The mutual density of $X,Y$ in ${|t|+|s|<1}$ is constant, are $X,Y$ independent?












0














Let $X,Y$ absolutely continuous random variables with density finctions $f_X,f_Y$. Assume that the mutual density $f_{X,Y}$ equals to a constant $c$ in ${(t,s)inmathbb{R}^2:|t|+|s|<1}$. Are $X,Y$ independent?





I guess I need to use $int_{-1}^0 int _{1+s}^{−1−s}f_{X,Y}(x,y)dxdy+int_0^1int_{1−s}^{s−1}f_{X,Y}(x,y)dxdy=c$, But I'm not sure how can it help me.



The defenition of independent random variables is random variables $X,Y$ such that $forall(t,s)in mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$.










share|cite|improve this question
























  • Your thoughts/work?
    – StubbornAtom
    Dec 26 '18 at 17:40






  • 1




    By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
    – StubbornAtom
    Dec 26 '18 at 18:28








  • 1




    Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
    – Just_to_Answer
    Dec 26 '18 at 18:34








  • 1




    and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
    – Just_to_Answer
    Dec 26 '18 at 18:36






  • 1




    An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
    – Dilip Sarwate
    Dec 26 '18 at 23:17


















0














Let $X,Y$ absolutely continuous random variables with density finctions $f_X,f_Y$. Assume that the mutual density $f_{X,Y}$ equals to a constant $c$ in ${(t,s)inmathbb{R}^2:|t|+|s|<1}$. Are $X,Y$ independent?





I guess I need to use $int_{-1}^0 int _{1+s}^{−1−s}f_{X,Y}(x,y)dxdy+int_0^1int_{1−s}^{s−1}f_{X,Y}(x,y)dxdy=c$, But I'm not sure how can it help me.



The defenition of independent random variables is random variables $X,Y$ such that $forall(t,s)in mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$.










share|cite|improve this question
























  • Your thoughts/work?
    – StubbornAtom
    Dec 26 '18 at 17:40






  • 1




    By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
    – StubbornAtom
    Dec 26 '18 at 18:28








  • 1




    Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
    – Just_to_Answer
    Dec 26 '18 at 18:34








  • 1




    and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
    – Just_to_Answer
    Dec 26 '18 at 18:36






  • 1




    An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
    – Dilip Sarwate
    Dec 26 '18 at 23:17
















0












0








0







Let $X,Y$ absolutely continuous random variables with density finctions $f_X,f_Y$. Assume that the mutual density $f_{X,Y}$ equals to a constant $c$ in ${(t,s)inmathbb{R}^2:|t|+|s|<1}$. Are $X,Y$ independent?





I guess I need to use $int_{-1}^0 int _{1+s}^{−1−s}f_{X,Y}(x,y)dxdy+int_0^1int_{1−s}^{s−1}f_{X,Y}(x,y)dxdy=c$, But I'm not sure how can it help me.



The defenition of independent random variables is random variables $X,Y$ such that $forall(t,s)in mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$.










share|cite|improve this question















Let $X,Y$ absolutely continuous random variables with density finctions $f_X,f_Y$. Assume that the mutual density $f_{X,Y}$ equals to a constant $c$ in ${(t,s)inmathbb{R}^2:|t|+|s|<1}$. Are $X,Y$ independent?





I guess I need to use $int_{-1}^0 int _{1+s}^{−1−s}f_{X,Y}(x,y)dxdy+int_0^1int_{1−s}^{s−1}f_{X,Y}(x,y)dxdy=c$, But I'm not sure how can it help me.



