Conditions on a bivariate distribution to be the distribution of $(X_1-X_0, X_1-X_2)$, $(X_2-X_0, X_2-X_1)$...












2












$begingroup$


Consider a bivariate probability distribution $P: mathbb{R}^2rightarrow [0,1]$. I have the following questions:



Are there necessary and sufficient conditions on the cumulative distribution function (CDF) associated with $P$ (joint or marginal) ensuring that
$$
exists text{ a random vector $(X_0,X_1,X_2)$ such that }
$$
$$
(X_1-X_0, X_1-X_2), (X_2-X_0, X_2-X_1), (X_0-X_1, X_0-X_2)
$$
$$
text{ have all probability distribution $P$? }
$$




Notice:



$(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ does not imply that some of the random variables among $X_1, X_2, X_0$ are degenerate. For example, $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ is implied by $(X_0, X_1, X_2)$ exchangeable.




My thoughts: among the necessary conditions, I would list the following: let $G$ be the CDF associated with $P$ and let $G_1,G_2$ be the two marginal CDFs. Then it should be that
$$
begin{cases}
G_1 text{ is symmetric around zero, i.e., $G_1(a)=1-G_1(-a)$ $forall a in mathbb{R}$}\
G_2 text{ is symmetric around zero, i.e., $G_2(a)=1-G_2(-a)$ $forall a in mathbb{R}$}\
end{cases}
$$



Are these conditions also sufficient? If not, what else should be added to get an exhaustive set of sufficient and necessary conditions?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:40










  • $begingroup$
    the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:41










  • $begingroup$
    @antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
    $endgroup$
    – STF
    Dec 13 '18 at 12:45






  • 1




    $begingroup$
    i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
    $endgroup$
    – antkam
    Dec 13 '18 at 13:30












  • $begingroup$
    I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
    $endgroup$
    – STF
    Dec 13 '18 at 13:36


















2












$begingroup$


Consider a bivariate probability distribution $P: mathbb{R}^2rightarrow [0,1]$. I have the following questions:



Are there necessary and sufficient conditions on the cumulative distribution function (CDF) associated with $P$ (joint or marginal) ensuring that
$$
exists text{ a random vector $(X_0,X_1,X_2)$ such that }
$$
$$
(X_1-X_0, X_1-X_2), (X_2-X_0, X_2-X_1), (X_0-X_1, X_0-X_2)
$$
$$
text{ have all probability distribution $P$? }
$$




Notice:



$(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ does not imply that some of the random variables among $X_1, X_2, X_0$ are degenerate. For example, $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ is implied by $(X_0, X_1, X_2)$ exchangeable.




My thoughts: among the necessary conditions, I would list the following: let $G$ be the CDF associated with $P$ and let $G_1,G_2$ be the two marginal CDFs. Then it should be that
$$
begin{cases}
G_1 text{ is symmetric around zero, i.e., $G_1(a)=1-G_1(-a)$ $forall a in mathbb{R}$}\
G_2 text{ is symmetric around zero, i.e., $G_2(a)=1-G_2(-a)$ $forall a in mathbb{R}$}\
end{cases}
$$



Are these conditions also sufficient? If not, what else should be added to get an exhaustive set of sufficient and necessary conditions?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:40










  • $begingroup$
    the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:41










  • $begingroup$
    @antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
    $endgroup$
    – STF
    Dec 13 '18 at 12:45






  • 1




    $begingroup$
    i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
    $endgroup$
    – antkam
    Dec 13 '18 at 13:30












  • $begingroup$
    I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
    $endgroup$
    – STF
    Dec 13 '18 at 13:36
















2












2








2


1



$begingroup$


Consider a bivariate probability distribution $P: mathbb{R}^2rightarrow [0,1]$. I have the following questions:



Are there necessary and sufficient conditions on the cumulative distribution function (CDF) associated with $P$ (joint or marginal) ensuring that
$$
exists text{ a random vector $(X_0,X_1,X_2)$ such that }
$$
$$
(X_1-X_0, X_1-X_2), (X_2-X_0, X_2-X_1), (X_0-X_1, X_0-X_2)
$$
$$
text{ have all probability distribution $P$? }
$$




Notice:



$(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ does not imply that some of the random variables among $X_1, X_2, X_0$ are degenerate. For example, $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ is implied by $(X_0, X_1, X_2)$ exchangeable.




