Find the marginal distribution of an point randomly chosen on an ellipse












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 8:58









Henry

100k480165




100k480165










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




















  • $begingroup$
    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












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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 $x$






share|cite|improve this 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 $x$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      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 $x$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        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 $x$






        share|cite|improve this answer









        $endgroup$



        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 $x$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 9:09









        HenryHenry

        100k480165




        100k480165






























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