# Limits of Integration for Joint Distribution problems

1. Jun 23, 2012

### 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 fxy(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?

2. Jun 23, 2012

### HallsofIvy

Staff Emeritus
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 uniform probability distribution, a constant over the region, whicy we can call "M", and so the integral over the entire region must be $$\int_{x=-1}^1\int_{y= 0}^{1- x^2} M dydx= 1$$ Which gives M= 4/3. That's what you did isn't it? Now, look at the "inner integral": $$\frac{4}{3}\int_{y= 0}^{1- x^2} dy$$ That is the integral with respect to y for fixed x. It is the probability of y for fixed x which is what is meant by "marginal probability". Now, look at the region described by "[itex]-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. For each y, 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 $x= +\sqrt{1- y}$.

That is, if you were to integrate
$$\int_{y= 0}^1\int_{x=-\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dxdy$$
You would get "1" again. Removing the outside integral gives the "marginal probabilty" of x for a given y:
$$\int_{x= -\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dx$$

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