My thoughts: among the necessary conditions, I would list the following: let $G$ be the CDF associated with $P$ and let $G_1,G_2$ be the two marginal CDFs. Then it should be that
$$
begin{cases}
G_1 text{ is symmetric around zero, i.e., $G_1(a)=1-G_1(-a)$ $forall a in mathbb{R}$}\
G_2 text{ is symmetric around zero, i.e., $G_2(a)=1-G_2(-a)$ $forall a in mathbb{R}$}\
end{cases}
$$



Are these conditions also sufficient? If not, what else should be added to get an exhaustive set of sufficient and necessary conditions?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$




Consider a bivariate probability distribution $P: mathbb{R}^2rightarrow [0,1]$. I have the following questions:



Are there necessary and sufficient conditions on the cumulative distribution function (CDF) associated with $P$ (joint or marginal) ensuring that
$$
exists text{ a random vector $(X_0,X_1,X_2)$ such that }
$$
$$
(X_1-X_0, X_1-X_2), (X_2-X_0, X_2-X_1), (X_0-X_1, X_0-X_2)
$$
$$
text{ have all probability distribution $P$? }
$$




Notice:



$(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ does not imply that some of the random variables among $X_1, X_2, X_0$ are degenerate. For example, $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ is implied by $(X_0, X_1, X_2)$ exchangeable.




My thoughts: among the necessary conditions, I would list the following: let $G$ be the CDF associated with $P$ and let $G_1,G_2$ be the two marginal CDFs. Then it should be that
$$
begin{cases}
G_1 text{ is symmetric around zero, i.e., $G_1(a)=1-G_1(-a)$ $forall a in mathbb{R}$}\
G_2 text{ is symmetric around zero, i.e., $G_2(a)=1-G_2(-a)$ $forall a in mathbb{R}$}\
end{cases}
$$



Are these conditions also sufficient? If not, what else should be added to get an exhaustive set of sufficient and necessary conditions?






probability-theory probability-distributions




probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:42









Did

247k23223460




247k23223460










asked Dec 6 '18 at 11:24









STFSTF

901420




901420












  • $begingroup$
    dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:40










  • $begingroup$
    the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:41










  • $begingroup$
    @antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
    $endgroup$
    – STF
    Dec 13 '18 at 12:45






  • 1




    $begingroup$
    i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
    $endgroup$
    – antkam
    Dec 13 '18 at 13:30












  • $begingroup$
    I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
    $endgroup$
    – STF
    Dec 13 '18 at 13:36




















  • $begingroup$
    dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:40










  • $begingroup$
    the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
    $endgroup$
    – antkam
    Dec 13 '18 at 12:41










  • $begingroup$
    @antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
    $endgroup$
    – STF
    Dec 13 '18 at 12:45






  • 1




    $begingroup$
    i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
    $endgroup$
    – antkam
    Dec 13 '18 at 13:30












  • $begingroup$
    I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
    $endgroup$
    – STF
    Dec 13 '18 at 13:36


















$begingroup$
dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
$endgroup$
– antkam
Dec 13 '18 at 12:40




$begingroup$
dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$?
$endgroup$
– antkam
Dec 13 '18 at 12:40












$begingroup$
the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
$endgroup$
– antkam
Dec 13 '18 at 12:41




$begingroup$
the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure?
$endgroup$
– antkam
Dec 13 '18 at 12:41












$begingroup$
@antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
$endgroup$
– STF
Dec 13 '18 at 12:45




$begingroup$
@antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
$endgroup$
– STF
Dec 13 '18 at 12:45




1




1




$begingroup$
i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
$endgroup$
– antkam
Dec 13 '18 at 13:30






$begingroup$
i mean if you cyclically substitute $X_0 rightarrow X_1 rightarrow X_2 rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :)
$endgroup$
– antkam
Dec 13 '18 at 13:30














$begingroup$
I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
$endgroup$
– STF
Dec 13 '18 at 13:36






$begingroup$
I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
$endgroup$
– STF
Dec 13 '18 at 13:36












1 Answer
1






active

oldest

votes


















0












$begingroup$

When you have a vector of random variables, or equivalently a random variable taking values in $mathbb R^2$, we can write it as $(U,V)$ where $U$ is the $x$-coordinate and $V$ is the $y$-coordinate of the random vector. So
$$G_1(u)=mathbb P(Ule u),$$
$$G_2(v)=mathbb P(Vle v).$$



Now, in general if $X$ and $Y$ are random variables and $F_X(x)=mathbb P(Xle x)$, $F_Y(y)=mathbb P(Yle y)$, then we write $Xsim Y$ if $F_X=F_Y$.




