Limits of Integration for Joint Distribution problems

[trap101]

Suppose that (X,Y) is uniformly distributed over the regiondefined by 0≤ y ≤ 1-x²
and -1≤ x ≤ 1.

a) find the marginal densities of X and Y


Attempted solution:

So first I have to find the joint density function which ends up being f_xy(x,y) = 3/4

and then from that I would solve for the marginal densities. Since there was a solution I was able to do these things, but my issue is finding the limits of integration.

in finding the joint density, why did they use -1 to 1 as the limit of integration and (1-x²) to find the joint density function. Then to find the marginal densities, they used 0 to 1-x² to find the marginal density of X and ± (1-y)^1/2 to find the marginal density of Y. How and why did these limits occur?

 

[HallsofIvy]

I assume you took a "multi-variable Calculus" course before this course. That is what is being used here. You are given that the region is "defined by 0≤ y ≤ 1-x2
and -1≤ x ≤ 1." To cover that region, x has to go from -1 to 1 and, for each x, y goes from 0 to [itex]1- x^2[/tex]. Since you are also given that this is a <b>uniform</b> probability distribution, a constant over the region, whicy we can call "M", and so the integral over the entire region must be <br /> $$\int_{x=-1}^1\int_{y= 0}^{1- x^2} M dydx= 1$$<br /> Which gives M= 4/3. That's what you did isn't it?<br /> <br /> Now, look at the "inner integral":<br /> $$\frac{4}{3}\int_{y= 0}^{1- x^2} dy$$<br /> That is the integral with respect to y for fixed x. It is the probability of y for fixed x which is what is <b>meant</b> by "marginal probability".<br /> <br /> Now, look at the region described by "$-1\le x\le 1$" and "$0\le y\le 1- x^2$. That is the region under the parabola $y= 1- x^2$ between x= -1 and x= 1. Clearly y can take on values as low as 0 or as high as 1. <b>For each y</b>, then x can go from the parabola on the left to the parabola on the right. Since the parabola is given by $y= 1- x^2$, $x^2= 1- y$ and $x= \pm\sqrt{1- y}$. The left side is, of course, $x= -\sqrt{1- y}$ and the right side is $x= +\sqrt{1- y}$. <br /> <br /> That is, if you were to integrate<br /> $$\int_{y= 0}^1\int_{x=-\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dxdy$$<br /> You would get "1" again. Removing the outside integral gives the "marginal probability" of x for a given y:<br /> $$\int_{x= -\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dx$$[/itex]