Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2exp(-lambda y)$.
$begingroup$
Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.
I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.
Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.
However, I struggle with two things:
- How to determine the range of $g(s,t)$ function?
- How to determine the lower and upper boundry pf the integral?
probability-theory statistics probability-distributions definite-integrals marginal-distribution
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
$endgroup$
|
show 3 more comments
$begingroup$
Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.
I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.
Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.
However, I struggle with two things:
- How to determine the range of $g(s,t)$ function?
- How to determine the lower and upper boundry pf the integral?
probability-theory statistics probability-distributions definite-integrals marginal-distribution
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
$endgroup$
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
1
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15
|
show 3 more comments
$begingroup$
Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.
I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.
Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.
However, I struggle with two things:
- How to determine the range of $g(s,t)$ function?
- How to determine the lower and upper boundry pf the integral?
probability-theory statistics probability-distributions definite-integrals marginal-distribution
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
$endgroup$
Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.
I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.
Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.
However, I struggle with two things:
- How to determine the range of $g(s,t)$ function?
- How to determine the lower and upper boundry pf the integral?
probability-theory statistics probability-distributions definite-integrals marginal-distribution
probability-theory statistics probability-distributions definite-integrals marginal-distribution
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
edited Dec 15 '18 at 10:04
Davide Giraudo
127k16153268
127k16153268
asked Dec 14 '18 at 21:57
user1607user1607
1718
asked Dec 14 '18 at 21:57
user1607user1607
1718
asked Dec 14 '18 at 21:57
user1607user1607
1718
1718
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
1
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15
|
show 3 more comments
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
1
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
1
1
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.
First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
$endgroup$
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%2f3039925%2ffind-the-density-function-xy-where-x-and-y-have-a-joint-density-fx-y%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
$begingroup$
The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.
First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
$endgroup$
add a comment |
$begingroup$
The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.
First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
$endgroup$
add a comment |
$begingroup$
The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.
First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
$endgroup$
The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.
First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
answered Dec 15 '18 at 10:15
J.G.J.G.
28.8k22845
28.8k22845
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%2f3039925%2ffind-the-density-function-xy-where-x-and-y-have-a-joint-density-fx-y%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
$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02
$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04
$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18
$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25
1
$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15