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Expectation of distance of two iid random variables
Clash Royale CLAN TAG#URR8PPP
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I'm reading up on 'energy distance' to use possibly as an analysis tool for images of different textures. I'm reading the paper "Energy statistics: A class of statistics based on distances" by Gábor J. Székely and Maria L. Rizzo.
They mention the Cramér distance, and the extension, Cramér–von Mises Smirnov distance and how they are not rotationally invariant for spaces of dim(F) > 1. This is the basic argument for why the energy distance is introduced.
The univariate Cramér-distance is defined as $$int_-infty^infty (F(x)_n-F(x)) dx$$
The univariate Cramér–von Mises Smirnov as $$int_-infty^infty (F(x)_n-F(x)) dF(x)$$
My questions regard the rotational invariance:
- What does rotationally invariant (in the multivariate case) mean here? If we identically rotate both cdfs. Fn (the empirical cdf) and F why would the C-MS distance not be the same? Do they mean only rotating one distribution at the time? And if so, what would this mean in practice? Fn is supposed to be approaching F asymptotically, right?
statistics probability-distributions
asked Jul 30 at 16:47
CupinaCoffee
106
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I'm reading up on 'energy distance' to use possibly as an analysis tool for images of different textures. I'm reading the paper "Energy statistics: A class of statistics based on distances" by Gábor J. Székely and Maria L. Rizzo.
They mention the Cramér distance, and the extension, Cramér–von Mises Smirnov distance and how they are not rotationally invariant for spaces of dim(F) > 1. This is the basic argument for why the energy distance is introduced.
The univariate Cramér-distance is defined as $$int_-infty^infty (F(x)_n-F(x)) dx$$
The univariate Cramér–von Mises Smirnov as $$int_-infty^infty (F(x)_n-F(x)) dF(x)$$
My questions regard the rotational invariance:
- What does rotationally invariant (in the multivariate case) mean here? If we identically rotate both cdfs. Fn (the empirical cdf) and F why would the C-MS distance not be the same? Do they mean only rotating one distribution at the time? And if so, what would this mean in practice? Fn is supposed to be approaching F asymptotically, right?
statistics probability-distributions
asked Jul 30 at 16:47
CupinaCoffee
106
Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10
add a comment |Â
up vote
0
down vote
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up vote
0
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favorite
I'm reading up on 'energy distance' to use possibly as an analysis tool for images of different textures. I'm reading the paper "Energy statistics: A class of statistics based on distances" by Gábor J. Székely and Maria L. Rizzo.
They mention the Cramér distance, and the extension, Cramér–von Mises Smirnov distance and how they are not rotationally invariant for spaces of dim(F) > 1. This is the basic argument for why the energy distance is introduced.
The univariate Cramér-distance is defined as $$int_-infty^infty (F(x)_n-F(x)) dx$$
The univariate Cramér–von Mises Smirnov as $$int_-infty^infty (F(x)_n-F(x)) dF(x)$$
My questions regard the rotational invariance:
- What does rotationally invariant (in the multivariate case) mean here? If we identically rotate both cdfs. Fn (the empirical cdf) and F why would the C-MS distance not be the same? Do they mean only rotating one distribution at the time? And if so, what would this mean in practice? Fn is supposed to be approaching F asymptotically, right?
statistics probability-distributions
asked Jul 30 at 16:47
CupinaCoffee
106
I'm reading up on 'energy distance' to use possibly as an analysis tool for images of different textures. I'm reading the paper "Energy statistics: A class of statistics based on distances" by Gábor J. Székely and Maria L. Rizzo.
They mention the Cramér distance, and the extension, Cramér–von Mises Smirnov distance and how they are not rotationally invariant for spaces of dim(F) > 1. This is the basic argument for why the energy distance is introduced.
The univariate Cramér-distance is defined as $$int_-infty^infty (F(x)_n-F(x)) dx$$
The univariate Cramér–von Mises Smirnov as $$int_-infty^infty (F(x)_n-F(x)) dF(x)$$
My questions regard the rotational invariance:
- What does rotationally invariant (in the multivariate case) mean here? If we identically rotate both cdfs. Fn (the empirical cdf) and F why would the C-MS distance not be the same? Do they mean only rotating one distribution at the time? And if so, what would this mean in practice? Fn is supposed to be approaching F asymptotically, right?
statistics probability-distributions
asked Jul 30 at 16:47
CupinaCoffee
106
edited Jul 30 at 17:11
asked Jul 30 at 16:47
CupinaCoffee
106
asked Jul 30 at 16:47
CupinaCoffee
106
asked Jul 30 at 16:47
CupinaCoffee
106
106
Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10
add a comment |Â
Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10
Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10
add a comment |Â
1 Answer
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Rotational invariance of a distance $D$ means that for random variables $X,Y$ and a unitary matrix $U$, that $D(X,Y)=D(UX,UY)$. Note that you are rotating the random variables, not the cdf. The cdfs $F_X$ of $X$ and $F_UX$ of $UX$ are not related in any simple way.
