Mixing
Distance between two Random Variables by comparing Cumulative Distribution Functions
Suppose $X$ and $Y$ are two random variables. Define the distance between $X$ and $Y$, $d(X, Y)$ as: $$d(X, Y) = int_{-infty}^{infty}left|mathbb{P}(X < t) - mathbb{P}(Y < t)right|dt$$
whenever this integral makes sense. Does this distance have a name? (Or, do you know of any similar constructions?) I am interested in examples for which the total variation distance is large but this distance is not so large.
probability probability-distributions random-variables
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
add a comment |
Suppose $X$ and $Y$ are two random variables. Define the distance between $X$ and $Y$, $d(X, Y)$ as: $$d(X, Y) = int_{-infty}^{infty}left|mathbb{P}(X < t) - mathbb{P}(Y < t)right|dt$$
whenever this integral makes sense. Does this distance have a name? (Or, do you know of any similar constructions?) I am interested in examples for which the total variation distance is large but this distance is not so large.
probability probability-distributions random-variables
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
add a comment |
Suppose $X$ and $Y$ are two random variables. Define the distance between $X$ and $Y$, $d(X, Y)$ as: $$d(X, Y) = int_{-infty}^{infty}left|mathbb{P}(X < t) - mathbb{P}(Y < t)right|dt$$
whenever this integral makes sense. Does this distance have a name? (Or, do you know of any similar constructions?) I am interested in examples for which the total variation distance is large but this distance is not so large.
probability probability-distributions random-variables
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
Suppose $X$ and $Y$ are two random variables. Define the distance between $X$ and $Y$, $d(X, Y)$ as: $$d(X, Y) = int_{-infty}^{infty}left|mathbb{P}(X < t) - mathbb{P}(Y < t)right|dt$$
whenever this integral makes sense. Does this distance have a name? (Or, do you know of any similar constructions?) I am interested in examples for which the total variation distance is large but this distance is not so large.
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
edited Jun 18 '13 at 14:27
AbleArcher
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
asked Jun 18 '13 at 4:07
AbleArcherAbleArcher
59149
59149
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
It seems that, for any distributions $mu$ and $nu$,
$$
d(mu,nu)=inf{mathbb E(|X-Y|)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
This is called the Wasserstein distance (for the $L^1$ distance) , or the Monge-Kantorovich-Rubinstein metric, or some other name.
By comparison, the total variation distance $d_{TV}$ is defined as
$$
d_{TV}(mu,nu)=inf{mathbb P(Xne Y)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
If $mu$ and $nu$ are measures on the integers, using the inequality $mathbb 1_{xne y}leqslant|x-y|$ for integers $(x,y)$, one sees that $d_{TV}leqslant d$ (but that no inequality $dleqslant ccdot d_{TV}$ can be valid).
For measures on the real line, no inequality $d_{TV}leqslant ccdot d$ can be valid, as the example of Dirac measses at $x$ and $y$ shows, when $x-yto0$.
answered Jun 19 '13 at 13:26
DidDid
246k23221456
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
|
show 2 more comments
It looks very close to what is called the total variation distance between two probability measures.
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
|
show 1 more comment
I do not know any special name for that function; in any case in order to satisfy the property
$$d(X,Y)=0 Leftrightarrow X=Y$$
one needs to consider the equivalent classes of continuous random variables which are equal in distribution. On other constructions: usually divergences are used to introduce distance-like measure of distances between random variables. You can check "Methods of Information Geometry" by Amari and Nagaoka for all constructions and definitions. If you are searching for a distance (in the pure mathematical sense), then probably you should have a look at Information Value and Hellinger distance.
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
add a comment |
In the paper "Calculation of the Wasserstein Distance Between Probability Distributions on the Line", it is shown that your distance is precisely the Wasserstein metric with $p=1$, if your space is the real line.
In particular, given two probability measures $P_X$ and $P_Y$ on $mathbb R$ (with corresponding CDFs $F_X$ and $F_Y$), the Wasserstein metric (with $p=1$) becomes:
$$
W_1(P_X,P_Y)=int_{-infty}^{+infty}|F_X(t)-F_Y(t)|,dt,
$$
which is exactly your measure. The result can be extended to probability measures defined on $mathbb R^n$.
