The mutual density of $X,Y$ in ${|t|+|s|<1}$ is constant, are $X,Y$ independent?
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
asked Dec 26 '18 at 17:25
J. Doe
976
|
show 3 more comments
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
asked Dec 26 '18 at 17:25
J. Doe
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
|
show 3 more comments
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
asked Dec 26 '18 at 17:25
J. Doe
976
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
probability random-variables density-function absolute-continuity
asked Dec 26 '18 at 17:25
J. Doe
976
asked Dec 26 '18 at 17:25
J. Doe
976
edited Dec 26 '18 at 18:04
asked Dec 26 '18 at 17:25
J. Doe
976
asked Dec 26 '18 at 17:25
J. Doe
976
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
|
show 3 more comments
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
|
show 3 more comments
1 Answer
1
active
oldest
votes
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:
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.
answered Dec 27 '18 at 7:28
StubbornAtom
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
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%2f3053126%2fthe-mutual-density-of-x-y-in-ts1-is-constant-are-x-y-independen%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
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.
answered Dec 27 '18 at 7:28
StubbornAtom
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
add a comment |
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:
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.
answered Dec 27 '18 at 7:28
StubbornAtom
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
add a comment |
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:
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.
answered Dec 27 '18 at 7:28
StubbornAtom
5,29211138
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:
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.
answered Dec 27 '18 at 7:28
StubbornAtom
5,29211138
answered Dec 27 '18 at 7:28
StubbornAtom
5,29211138
answered Dec 27 '18 at 7:28
StubbornAtom
5,29211138
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
add a comment |
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
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%2f3053126%2fthe-mutual-density-of-x-y-in-ts1-is-constant-are-x-y-independen%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
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