Mixing
Finding the law of X - Y when X and Y are not independent
$begingroup$
I have the following joint density: $$ f_{X,Y}(x,y) =frac{2}{(1+x)^3} (0<y<x) $$
I want to find the density of X - Y. Now, I found that $f_X(x) = frac{1}{x} (0<x<infty)$ and $ f_Y(y) = mathbb{1} (0,x)$, so X and Y are not independent.
I want to use the following formula that gives me the density function of Z = X + Y:
$$ f_Z(z) = int f_{X,Y}(x,z-x)dx $$
However I am not sure how should I change it to get the density of X - Y. I tried using $x - z$ instead of $z-x$ but it does not work.
Any hint?
probability density-function
asked Jan 26 at 12:39
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
I have the following joint density: $$ f_{X,Y}(x,y) =frac{2}{(1+x)^3} (0<y<x) $$
I want to find the density of X - Y. Now, I found that $f_X(x) = frac{1}{x} (0<x<infty)$ and $ f_Y(y) = mathbb{1} (0,x)$, so X and Y are not independent.
I want to use the following formula that gives me the density function of Z = X + Y:
$$ f_Z(z) = int f_{X,Y}(x,z-x)dx $$
However I am not sure how should I change it to get the density of X - Y. I tried using $x - z$ instead of $z-x$ but it does not work.
Any hint?
probability density-function
asked Jan 26 at 12:39
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
I have the following joint density: $$ f_{X,Y}(x,y) =frac{2}{(1+x)^3} (0<y<x) $$
I want to find the density of X - Y. Now, I found that $f_X(x) = frac{1}{x} (0<x<infty)$ and $ f_Y(y) = mathbb{1} (0,x)$, so X and Y are not independent.
I want to use the following formula that gives me the density function of Z = X + Y:
$$ f_Z(z) = int f_{X,Y}(x,z-x)dx $$
However I am not sure how should I change it to get the density of X - Y. I tried using $x - z$ instead of $z-x$ but it does not work.
Any hint?
probability density-function
asked Jan 26 at 12:39
qcc101qcc101
627213
$endgroup$
I have the following joint density: $$ f_{X,Y}(x,y) =frac{2}{(1+x)^3} (0<y<x) $$
I want to find the density of X - Y. Now, I found that $f_X(x) = frac{1}{x} (0<x<infty)$ and $ f_Y(y) = mathbb{1} (0,x)$, so X and Y are not independent.
I want to use the following formula that gives me the density function of Z = X + Y:
$$ f_Z(z) = int f_{X,Y}(x,z-x)dx $$
However I am not sure how should I change it to get the density of X - Y. I tried using $x - z$ instead of $z-x$ but it does not work.
Any hint?
probability density-function
probability density-function
asked Jan 26 at 12:39
qcc101qcc101
627213
asked Jan 26 at 12:39
qcc101qcc101
627213
asked Jan 26 at 12:39
qcc101qcc101
627213
asked Jan 26 at 12:39
qcc101qcc101
627213
asked Jan 26 at 12:39
qcc101qcc101
627213
627213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No need to write down marginal densities. $P(X-Yleq t) =int _0^{infty} int _y ^{y+t} frac 2 {(1+x)^{3}} dx dy=1-frac 1 {1+t}$ for $t geq 0$ and $0$ for $t <0$. Hence the density of $X-Y$ is $frac 1 {(1+t)^{2}}, 0<t<infty$. [ To get the limits for integration note that When $t >0$, $x-y < t$ and $0 <y<x$ are equivalent to $y<x<y+t$ and $ x >0$
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
$endgroup$
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
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%2f3088200%2ffinding-the-law-of-x-y-when-x-and-y-are-not-independent%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
$begingroup$
No need to write down marginal densities. $P(X-Yleq t) =int _0^{infty} int _y ^{y+t} frac 2 {(1+x)^{3}} dx dy=1-frac 1 {1+t}$ for $t geq 0$ and $0$ for $t <0$. Hence the density of $X-Y$ is $frac 1 {(1+t)^{2}}, 0<t<infty$. [ To get the limits for integration note that When $t >0$, $x-y < t$ and $0 <y<x$ are equivalent to $y<x<y+t$ and $ x >0$
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
$endgroup$
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
add a comment |
$begingroup$
No need to write down marginal densities. $P(X-Yleq t) =int _0^{infty} int _y ^{y+t} frac 2 {(1+x)^{3}} dx dy=1-frac 1 {1+t}$ for $t geq 0$ and $0$ for $t <0$. Hence the density of $X-Y$ is $frac 1 {(1+t)^{2}}, 0<t<infty$. [ To get the limits for integration note that When $t >0$, $x-y < t$ and $0 <y<x$ are equivalent to $y<x<y+t$ and $ x >0$
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
$endgroup$
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
add a comment |
$begingroup$
No need to write down marginal densities. $P(X-Yleq t) =int _0^{infty} int _y ^{y+t} frac 2 {(1+x)^{3}} dx dy=1-frac 1 {1+t}$ for $t geq 0$ and $0$ for $t <0$. Hence the density of $X-Y$ is $frac 1 {(1+t)^{2}}, 0<t<infty$. [ To get the limits for integration note that When $t >0$, $x-y < t$ and $0 <y<x$ are equivalent to $y<x<y+t$ and $ x >0$
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
$endgroup$
No need to write down marginal densities. $P(X-Yleq t) =int _0^{infty} int _y ^{y+t} frac 2 {(1+x)^{3}} dx dy=1-frac 1 {1+t}$ for $t geq 0$ and $0$ for $t <0$. Hence the density of $X-Y$ is $frac 1 {(1+t)^{2}}, 0<t<infty$. [ To get the limits for integration note that When $t >0$, $x-y < t$ and $0 <y<x$ are equivalent to $y<x<y+t$ and $ x >0$
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
answered Jan 26 at 12:57
Kavi Rama MurthyKavi Rama Murthy
69k53169
69k53169
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
add a comment |
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
$begingroup$
I see. Can I say that X - Y and Y are not independent even though they have the same law because X and Y are not independent?
$endgroup$
– qcc101
Jan 26 at 13:28
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%2f3088200%2ffinding-the-law-of-x-y-when-x-and-y-are-not-independent%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
