joint PDF of square of 2 dependent random variables & independence of squares of those random variables
Say we are given some joint PDF $f_{X,Y}(x, y)$ over a set: $-1 < X, Y < 1$, and we know that $X, Y$ are NOT independent (we could have verified this by finding the marginal PDFs and showing that at some point in the set, the product of the marginal PDFs does not equal the value at that point of the joint PDF).
Now we want to show the squares of these variables $X^2, Y^2$ are independent over the same set. I know how to find the PMFs of $X^2$ and $Y^2$ individually (from their marginal distributions), but I'm not sure how to show they are independent -- wouldn't this require finding the joint PDF of $X^2$ and $Y^2$? Could someone please walk through how to find that PMF (my intuition is telling me it would somehow involve the joint PDF of just $X, Y$, but I'm not sure how to manipulate it)?
Out of curiosity, is there a (reasonably easy) way to show they are independent without finding the joint PDF of $X^2$ and $Y^2$? If so, what is it?
Edit: $X, Y$ are continuous.
probability probability-theory statistics probability-distributions
asked Nov 26 '18 at 17:41
0k33
12010
|
show 1 more comment
Say we are given some joint PDF $f_{X,Y}(x, y)$ over a set: $-1 < X, Y < 1$, and we know that $X, Y$ are NOT independent (we could have verified this by finding the marginal PDFs and showing that at some point in the set, the product of the marginal PDFs does not equal the value at that point of the joint PDF).
Now we want to show the squares of these variables $X^2, Y^2$ are independent over the same set. I know how to find the PMFs of $X^2$ and $Y^2$ individually (from their marginal distributions), but I'm not sure how to show they are independent -- wouldn't this require finding the joint PDF of $X^2$ and $Y^2$? Could someone please walk through how to find that PMF (my intuition is telling me it would somehow involve the joint PDF of just $X, Y$, but I'm not sure how to manipulate it)?
Out of curiosity, is there a (reasonably easy) way to show they are independent without finding the joint PDF of $X^2$ and $Y^2$? If so, what is it?
Edit: $X, Y$ are continuous.
probability probability-theory statistics probability-distributions
asked Nov 26 '18 at 17:41
0k33
12010
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
1
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
1
Related.
– Did
Nov 26 '18 at 23:02
|
show 1 more comment
Say we are given some joint PDF $f_{X,Y}(x, y)$ over a set: $-1 < X, Y < 1$, and we know that $X, Y$ are NOT independent (we could have verified this by finding the marginal PDFs and showing that at some point in the set, the product of the marginal PDFs does not equal the value at that point of the joint PDF).
Now we want to show the squares of these variables $X^2, Y^2$ are independent over the same set. I know how to find the PMFs of $X^2$ and $Y^2$ individually (from their marginal distributions), but I'm not sure how to show they are independent -- wouldn't this require finding the joint PDF of $X^2$ and $Y^2$? Could someone please walk through how to find that PMF (my intuition is telling me it would somehow involve the joint PDF of just $X, Y$, but I'm not sure how to manipulate it)?
Out of curiosity, is there a (reasonably easy) way to show they are independent without finding the joint PDF of $X^2$ and $Y^2$? If so, what is it?
Edit: $X, Y$ are continuous.
probability probability-theory statistics probability-distributions
asked Nov 26 '18 at 17:41
0k33
12010
Say we are given some joint PDF $f_{X,Y}(x, y)$ over a set: $-1 < X, Y < 1$, and we know that $X, Y$ are NOT independent (we could have verified this by finding the marginal PDFs and showing that at some point in the set, the product of the marginal PDFs does not equal the value at that point of the joint PDF).
Now we want to show the squares of these variables $X^2, Y^2$ are independent over the same set. I know how to find the PMFs of $X^2$ and $Y^2$ individually (from their marginal distributions), but I'm not sure how to show they are independent -- wouldn't this require finding the joint PDF of $X^2$ and $Y^2$? Could someone please walk through how to find that PMF (my intuition is telling me it would somehow involve the joint PDF of just $X, Y$, but I'm not sure how to manipulate it)?
Out of curiosity, is there a (reasonably easy) way to show they are independent without finding the joint PDF of $X^2$ and $Y^2$? If so, what is it?
Edit: $X, Y$ are continuous.
probability probability-theory statistics probability-distributions
probability probability-theory statistics probability-distributions
asked Nov 26 '18 at 17:41
0k33
12010
asked Nov 26 '18 at 17:41
0k33
12010
edited Nov 26 '18 at 19:06
asked Nov 26 '18 at 17:41
0k33
12010
asked Nov 26 '18 at 17:41
0k33
12010
asked Nov 26 '18 at 17:41
0k33
12010
12010
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
1
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
1
Related.
– Did
Nov 26 '18 at 23:02
|
show 1 more comment
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
1
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
1
Related.
