# Marginals and conditional of a uniform distribution

• May 13th 2009, 01:09 PM
gablar
Marginals and conditional of a uniform distribution
Hello, I have been going over this problem for a while. I might have a correct solution but I'm not sure about it. Can anyone verify if it is correct? if is not, can you point me in the right direction. Thanks in advanced.

Suppose that (X,Y) are uniformily distributed over the interval:

$\displaystyle 0 \leq y \leq 1-x^2 , y-1 \leq x \leq 1$

I'm unsure about the limits of integration, This I think is the graph
http://g.imagehost.org/0247/graph.jpg

a) Find the marginals of X and Y:

To find $\displaystyle f_{x}(x)$ I think I have to use:

$\displaystyle \int_0^{1-x^2}\frac{1}{b-a}dy$

and obtain

$\displaystyle \frac{1-x^2}{b-a}$

for $\displaystyle f_{y}(y)$

$\displaystyle \int_{y-1}^{1}\frac{1}{b-a}dx$

and obtain

$\displaystyle \frac{2-1}{b-a}$

Again I'm unsure about the limits of integration

For the second part

b)Find both conditional density functions

$\displaystyle f(x|y)= \frac{\int_{-1}^{0}\int_{x-1}^{1-x^2} \frac{1}{b-a} dy dx}{\frac{2-1}{b-a}}$

All I have left to do is integrate and simplify to obtain the conditional density . Can someone verify the limits of integration again? I will not keep posting the details because if the analysis is right I know the rest and if the analysis is wrong , well, I don't have much time. (Worried) Any help will be greatly appreciated.
• May 14th 2009, 04:36 AM
The Second Solution
Quote:

Originally Posted by gablar
Hello, I have been going over this problem for a while. I might have a correct solution but I'm not sure about it. Can anyone verify if it is correct? if is not, can you point me in the right direction. Thanks in advanced.

Suppose that (X,Y) are uniformily distributed over the interval:

$\displaystyle 0 \leq Y \leq 1-x^2 , y-1 \leq X \leq 1$

I'm unsure about the limits of integration, This I think is the graph
http://g.imagehost.org/0247/graph.jpg

a) Find the marginals of X and Y:

To find $\displaystyle f_{x}(x)$ I think I have to use:

$\displaystyle \int_0^{1-x^2}\frac{1}{b-a}dy$

and obtain

$\displaystyle \frac{1-x^2}{b-a}$

for $\displaystyle f_{y}(y)$

$\displaystyle \int_{y-1}^{1}\frac{1}{b-a}dx$

and obtain

$\displaystyle \frac{2-1}{b-a}$

Again I'm unsure about the limits of integration

For the second part

b)Find both conditional density functions

$\displaystyle f(x|y)= \frac{\int_{-1}^{0}\int_{x-1}^{1-x^2} \frac{1}{b-a} dy dx}{\frac{2-1}{b-a}}$

All I have left to do is integrate and simplify to obtain the conditional density . Can someone verify the limits of integration again? I will not keep posting the details because if the analysis is right I know the rest and if the analysis is wrong , well, I don't have much time. (Worried) Any help will be greatly appreciated.

The bounds for the region over which the pdf f(x, y) is uniform do not make sense to me. Are you sure they are correct?
• May 14th 2009, 04:54 AM
gablar
I changed X and Y to small case as the original problem states. Othewise the bounds are the ones given.

I also found the boundaries a little weird. Is it correct that the area of integration is in the 4th cuadrant?