Moreover, if your space is bounded in $mathbb R$, $W_1$ metrizes weak convergence. That is, letting $rightharpoonup$ denote weak convergence, we have:
$$
P_nrightharpoonup Pqquadtext{if and only if}qquad lim_{nrightarrowinfty}W_1(P_n,P)=0.
$$
answered Nov 22 '18 at 11:02
CoralioCoralio
62
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f423327%2fdistance-between-two-random-variables-by-comparing-cumulative-distribution-funct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
It seems that, for any distributions $mu$ and $nu$,
$$
d(mu,nu)=inf{mathbb E(|X-Y|)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
This is called the Wasserstein distance (for the $L^1$ distance) , or the Monge-Kantorovich-Rubinstein metric, or some other name.
By comparison, the total variation distance $d_{TV}$ is defined as
$$
d_{TV}(mu,nu)=inf{mathbb P(Xne Y)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
If $mu$ and $nu$ are measures on the integers, using the inequality $mathbb 1_{xne y}leqslant|x-y|$ for integers $(x,y)$, one sees that $d_{TV}leqslant d$ (but that no inequality $dleqslant ccdot d_{TV}$ can be valid).
For measures on the real line, no inequality $d_{TV}leqslant ccdot d$ can be valid, as the example of Dirac measses at $x$ and $y$ shows, when $x-yto0$.
answered Jun 19 '13 at 13:26
DidDid
246k23221456
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
|
show 2 more comments
It seems that, for any distributions $mu$ and $nu$,
$$
d(mu,nu)=inf{mathbb E(|X-Y|)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
This is called the Wasserstein distance (for the $L^1$ distance) , or the Monge-Kantorovich-Rubinstein metric, or some other name.
By comparison, the total variation distance $d_{TV}$ is defined as
$$
d_{TV}(mu,nu)=inf{mathbb P(Xne Y)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
If $mu$ and $nu$ are measures on the integers, using the inequality $mathbb 1_{xne y}leqslant|x-y|$ for integers $(x,y)$, one sees that $d_{TV}leqslant d$ (but that no inequality $dleqslant ccdot d_{TV}$ can be valid).
For measures on the real line, no inequality $d_{TV}leqslant ccdot d$ can be valid, as the example of Dirac measses at $x$ and $y$ shows, when $x-yto0$.
answered Jun 19 '13 at 13:26
DidDid
246k23221456
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
|
show 2 more comments
It seems that, for any distributions $mu$ and $nu$,
$$
d(mu,nu)=inf{mathbb E(|X-Y|)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
This is called the Wasserstein distance (for the $L^1$ distance) , or the Monge-Kantorovich-Rubinstein metric, or some other name.
By comparison, the total variation distance $d_{TV}$ is defined as
$$
d_{TV}(mu,nu)=inf{mathbb P(Xne Y)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
If $mu$ and $nu$ are measures on the integers, using the inequality $mathbb 1_{xne y}leqslant|x-y|$ for integers $(x,y)$, one sees that $d_{TV}leqslant d$ (but that no inequality $dleqslant ccdot d_{TV}$ can be valid).
For measures on the real line, no inequality $d_{TV}leqslant ccdot d$ can be valid, as the example of Dirac measses at $x$ and $y$ shows, when $x-yto0$.
answered Jun 19 '13 at 13:26
DidDid
246k23221456
It seems that, for any distributions $mu$ and $nu$,
$$
d(mu,nu)=inf{mathbb E(|X-Y|)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
This is called the Wasserstein distance (for the $L^1$ distance) , or the Monge-Kantorovich-Rubinstein metric, or some other name.
By comparison, the total variation distance $d_{TV}$ is defined as
$$
d_{TV}(mu,nu)=inf{mathbb P(Xne Y)mid mathbb P_X=mu,mathbb P_Y=nu}.
$$
If $mu$ and $nu$ are measures on the integers, using the inequality $mathbb 1_{xne y}leqslant|x-y|$ for integers $(x,y)$, one sees that $d_{TV}leqslant d$ (but that no inequality $dleqslant ccdot d_{TV}$ can be valid).
For measures on the real line, no inequality $d_{TV}leqslant ccdot d$ can be valid, as the example of Dirac measses at $x$ and $y$ shows, when $x-yto0$.
answered Jun 19 '13 at 13:26
DidDid
246k23221456
answered Jun 19 '13 at 13:26
DidDid
246k23221456
answered Jun 19 '13 at 13:26
DidDid
246k23221456
answered Jun 19 '13 at 13:26
DidDid
246k23221456
246k23221456
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
|
show 2 more comments
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
Thanks for the info! I am unsure whether the Wasserstein distance is equivalent to the distance I alluded to (is this what you were suggesting by "it seems that"?)
– AbleArcher
Jun 20 '13 at 3:55
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
If I were you, I would try hard to show that your $d$ is indeed the Wasserstein distance.