– Did
Nov 26 '18 at 23:02
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
1
1
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
1
1
Related.
– Did
Nov 26 '18 at 23:02
Related.
– Did
Nov 26 '18 at 23:02
|
show 1 more comment
1 Answer
1
active
oldest
votes
Theorem. For two absolutely continuous random variables $Z$ and $R$ to be independent, it is necessary and sufficient, that their corresponding densities $f_Z$ and $f_R$ should satisfy that the function $(z, r) mapsto f_Z(z) f_R(r)$ is a density of the random vector $(Z, R).$
Proof. Is done by the usual way; to be found in any book of probability based on measure theory. Q.E.D.
answered Nov 26 '18 at 18:30
Will M.
2,377314
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
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%2f3014648%2fjoint-pdf-of-square-of-2-dependent-random-variables-independence-of-squares-of%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
Theorem. For two absolutely continuous random variables $Z$ and $R$ to be independent, it is necessary and sufficient, that their corresponding densities $f_Z$ and $f_R$ should satisfy that the function $(z, r) mapsto f_Z(z) f_R(r)$ is a density of the random vector $(Z, R).$
Proof. Is done by the usual way; to be found in any book of probability based on measure theory. Q.E.D.
answered Nov 26 '18 at 18:30
Will M.
2,377314
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
add a comment |
Theorem. For two absolutely continuous random variables $Z$ and $R$ to be independent, it is necessary and sufficient, that their corresponding densities $f_Z$ and $f_R$ should satisfy that the function $(z, r) mapsto f_Z(z) f_R(r)$ is a density of the random vector $(Z, R).$
Proof. Is done by the usual way; to be found in any book of probability based on measure theory. Q.E.D.
answered Nov 26 '18 at 18:30
Will M.
2,377314
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
add a comment |
Theorem. For two absolutely continuous random variables $Z$ and $R$ to be independent, it is necessary and sufficient, that their corresponding densities $f_Z$ and $f_R$ should satisfy that the function $(z, r) mapsto f_Z(z) f_R(r)$ is a density of the random vector $(Z, R).$
Proof. Is done by the usual way; to be found in any book of probability based on measure theory. Q.E.D.
answered Nov 26 '18 at 18:30
Will M.
2,377314
Theorem. For two absolutely continuous random variables $Z$ and $R$ to be independent, it is necessary and sufficient, that their corresponding densities $f_Z$ and $f_R$ should satisfy that the function $(z, r) mapsto f_Z(z) f_R(r)$ is a density of the random vector $(Z, R).$
Proof. Is done by the usual way; to be found in any book of probability based on measure theory. Q.E.D.
answered Nov 26 '18 at 18:30
Will M.
2,377314
answered Nov 26 '18 at 18:30
Will M.
2,377314
answered Nov 26 '18 at 18:30
Will M.
2,377314
answered Nov 26 '18 at 18:30
Will M.
2,377314
2,377314
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
add a comment |
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
Did you read the question at all?
– Did
Nov 26 '18 at 18:33
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Will M. Sure. I'm familiar with that result and in fact used it to show that $X, Y$ are dependent. My question primarily concerns how to find the joint PDF of $X^2$ and $Y^2$.
– 0k33
Nov 26 '18 at 18:57
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
@Did I Did (see what I Did there?) and this part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 4:25
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
If you did read the question, how comes your post does not even begin to scratch its surface?
– Did
Nov 27 '18 at 7:28
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
Oops, I think I Did not explain myself correctly. This part is confusing: "I know how to find the PMFs of X2 and Y2 individually (from their marginal distributions), but I'm not sure how to show they are independent"
– Will M.
Nov 27 '18 at 17:20
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%2f3014648%2fjoint-pdf-of-square-of-2-dependent-random-variables-independence-of-squares-of%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
For discrete random variables X and Y?
– Did
Nov 26 '18 at 18:34
@Did no, continuous!
– 0k33
Nov 26 '18 at 19:00
@Did where I'm stumped is I can't seem to use the standard change-of-variables technique involving the Jacobian, etc. to obtain the joint pdf of the squares, as the square function isn't one-to-one on the interval [-1, 1] (I forgot to add that in the original statement, but have added it now -- sorry for the lack of detail).
– 0k33
Nov 26 '18 at 19:04
1
The method you mention yields $$f_{X^2,Y^2}(u,v)=frac1{sqrt{4uv}}g(u,v)mathbf 1_{u>0,v>0}$$ where $$g(u,v)=f_{X,Y}(sqrt u,sqrt v)+f_{X,Y}(sqrt u,-sqrt v)+f_{X,Y}(-sqrt u,sqrt v)+f_{X,Y}(-sqrt u,-sqrt v)$$
– Did
Nov 26 '18 at 22:51
1
Related.
– Did
Nov 26 '18 at 23:02