Find the marginal distribution of an point randomly chosen on an ellipse
$begingroup$
This exercise comes from rice 3.6 and states: A point is chosen randomly in the interior of an ellipse:
$$ frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$
Find the marginal densities of the $x$ and $y$ coordinates of the point.
I was aware that I would need to use integration to find the marginal distribution from a joint distribution, but I was unsure how to start. I found a post here that outlines the solution as follows:
So this ought to be uniformly distributed, thus the density function for $( x , y )$ is $f_{x,y} =frac{ 1}{ pi a b} $ (where $pi ab$ is the area of the ellipse)
the limits of integration are $− frac{ b}{a} sqrt{a^2 − x^2}$ and $frac{ b}{a}
sqrt{a^2−x^2}$
How are the limits of integration determined?
why is $f_{x,y} = frac{ 1}{ pi a b}$ ?
integration statistics probability-distributions uniform-distribution bivariate-distributions
edited Dec 11 '18 at 8:58
Henry
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
This exercise comes from rice 3.6 and states: A point is chosen randomly in the interior of an ellipse:
$$ frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$
Find the marginal densities of the $x$ and $y$ coordinates of the point.
I was aware that I would need to use integration to find the marginal distribution from a joint distribution, but I was unsure how to start. I found a post here that outlines the solution as follows:
So this ought to be uniformly distributed, thus the density function for $( x , y )$ is $f_{x,y} =frac{ 1}{ pi a b} $ (where $pi ab$ is the area of the ellipse)
the limits of integration are $− frac{ b}{a} sqrt{a^2 − x^2}$ and $frac{ b}{a}
sqrt{a^2−x^2}$
How are the limits of integration determined?
why is $f_{x,y} = frac{ 1}{ pi a b}$ ?
integration statistics probability-distributions uniform-distribution bivariate-distributions
edited Dec 11 '18 at 8:58
Henry
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
$endgroup$
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15
add a comment |
$begingroup$
This exercise comes from rice 3.6 and states: A point is chosen randomly in the interior of an ellipse:
$$ frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$
Find the marginal densities of the $x$ and $y$ coordinates of the point.
I was aware that I would need to use integration to find the marginal distribution from a joint distribution, but I was unsure how to start. I found a post here that outlines the solution as follows:
So this ought to be uniformly distributed, thus the density function for $( x , y )$ is $f_{x,y} =frac{ 1}{ pi a b} $ (where $pi ab$ is the area of the ellipse)
the limits of integration are $− frac{ b}{a} sqrt{a^2 − x^2}$ and $frac{ b}{a}
sqrt{a^2−x^2}$
How are the limits of integration determined?
why is $f_{x,y} = frac{ 1}{ pi a b}$ ?
integration statistics probability-distributions uniform-distribution bivariate-distributions
edited Dec 11 '18 at 8:58
Henry
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
$endgroup$
This exercise comes from rice 3.6 and states: A point is chosen randomly in the interior of an ellipse:
$$ frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$
Find the marginal densities of the $x$ and $y$ coordinates of the point.
I was aware that I would need to use integration to find the marginal distribution from a joint distribution, but I was unsure how to start. I found a post here that outlines the solution as follows:
So this ought to be uniformly distributed, thus the density function for $( x , y )$ is $f_{x,y} =frac{ 1}{ pi a b} $ (where $pi ab$ is the area of the ellipse)
the limits of integration are $− frac{ b}{a} sqrt{a^2 − x^2}$ and $frac{ b}{a}
sqrt{a^2−x^2}$
How are the limits of integration determined?
why is $f_{x,y} = frac{ 1}{ pi a b}$ ?
integration statistics probability-distributions uniform-distribution bivariate-distributions
integration statistics probability-distributions uniform-distribution bivariate-distributions
edited Dec 11 '18 at 8:58
Henry
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
edited Dec 11 '18 at 8:58
Henry
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
edited Dec 11 '18 at 8:58
Henry
100k481167
edited Dec 11 '18 at 8:58
Henry
100k481167
edited Dec 11 '18 at 8:58
Henry
100k481167
100k481167
asked Dec 11 '18 at 8:51
user1607user1607
1718
asked Dec 11 '18 at 8:51
user1607user1607
1718
asked Dec 11 '18 at 8:51
user1607user1607
1718
1718
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15
add a comment |
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $pi a b, d = 1$, and thus the density must be $d= frac1{pi a b}$
Within the ellipse you have $frac{x^2}{a^2} + frac{y^2}{b^2} le 1$, i.e. $y^2 le frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− frac{ b}{a} sqrt{a^2 − x^2} le y le frac{ b}{a} sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
$endgroup$
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%2f3035081%2ffind-the-marginal-distribution-of-an-point-randomly-chosen-on-an-ellipse%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 distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $pi a b, d = 1$, and thus the density must be $d= frac1{pi a b}$
Within the ellipse you have $frac{x^2}{a^2} + frac{y^2}{b^2} le 1$, i.e. $y^2 le frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− frac{ b}{a} sqrt{a^2 − x^2} le y le frac{ b}{a} sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
$endgroup$
add a comment |
$begingroup$
The distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $pi a b, d = 1$, and thus the density must be $d= frac1{pi a b}$
Within the ellipse you have $frac{x^2}{a^2} + frac{y^2}{b^2} le 1$, i.e. $y^2 le frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− frac{ b}{a} sqrt{a^2 − x^2} le y le frac{ b}{a} sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
$endgroup$
add a comment |
$begingroup$
The distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $pi a b, d = 1$, and thus the density must be $d= frac1{pi a b}$
Within the ellipse you have $frac{x^2}{a^2} + frac{y^2}{b^2} le 1$, i.e. $y^2 le frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− frac{ b}{a} sqrt{a^2 − x^2} le y le frac{ b}{a} sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
$endgroup$
The distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $pi a b, d = 1$, and thus the density must be $d= frac1{pi a b}$
Within the ellipse you have $frac{x^2}{a^2} + frac{y^2}{b^2} le 1$, i.e. $y^2 le frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− frac{ b}{a} sqrt{a^2 − x^2} le y le frac{ b}{a} sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
answered Dec 11 '18 at 9:09
HenryHenry
100k481167
100k481167
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%2f3035081%2ffind-the-marginal-distribution-of-an-point-randomly-chosen-on-an-ellipse%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
$begingroup$
It is a bit sloppy to say that the density equals $1/pi ab$. It only takes that value on the set $E={(x,y)mid x^2/a^2+y^2/b^2=1}$. On $mathbb R^2-E$ it takes value $0$.
$endgroup$
– drhab
Dec 11 '18 at 9:15