The defenition of independent random variables is random variables $X,Y$ such that $forall(t,s)in mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$.







probability random-variables density-function absolute-continuity






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share|cite|improve this question













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edited Dec 26 '18 at 18:04

























asked Dec 26 '18 at 17:25









J. Doe

976




976












  • Your thoughts/work?
    – StubbornAtom
    Dec 26 '18 at 17:40






  • 1




    By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
    – StubbornAtom
    Dec 26 '18 at 18:28








  • 1




    Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
    – Just_to_Answer
    Dec 26 '18 at 18:34








  • 1




    and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
    – Just_to_Answer
    Dec 26 '18 at 18:36






  • 1




    An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
    – Dilip Sarwate
    Dec 26 '18 at 23:17




















  • Your thoughts/work?
    – StubbornAtom
    Dec 26 '18 at 17:40






  • 1




    By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
    – StubbornAtom
    Dec 26 '18 at 18:28








  • 1




    Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
    – Just_to_Answer
    Dec 26 '18 at 18:34








  • 1




    and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
    – Just_to_Answer
    Dec 26 '18 at 18:36






  • 1




    An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
    – Dilip Sarwate
    Dec 26 '18 at 23:17


















Your thoughts/work?
– StubbornAtom
Dec 26 '18 at 17:40




Your thoughts/work?
– StubbornAtom
Dec 26 '18 at 17:40




1




1




By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
– StubbornAtom
Dec 26 '18 at 18:28






By sketching a picture of the region $S={(x,y):|x|+|y|<1}$ you will see that Support$(X_1)times$ Support$(X_2)ne S$, violating a necessary condition of independence of $X_1$ and $X_2$.
– StubbornAtom
Dec 26 '18 at 18:28






1




1




Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
– Just_to_Answer
Dec 26 '18 at 18:34






Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/…
– Just_to_Answer
Dec 26 '18 at 18:34






1




1




and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
– Just_to_Answer
Dec 26 '18 at 18:36




and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero.
– Just_to_Answer
Dec 26 '18 at 18:36




1




1




An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
– Dilip Sarwate
Dec 26 '18 at 23:17






An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane.
– Dilip Sarwate
Dec 26 '18 at 23:17












1 Answer
1






active

oldest

votes


















1














The joint support of $(X,Y)$ is the set



$$S=left{(x,y)inmathbb R^2:|x|+|y|le 1right}$$



Sketch the region $S$. It should look like this picture:



enter image description here



Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,



$$S_1={xinmathbb R:-1le xle 1}={yinmathbb R:-1le yle 1}=S_2$$



A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.




That is, $X$ and $Y$ are independent only if $text{supp}(X,Y)=text{supp}(X)times text{supp}(Y)$.




[Simplest example: Consider $(X,Y)$ uniform on the unit square.]



Here of course $Sne S_1times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.



Equivalently, observe that $$P{0.5le Xle 1,0.5le Yle 1}=0ne P{0.5le Xle 1}P{0.5le Yle 1}$$



So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.



Also see this relevant answer from Dilip Sarwate.






share|cite|improve this answer





















  • I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
    – J. Doe
    Dec 27 '18 at 12:22












  • @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
    – StubbornAtom
    Dec 27 '18 at 12:34












  • @J.Doe "contradict the support of $f_{X,Y}$" ??
    – StubbornAtom
    Dec 27 '18 at 12:40











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1 Answer
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1














The joint support of $(X,Y)$ is the set



$$S=left{(x,y)inmathbb R^2:|x|+|y|le 1right}$$



Sketch the region $S$. It should look like this picture:



enter image description here



Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,



$$S_1={xinmathbb R:-1le xle 1}={yinmathbb R:-1le yle 1}=S_2$$



A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.




That is, $X$ and $Y$ are independent only if $text{supp}(X,Y)=text{supp}(X)times text{supp}(Y)$.




[Simplest example: Consider $(X,Y)$ uniform on the unit square.]



Here of course $Sne S_1times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.



Equivalently, observe that $$P{0.5le Xle 1,0.5le Yle 1}=0ne P{0.5le Xle 1}P{0.5le Yle 1}$$



So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.



