1. solved

Hey, I am doing a homework exercise and I think im getting the wrong answer. I have to work out a marginal distribution from a joint distribution. So I use the definition of marginal distribution for the continous case.

$f_y(y)=\int_{X}f_{X,Y}(x,y) dx$. When I evalute my integral it gives me infinite, could this be right? Are the other ways to do it? At first the problem gave me to random variables X and Y, with density functions $\frac{-x^2}{75}-\frac{6x}{75} + \frac{8}{15}$ when 0<=x<=3 and Y distributed uniformly between half of X and two times X. To find the joint distribution I used the following formula

Where $f_{Y|X}(y|x)= Y$ when x is choosen and $f_{X}(x)= X$.

The integral im evaluating to get the marginal distribution is

Your help is greatly apreciated.

2. Re: Marginal distribution

Hey gordo151091.

You have f(Y|X)(y|x) = 1/(2x - 0.5x) since distribution of a uniform is 1/(b-a) given U(a,b). However this doesn't work when x = 0. Anyway, using fy(y) = integral over X f(x,y)(x,y) = Integral over x f(y|x)*fx(x) = 1/(2x - 0.5x) * (-x^2/75 - 6x/75 + 8/15).

We should check our pdf as well. Integral over x f(x)dx = [-x^3/225 - 3x^2/75 + 8/15x]{3,0} = -27/225 - 27/75 + 24/15 = 1.12 != 1. This means given my assumptions, you have an invalid PDF function.