Besides the conditions you give,



namely: if $(U,V)$ is a random variable on $R^2$ as desired then $Usim -U$ and $Vsim -V$, where $sim$ denotes "has the same distribution as",



there's also




$$V-Usim (X_0-X_2)-(X_0-X_1)= X_1-X_2sim V$$




And note that $Vsim -V$, $Usim -U$ does not imply $V-Usim V$, e.g., take $U$, $V$ to be independent standard normal $N(0,1)$ random variables:
$$mathrm{Var}(V-U)=mathrm{Var}+mathrm{Var}(U) = 2>1=mathrm{Var}(V)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
    $endgroup$
    – STF
    Dec 23 '18 at 9:38












  • $begingroup$
    Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
    $endgroup$
    – STF
    Dec 23 '18 at 9:42












  • $begingroup$
    Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
    $endgroup$
    – STF
    Dec 23 '18 at 9:47










  • $begingroup$
    @STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
    $endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23 '18 at 11:00












  • $begingroup$
    Thanks. (2) Can you make a parametric example of $V-Usim V$?
    $endgroup$
    – STF
    Dec 23 '18 at 11:08













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

When you have a vector of random variables, or equivalently a random variable taking values in $mathbb R^2$, we can write it as $(U,V)$ where $U$ is the $x$-coordinate and $V$ is the $y$-coordinate of the random vector. So
$$G_1(u)=mathbb P(Ule u),$$
$$G_2(v)=mathbb P(Vle v).$$



Now, in general if $X$ and $Y$ are random variables and $F_X(x)=mathbb P(Xle x)$, $F_Y(y)=mathbb P(Yle y)$, then we write $Xsim Y$ if $F_X=F_Y$.




Besides the conditions you give,



namely: if $(U,V)$ is a random variable on $R^2$ as desired then $Usim -U$ and $Vsim -V$, where $sim$ denotes "has the same distribution as",



there's also




$$V-Usim (X_0-X_2)-(X_0-X_1)= X_1-X_2sim V$$




And note that $Vsim -V$, $Usim -U$ does not imply $V-Usim V$, e.g., take $U$, $V$ to be independent standard normal $N(0,1)$ random variables:
$$mathrm{Var}(V-U)=mathrm{Var}+mathrm{Var}(U) = 2>1=mathrm{Var}(V)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
    $endgroup$
    – STF
    Dec 23 '18 at 9:38












  • $begingroup$
    Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
    $endgroup$
    – STF
    Dec 23 '18 at 9:42












  • $begingroup$
    Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
    $endgroup$
    – STF
    Dec 23 '18 at 9:47










  • $begingroup$
    @STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
    $endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23 '18 at 11:00












  • $begingroup$
    Thanks. (2) Can you make a parametric example of $V-Usim V$?
    $endgroup$
    – STF
    Dec 23 '18 at 11:08


















0












$begingroup$

When you have a vector of random variables, or equivalently a random variable taking values in $mathbb R^2$, we can write it as $(U,V)$ where $U$ is the $x$-coordinate and $V$ is the $y$-coordinate of the random vector. So
$$G_1(u)=mathbb P(Ule u),$$
$$G_2(v)=mathbb P(Vle v).$$



Now, in general if $X$ and $Y$ are random variables and $F_X(x)=mathbb P(Xle x)$, $F_Y(y)=mathbb P(Yle y)$, then we write $Xsim Y$ if $F_X=F_Y$.




Besides the conditions you give,



namely: if $(U,V)$ is a random variable on $R^2$ as desired then $Usim -U$ and $Vsim -V$, where $sim$ denotes "has the same distribution as",



there's also




$$V-Usim (X_0-X_2)-(X_0-X_1)= X_1-X_2sim V$$




And note that $Vsim -V$, $Usim -U$ does not imply $V-Usim V$, e.g., take $U$, $V$ to be independent standard normal $N(0,1)$ random variables:
$$mathrm{Var}(V-U)=mathrm{Var}+mathrm{Var}(U) = 2>1=mathrm{Var}(V)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
    $endgroup$
    – STF
    Dec 23 '18 at 9:38












  • $begingroup$
    Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
    $endgroup$
    – STF
    Dec 23 '18 at 9:42












  • $begingroup$
    Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
    $endgroup$
    – STF
    Dec 23 '18 at 9:47










  • $begingroup$
    @STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
    $endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23 '18 at 11:00












  • $begingroup$
    Thanks. (2) Can you make a parametric example of $V-Usim V$?
    $endgroup$
    – STF
    Dec 23 '18 at 11:08
















0












0








0





$begingroup$

When you have a vector of random variables, or equivalently a random variable taking values in $mathbb R^2$, we can write it as $(U,V)$ where $U$ is the $x$-coordinate and $V$ is the $y$-coordinate of the random vector. So
$$G_1(u)=mathbb P(Ule u),$$
$$G_2(v)=mathbb P(Vle v).$$



Now, in general if $X$ and $Y$ are random variables and $F_X(x)=mathbb P(Xle x)$, $F_Y(y)=mathbb P(Yle y)$, then we write $Xsim Y$ if $F_X=F_Y$.




