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
100k480165
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
100k480165
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
100k480165
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
100k480165
asked Dec 11 '18 at 8:51
user1607user1607
1718
edited Dec 11 '18 at 8:58
Henry
100k480165
asked Dec 11 '18 at 8:51
user1607user1607
1718
edited Dec 11 '18 at 8:58
Henry
100k480165
edited Dec 11 '18 at 8:58
Henry
100k480165
edited Dec 11 '18 at 8:58
Henry
100k480165
100k480165
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
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$begingroup$
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
100k480165
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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
100k480165
$endgroup$
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
100k480165
$endgroup$
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
100k480165
$endgroup$
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
100k480165
answered Dec 11 '18 at 9:09
HenryHenry
100k480165
answered Dec 11 '18 at 9:09
HenryHenry
100k480165
answered Dec 11 '18 at 9:09
HenryHenry
100k480165
100k480165
add a comment |
add a comment |
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$begingroup$
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