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 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 &quot;M&quot;, 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&#039;s what you did isn&#039;t it?<br /> <br /> Now, look at the &quot;inner integral&quot;:<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 &quot;marginal probability&quot;.<br /> <br /> Now, look at the region described by &quot;-1\le x\le 1&quot; and &quot;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 &quot;1&quot; again. Removing the outside integral gives the &quot;marginal probability&quot; of x for a given y:<br /> \int_{x= -\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dx