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 $f_{x,y} = \frac{ 1}{ \pi a b}$ ?
Best Answer
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 $x$