Also see this relevant answer from Dilip Sarwate.






share|cite|improve this answer





















  • I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
    – J. Doe
    Dec 27 '18 at 12:22












  • @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
    – StubbornAtom
    Dec 27 '18 at 12:34












  • @J.Doe "contradict the support of $f_{X,Y}$" ??
    – StubbornAtom
    Dec 27 '18 at 12:40
















1














The joint support of $(X,Y)$ is the set



$$S=left{(x,y)inmathbb R^2:|x|+|y|le 1right}$$



Sketch the region $S$. It should look like this picture:



enter image description here



Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,



$$S_1={xinmathbb R:-1le xle 1}={yinmathbb R:-1le yle 1}=S_2$$



A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.




That is, $X$ and $Y$ are independent only if $text{supp}(X,Y)=text{supp}(X)times text{supp}(Y)$.




[Simplest example: Consider $(X,Y)$ uniform on the unit square.]



Here of course $Sne S_1times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.



Equivalently, observe that $$P{0.5le Xle 1,0.5le Yle 1}=0ne P{0.5le Xle 1}P{0.5le Yle 1}$$



So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.



Also see this relevant answer from Dilip Sarwate.






share|cite|improve this answer





















  • I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
    – J. Doe
    Dec 27 '18 at 12:22












  • @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
    – StubbornAtom
    Dec 27 '18 at 12:34












  • @J.Doe "contradict the support of $f_{X,Y}$" ??
    – StubbornAtom
    Dec 27 '18 at 12:40














1












1








1






The joint support of $(X,Y)$ is the set



$$S=left{(x,y)inmathbb R^2:|x|+|y|le 1right}$$



Sketch the region $S$. It should look like this picture:



enter image description here



Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,



$$S_1={xinmathbb R:-1le xle 1}={yinmathbb R:-1le yle 1}=S_2$$



A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.




That is, $X$ and $Y$ are independent only if $text{supp}(X,Y)=text{supp}(X)times text{supp}(Y)$.




[Simplest example: Consider $(X,Y)$ uniform on the unit square.]



Here of course $Sne S_1times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.



Equivalently, observe that $$P{0.5le Xle 1,0.5le Yle 1}=0ne P{0.5le Xle 1}P{0.5le Yle 1}$$



So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.



Also see this relevant answer from Dilip Sarwate.






share|cite|improve this answer












The joint support of $(X,Y)$ is the set



$$S=left{(x,y)inmathbb R^2:|x|+|y|le 1right}$$



Sketch the region $S$. It should look like this picture:



enter image description here



Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,



$$S_1={xinmathbb R:-1le xle 1}={yinmathbb R:-1le yle 1}=S_2$$



A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.




That is, $X$ and $Y$ are independent only if $text{supp}(X,Y)=text{supp}(X)times text{supp}(Y)$.




[Simplest example: Consider $(X,Y)$ uniform on the unit square.]



Here of course $Sne S_1times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.



Equivalently, observe that $$P{0.5le Xle 1,0.5le Yle 1}=0ne P{0.5le Xle 1}P{0.5le Yle 1}$$



So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.



Also see this relevant answer from Dilip Sarwate.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 7:28









StubbornAtom

5,29211138




5,29211138












  • I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
    – J. Doe
    Dec 27 '18 at 12:22












  • @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
    – StubbornAtom
    Dec 27 '18 at 12:34












  • @J.Doe "contradict the support of $f_{X,Y}$" ??
    – StubbornAtom
    Dec 27 '18 at 12:40


















  • I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
    – J. Doe
    Dec 27 '18 at 12:22












  • @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
    – StubbornAtom
    Dec 27 '18 at 12:34












  • @J.Doe "contradict the support of $f_{X,Y}$" ??
    – StubbornAtom
    Dec 27 '18 at 12:40
















I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
– J. Doe
Dec 27 '18 at 12:22






I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom
– J. Doe
Dec 27 '18 at 12:22














@J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
– StubbornAtom
Dec 27 '18 at 12:34






@J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$.
– StubbornAtom
Dec 27 '18 at 12:34














@J.Doe "contradict the support of $f_{X,Y}$" ??
– StubbornAtom
Dec 27 '18 at 12:40




@J.Doe "contradict the support of $f_{X,Y}$" ??
– StubbornAtom
Dec 27 '18 at 12:40


















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