– Did
Jun 20 '13 at 5:28
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
They may be equivalent, but I don't think they are the same, as evaluating them on a uniform distribution on $[0, 1)$ and $[1, 2)$ shows.
– AbleArcher
Jun 20 '13 at 14:43
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
For Uni(0,1) and Uni(1,2), one gets distance 1 with both formulas...
– Did
Jun 20 '13 at 16:24
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
You're right. My error.
– AbleArcher
Jun 20 '13 at 16:31
|
show 2 more comments
It looks very close to what is called the total variation distance between two probability measures.
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
|
show 1 more comment
It looks very close to what is called the total variation distance between two probability measures.
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
|
show 1 more comment
It looks very close to what is called the total variation distance between two probability measures.
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
It looks very close to what is called the total variation distance between two probability measures.
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
answered Jun 18 '13 at 6:16
Caran-d'AcheCaran-d'Ache
3,0931924
3,0931924
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
|
show 1 more comment
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
Total variation distance seems similar, but may be quite different. Take the example of two random variables that have disjoint, but interlaced, supports. Total variation will be as large as possible, but the distance described above will be relatively small (compared to supports that are widely separated on R).
– AbleArcher
Jun 18 '13 at 14:36
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang What do you mean with "disjoint but $interlaced$" supports?
– Avitus
Jun 18 '13 at 14:47
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
@thethuthinnang Do you mean, that the support of each (or one of) random variable is disjoint but for both of them they interlace? Well, this is as close as I have seen in literature (I mean the total variation distance).
– Caran-d'Ache
Jun 18 '13 at 16:27
1
1
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I think he means something like a situation where the supports of $X$ and $Y$ are $bigcup_{n=1}^infty (2n+1,2n+2)$ and $bigcup_{n=1}^infty (2n,2n+1)$ respectively.
– Rookatu
Jun 18 '13 at 19:01
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
I mean exactly the sort of situation Rookatu describes. If you add further that X is n on (2n + 1, 2n + 2), and Y is n on (2n, 2n + 1), with the measures of those intervals being the same under the corresponding probability measures, then we have a situation where two random variables are very similar, but total variation sees them as being as different as possible.
– AbleArcher
Jun 18 '13 at 21:14
|
show 1 more comment
I do not know any special name for that function; in any case in order to satisfy the property
$$d(X,Y)=0 Leftrightarrow X=Y$$
one needs to consider the equivalent classes of continuous random variables which are equal in distribution. On other constructions: usually divergences are used to introduce distance-like measure of distances between random variables. You can check "Methods of Information Geometry" by Amari and Nagaoka for all constructions and definitions. If you are searching for a distance (in the pure mathematical sense), then probably you should have a look at Information Value and Hellinger distance.
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
add a comment |
I do not know any special name for that function; in any case in order to satisfy the property
$$d(X,Y)=0 Leftrightarrow X=Y$$
one needs to consider the equivalent classes of continuous random variables which are equal in distribution. On other constructions: usually divergences are used to introduce distance-like measure of distances between random variables. You can check "Methods of Information Geometry" by Amari and Nagaoka for all constructions and definitions. If you are searching for a distance (in the pure mathematical sense), then probably you should have a look at Information Value and Hellinger distance.
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
add a comment |
I do not know any special name for that function; in any case in order to satisfy the property
$$d(X,Y)=0 Leftrightarrow X=Y$$
one needs to consider the equivalent classes of continuous random variables which are equal in distribution. On other constructions: usually divergences are used to introduce distance-like measure of distances between random variables. You can check "Methods of Information Geometry" by Amari and Nagaoka for all constructions and definitions. If you are searching for a distance (in the pure mathematical sense), then probably you should have a look at Information Value and Hellinger distance.
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
I do not know any special name for that function; in any case in order to satisfy the property
$$d(X,Y)=0 Leftrightarrow X=Y$$
one needs to consider the equivalent classes of continuous random variables which are equal in distribution. On other constructions: usually divergences are used to introduce distance-like measure of distances between random variables. You can check "Methods of Information Geometry" by Amari and Nagaoka for all constructions and definitions. If you are searching for a distance (in the pure mathematical sense), then probably you should have a look at Information Value and Hellinger distance.
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
answered Jun 18 '13 at 7:46
AvitusAvitus
11.6k11840
11.6k11840
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
add a comment |
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
This is a distance, but between distributions, not between random variables.
– Did
Jun 18 '13 at 9:51
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
@Did: sorry, but I can not see that point; in the question it is stated that $X$, $Y$ are random variables, and the distance is $d(X,Y)$.
– Avitus
Jun 18 '13 at 9:56
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
I know what is stated in the question (since I can read). My comment explains what should have been written.
