
begin{align*}
x_1&={varrho sin(theta)}^2\
x_1&={varrho cos(theta)}^2\
end{align*}
we get that
$$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
and
$$Y=frac{varrho^2}{varrho^2+x_3}$$
are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
$endgroup$
It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
begin{align*}
x_1&={varrho sin(theta)}^2\
x_1&={varrho cos(theta)}^2\
end{align*}
we get that
$$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
and
$$Y=frac{varrho^2}{varrho^2+x_3}$$
are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
edited Jan 14 at 7:08
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
answered Jan 13 at 15:31
Xi'anXi'an
59.6k897368
59.6k897368
add a comment |
add a comment |
$begingroup$
A geometrical interpretation/intuition
You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.
The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.
See here a similar (but much simpler) calculation.
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
$endgroup$
add a comment |
$begingroup$
A geometrical interpretation/intuition
You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.
The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.
See here a similar (but much simpler) calculation.
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
$endgroup$
add a comment |
$begingroup$
A geometrical interpretation/intuition
You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.
The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.
See here a similar (but much simpler) calculation.
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
$endgroup$
A geometrical interpretation/intuition
You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.
The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.
See here a similar (but much simpler) calculation.
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
edited Jan 15 at 15:09
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
answered Jan 15 at 14:33
Martijn WeteringsMartijn Weterings
14.8k2164
14.8k2164
add a comment |
add a comment |
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