Mixing
short question about independence of random variables (specific example )
$begingroup$
Okey imagine that you have two random variables $ X,Y sim Bin(1,frac{1}{2})$, which are independent. Define $Z_1, Z_2$ with $ Z_1 := | X - Y | $ and $Z_2 = X + Y$. Show that $Z_1$ and $Z_2$ are not independent.
I already solved this exercise. So why I ask this question? A friend of mine asked me today if his solution is correct. To be honest I didn't know, so I've deciced to ask you. He said that:
it holds that $ Z_1 + Z_2 = 2X $ ,$mbox{ }$ $ Z_1 - Z_2 = - 2Y $ if $ X geq Y $ . and $Z_1+Z_2 =2Y$ , $Z_1 - Z_2 = -2X $ if $ X < Y $. Then he said that we get $Z_1 = |Z_2 - 2X | = | Z_2 - 2Y |$ so $Z_1 = pm(Z_2 - 2X) = mp (Z_2 - 2Y)$, which show that $Z_1$ and $Z_2$ are not independent? So is this really correct?
probability random-variables independence
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Okey imagine that you have two random variables $ X,Y sim Bin(1,frac{1}{2})$, which are independent. Define $Z_1, Z_2$ with $ Z_1 := | X - Y | $ and $Z_2 = X + Y$. Show that $Z_1$ and $Z_2$ are not independent.
I already solved this exercise. So why I ask this question? A friend of mine asked me today if his solution is correct. To be honest I didn't know, so I've deciced to ask you. He said that:
it holds that $ Z_1 + Z_2 = 2X $ ,$mbox{ }$ $ Z_1 - Z_2 = - 2Y $ if $ X geq Y $ . and $Z_1+Z_2 =2Y$ , $Z_1 - Z_2 = -2X $ if $ X < Y $. Then he said that we get $Z_1 = |Z_2 - 2X | = | Z_2 - 2Y |$ so $Z_1 = pm(Z_2 - 2X) = mp (Z_2 - 2Y)$, which show that $Z_1$ and $Z_2$ are not independent? So is this really correct?
probability random-variables independence
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Okey imagine that you have two random variables $ X,Y sim Bin(1,frac{1}{2})$, which are independent. Define $Z_1, Z_2$ with $ Z_1 := | X - Y | $ and $Z_2 = X + Y$. Show that $Z_1$ and $Z_2$ are not independent.
I already solved this exercise. So why I ask this question? A friend of mine asked me today if his solution is correct. To be honest I didn't know, so I've deciced to ask you. He said that:
it holds that $ Z_1 + Z_2 = 2X $ ,$mbox{ }$ $ Z_1 - Z_2 = - 2Y $ if $ X geq Y $ . and $Z_1+Z_2 =2Y$ , $Z_1 - Z_2 = -2X $ if $ X < Y $. Then he said that we get $Z_1 = |Z_2 - 2X | = | Z_2 - 2Y |$ so $Z_1 = pm(Z_2 - 2X) = mp (Z_2 - 2Y)$, which show that $Z_1$ and $Z_2$ are not independent? So is this really correct?
probability random-variables independence
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
Okey imagine that you have two random variables $ X,Y sim Bin(1,frac{1}{2})$, which are independent. Define $Z_1, Z_2$ with $ Z_1 := | X - Y | $ and $Z_2 = X + Y$. Show that $Z_1$ and $Z_2$ are not independent.
I already solved this exercise. So why I ask this question? A friend of mine asked me today if his solution is correct. To be honest I didn't know, so I've deciced to ask you. He said that:
it holds that $ Z_1 + Z_2 = 2X $ ,$mbox{ }$ $ Z_1 - Z_2 = - 2Y $ if $ X geq Y $ . and $Z_1+Z_2 =2Y$ , $Z_1 - Z_2 = -2X $ if $ X < Y $. Then he said that we get $Z_1 = |Z_2 - 2X | = | Z_2 - 2Y |$ so $Z_1 = pm(Z_2 - 2X) = mp (Z_2 - 2Y)$, which show that $Z_1$ and $Z_2$ are not independent? So is this really correct?
probability random-variables independence
probability random-variables independence
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
asked Jan 22 at 23:08
RukiaKuchikiRukiaKuchiki
337211
337211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
His argument is not correct. I don't see how he got $Z_1=|Z_2-2X|=|Z_2-2Y|$. Even if this is true (in the last line) how did he jump to the conclusion that $Z_1$ and $Z_2$ are not independent?
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
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%2f3083849%2fshort-question-about-independence-of-random-variables-specific-example%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
$begingroup$
His argument is not correct. I don't see how he got $Z_1=|Z_2-2X|=|Z_2-2Y|$. Even if this is true (in the last line) how did he jump to the conclusion that $Z_1$ and $Z_2$ are not independent?
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
add a comment |
$begingroup$
His argument is not correct. I don't see how he got $Z_1=|Z_2-2X|=|Z_2-2Y|$. Even if this is true (in the last line) how did he jump to the conclusion that $Z_1$ and $Z_2$ are not independent?
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
add a comment |
$begingroup$
His argument is not correct. I don't see how he got $Z_1=|Z_2-2X|=|Z_2-2Y|$. Even if this is true (in the last line) how did he jump to the conclusion that $Z_1$ and $Z_2$ are not independent?
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
His argument is not correct. I don't see how he got $Z_1=|Z_2-2X|=|Z_2-2Y|$. Even if this is true (in the last line) how did he jump to the conclusion that $Z_1$ and $Z_2$ are not independent?
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 22 at 23:31
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
66.6k52867
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3083849%2fshort-question-about-independence-of-random-variables-specific-example%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