– Did
Jun 18 '13 at 9:58
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
@Did happy to $read$ that!
– Avitus
Jun 18 '13 at 10:04
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
I am interested in comparing distributions, not random variables. Feel free to edit the question to reflect that. I didn't know how to word it.
– AbleArcher
Jun 18 '13 at 14:43
add a comment |
In the paper "Calculation of the Wasserstein Distance Between Probability Distributions on the Line", it is shown that your distance is precisely the Wasserstein metric with $p=1$, if your space is the real line.
In particular, given two probability measures $P_X$ and $P_Y$ on $mathbb R$ (with corresponding CDFs $F_X$ and $F_Y$), the Wasserstein metric (with $p=1$) becomes:
$$
W_1(P_X,P_Y)=int_{-infty}^{+infty}|F_X(t)-F_Y(t)|,dt,
$$
which is exactly your measure. The result can be extended to probability measures defined on $mathbb R^n$.
Moreover, if your space is bounded in $mathbb R$, $W_1$ metrizes weak convergence. That is, letting $rightharpoonup$ denote weak convergence, we have:
$$
P_nrightharpoonup Pqquadtext{if and only if}qquad lim_{nrightarrowinfty}W_1(P_n,P)=0.
$$
answered Nov 22 '18 at 11:02
CoralioCoralio
62
add a comment |
In the paper "Calculation of the Wasserstein Distance Between Probability Distributions on the Line", it is shown that your distance is precisely the Wasserstein metric with $p=1$, if your space is the real line.
In particular, given two probability measures $P_X$ and $P_Y$ on $mathbb R$ (with corresponding CDFs $F_X$ and $F_Y$), the Wasserstein metric (with $p=1$) becomes:
$$
W_1(P_X,P_Y)=int_{-infty}^{+infty}|F_X(t)-F_Y(t)|,dt,
$$
which is exactly your measure. The result can be extended to probability measures defined on $mathbb R^n$.
Moreover, if your space is bounded in $mathbb R$, $W_1$ metrizes weak convergence. That is, letting $rightharpoonup$ denote weak convergence, we have:
$$
P_nrightharpoonup Pqquadtext{if and only if}qquad lim_{nrightarrowinfty}W_1(P_n,P)=0.
$$
answered Nov 22 '18 at 11:02
CoralioCoralio
62
add a comment |
In the paper "Calculation of the Wasserstein Distance Between Probability Distributions on the Line", it is shown that your distance is precisely the Wasserstein metric with $p=1$, if your space is the real line.
In particular, given two probability measures $P_X$ and $P_Y$ on $mathbb R$ (with corresponding CDFs $F_X$ and $F_Y$), the Wasserstein metric (with $p=1$) becomes:
$$
W_1(P_X,P_Y)=int_{-infty}^{+infty}|F_X(t)-F_Y(t)|,dt,
$$
which is exactly your measure. The result can be extended to probability measures defined on $mathbb R^n$.
Moreover, if your space is bounded in $mathbb R$, $W_1$ metrizes weak convergence. That is, letting $rightharpoonup$ denote weak convergence, we have:
$$
P_nrightharpoonup Pqquadtext{if and only if}qquad lim_{nrightarrowinfty}W_1(P_n,P)=0.
$$
answered Nov 22 '18 at 11:02
CoralioCoralio
62
In the paper "Calculation of the Wasserstein Distance Between Probability Distributions on the Line", it is shown that your distance is precisely the Wasserstein metric with $p=1$, if your space is the real line.
In particular, given two probability measures $P_X$ and $P_Y$ on $mathbb R$ (with corresponding CDFs $F_X$ and $F_Y$), the Wasserstein metric (with $p=1$) becomes:
$$
W_1(P_X,P_Y)=int_{-infty}^{+infty}|F_X(t)-F_Y(t)|,dt,
$$
which is exactly your measure. The result can be extended to probability measures defined on $mathbb R^n$.
Moreover, if your space is bounded in $mathbb R$, $W_1$ metrizes weak convergence. That is, letting $rightharpoonup$ denote weak convergence, we have:
$$
P_nrightharpoonup Pqquadtext{if and only if}qquad lim_{nrightarrowinfty}W_1(P_n,P)=0.
$$
answered Nov 22 '18 at 11:02
CoralioCoralio
62
answered Nov 22 '18 at 11:02
CoralioCoralio
62
answered Nov 22 '18 at 11:02
CoralioCoralio
62
answered Nov 22 '18 at 11:02
CoralioCoralio
62
62
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f423327%2fdistance-between-two-random-variables-by-comparing-cumulative-distribution-funct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