For example, suppose that $X$ is a point mass at $(1,2)$. Then $F_X(x)=1$ if $x$ is to the north east of $(1,2)$, and $F_X(x)=0$ otherwise. On the other hand, let $U$ be rotation by $90^circ$. Then $UX$ is a point mass at $(-2,1)$, and the cdf $F_UX$ is one to the north and east of $(-2,1)$. Notice that this is not a rotation of $F_X$.
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Rotational invariance of a distance $D$ means that for random variables $X,Y$ and a unitary matrix $U$, that $D(X,Y)=D(UX,UY)$. Note that you are rotating the random variables, not the cdf. The cdfs $F_X$ of $X$ and $F_UX$ of $UX$ are not related in any simple way.
For example, suppose that $X$ is a point mass at $(1,2)$. Then $F_X(x)=1$ if $x$ is to the north east of $(1,2)$, and $F_X(x)=0$ otherwise. On the other hand, let $U$ be rotation by $90^circ$. Then $UX$ is a point mass at $(-2,1)$, and the cdf $F_UX$ is one to the north and east of $(-2,1)$. Notice that this is not a rotation of $F_X$.
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
add a comment |Â
up vote
0
down vote
accepted
Rotational invariance of a distance $D$ means that for random variables $X,Y$ and a unitary matrix $U$, that $D(X,Y)=D(UX,UY)$. Note that you are rotating the random variables, not the cdf. The cdfs $F_X$ of $X$ and $F_UX$ of $UX$ are not related in any simple way.
For example, suppose that $X$ is a point mass at $(1,2)$. Then $F_X(x)=1$ if $x$ is to the north east of $(1,2)$, and $F_X(x)=0$ otherwise. On the other hand, let $U$ be rotation by $90^circ$. Then $UX$ is a point mass at $(-2,1)$, and the cdf $F_UX$ is one to the north and east of $(-2,1)$. Notice that this is not a rotation of $F_X$.
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Rotational invariance of a distance $D$ means that for random variables $X,Y$ and a unitary matrix $U$, that $D(X,Y)=D(UX,UY)$. Note that you are rotating the random variables, not the cdf. The cdfs $F_X$ of $X$ and $F_UX$ of $UX$ are not related in any simple way.
For example, suppose that $X$ is a point mass at $(1,2)$. Then $F_X(x)=1$ if $x$ is to the north east of $(1,2)$, and $F_X(x)=0$ otherwise. On the other hand, let $U$ be rotation by $90^circ$. Then $UX$ is a point mass at $(-2,1)$, and the cdf $F_UX$ is one to the north and east of $(-2,1)$. Notice that this is not a rotation of $F_X$.
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
Rotational invariance of a distance $D$ means that for random variables $X,Y$ and a unitary matrix $U$, that $D(X,Y)=D(UX,UY)$. Note that you are rotating the random variables, not the cdf. The cdfs $F_X$ of $X$ and $F_UX$ of $UX$ are not related in any simple way.
For example, suppose that $X$ is a point mass at $(1,2)$. Then $F_X(x)=1$ if $x$ is to the north east of $(1,2)$, and $F_X(x)=0$ otherwise. On the other hand, let $U$ be rotation by $90^circ$. Then $UX$ is a point mass at $(-2,1)$, and the cdf $F_UX$ is one to the north and east of $(-2,1)$. Notice that this is not a rotation of $F_X$.
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
answered Jul 30 at 17:15
Mike Earnest
14.8k11644
14.8k11644
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
add a comment |Â
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
So, very non-rigorous; Rotation of the possibly outcomes of a random variable changes the Cramér distance in 'unpredictable' ways (otherwise you could just add some function dependent on the rotation to keep the expression for the Cramer distance rotationally invariant). A follow-up, in the asymptotic case, (n -> $infty$): F$_n$ = F, right? So, even if we rotate the random variables, at n = $infty$ it's rotationally invariant?
– CupinaCoffee
Jul 30 at 19:53
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
@CupinaCoffee I do not know enough to answer your follow-up.
– Mike Earnest
Jul 30 at 23:50
add a comment |Â
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Sorry for the dumb question, but how is the cdf of a multivariate distribution defined here? In particular, so that it is univariate (given the integral)?
– Clement C.
Jul 30 at 17:03
The one I've written is the univariate one. The multivariate is as you'd expect. I will edit to reflect it. My bad :). Thanks for pointing it out!
– CupinaCoffee
Jul 30 at 17:08
That'd be a good idea, since for the univariate case "rotationally invariant" is not that meaningful.
– Clement C.
Jul 30 at 17:10