Besides the conditions you give,



namely: if $(U,V)$ is a random variable on $R^2$ as desired then $Usim -U$ and $Vsim -V$, where $sim$ denotes "has the same distribution as",



there's also




$$V-Usim (X_0-X_2)-(X_0-X_1)= X_1-X_2sim V$$




And note that $Vsim -V$, $Usim -U$ does not imply $V-Usim V$, e.g., take $U$, $V$ to be independent standard normal $N(0,1)$ random variables:
$$mathrm{Var}(V-U)=mathrm{Var}+mathrm{Var}(U) = 2>1=mathrm{Var}(V)$$






share|cite|improve this answer











$endgroup$



When you have a vector of random variables, or equivalently a random variable taking values in $mathbb R^2$, we can write it as $(U,V)$ where $U$ is the $x$-coordinate and $V$ is the $y$-coordinate of the random vector. So
$$G_1(u)=mathbb P(Ule u),$$
$$G_2(v)=mathbb P(Vle v).$$



Now, in general if $X$ and $Y$ are random variables and $F_X(x)=mathbb P(Xle x)$, $F_Y(y)=mathbb P(Yle y)$, then we write $Xsim Y$ if $F_X=F_Y$.




Besides the conditions you give,



namely: if $(U,V)$ is a random variable on $R^2$ as desired then $Usim -U$ and $Vsim -V$, where $sim$ denotes "has the same distribution as",



there's also




$$V-Usim (X_0-X_2)-(X_0-X_1)= X_1-X_2sim V$$




And note that $Vsim -V$, $Usim -U$ does not imply $V-Usim V$, e.g., take $U$, $V$ to be independent standard normal $N(0,1)$ random variables:
$$mathrm{Var}(V-U)=mathrm{Var}+mathrm{Var}(U) = 2>1=mathrm{Var}(V)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 23 '18 at 17:27

























answered Dec 22 '18 at 8:04









Bjørn Kjos-HanssenBjørn Kjos-Hanssen

2,086818




2,086818












  • $begingroup$
    Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
    $endgroup$
    – STF
    Dec 23 '18 at 9:38












  • $begingroup$
    Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
    $endgroup$
    – STF
    Dec 23 '18 at 9:42












  • $begingroup$
    Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
    $endgroup$
    – STF
    Dec 23 '18 at 9:47










  • $begingroup$
    @STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
    $endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23 '18 at 11:00












  • $begingroup$
    Thanks. (2) Can you make a parametric example of $V-Usim V$?
    $endgroup$
    – STF
    Dec 23 '18 at 11:08




















  • $begingroup$
    Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
    $endgroup$
    – STF
    Dec 23 '18 at 9:38












  • $begingroup$
    Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
    $endgroup$
    – STF
    Dec 23 '18 at 9:42












  • $begingroup$
    Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
    $endgroup$
    – STF
    Dec 23 '18 at 9:47










  • $begingroup$
    @STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
    $endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23 '18 at 11:00












  • $begingroup$
    Thanks. (2) Can you make a parametric example of $V-Usim V$?
    $endgroup$
    – STF
    Dec 23 '18 at 11:08


















$begingroup$
Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
$endgroup$
– STF
Dec 23 '18 at 9:38






$begingroup$
Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
$endgroup$
– STF
Dec 23 '18 at 9:38














$begingroup$
Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
$endgroup$
– STF
Dec 23 '18 at 9:42






$begingroup$
Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ $Rightarrow$ $begin{cases}(X_1-X_2)sim( X_2-X_1)sim (X_0-X_2)\ (X_1-X_0)sim (X_2-X_0) sim (X_0-X_1) end{cases}$ $Rightarrow$ ...?
$endgroup$
– STF
Dec 23 '18 at 9:42














$begingroup$
Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
$endgroup$
– STF
Dec 23 '18 at 9:47




$begingroup$
Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)sim (X_2-X_0, X_2-X_1)sim (X_0-X_1, X_0-X_2)$.
$endgroup$
– STF
Dec 23 '18 at 9:47












$begingroup$
@STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 23 '18 at 11:00






$begingroup$
@STF I agree with you regarding (2). $V-Usim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)sim (X,Y)$ then $Usim X$ and $Vsim Y$.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 23 '18 at 11:00














$begingroup$
Thanks. (2) Can you make a parametric example of $V-Usim V$?
$endgroup$
– STF
Dec 23 '18 at 11:08






$begingroup$
Thanks. (2) Can you make a parametric example of $V-Usim V$?
$endgroup$
– STF
Dec 23 '18 at 11:08




















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