# marginal distribution

• Oct 19th 2010, 09:47 PM
chutiya
marginal distribution
I have a pdf of two random variables given as $\displaystyle f(x,y) = \frac{K}{x}$ if $\displaystyle 0<y<x \mbox{; and ;} 0<x<2$, and 0 otherwise.

I found out the value of K by integrating the pdf equal to 1, that is:

$\displaystyle \int_0^2 \int_0^x \frac{K}{x} \mbox{dy} \mbox{dx} = 1$

and got k = 1/2

then I found of the marginal pdf of x by:

$\displaystyle f_{X}(x) = \int_0^x \frac{1}{2x} dy = \frac{1}{2}$

but when I tried to find the marginal pdf of y by doing:

$\displaystyle f_{Y}(x) = \int_0^2 \frac{1}{2x} dx$, I am getting infinity!!

Is that correct or did I make a mistake while finding the marginal pdf of Y??

Thanks.
• Oct 19th 2010, 10:08 PM
matheagle
the lower bound on that last integral is y
The region is 0<y<x<2.
• Oct 19th 2010, 10:19 PM
chutiya
Thank you Matheagle!

I did not grasp this material completely.

So $\displaystyle f_{Y}(y) = \int_0^x \frac{1}{2x} dx$ is what I need to evaluate?
• Oct 19th 2010, 10:21 PM
matheagle
you cannot integrate x from 0 to itself, key word is ITSELF, 0<x<x does not make sense.

You want $\displaystyle {1\over 2}\int_y^2{dx\over x}$ where 0<y<2.

which is a valid, but strange density, I checked.
• Oct 19th 2010, 10:29 PM
chutiya
Quote:

Originally Posted by matheagle
you cannot integrate x from 0 to itself, key word is ITSELF, 0<x<x does not make sense.

You want $\displaystyle {1\over 2}\int_y^2{dx\over x}$ where 0<y<2.

which is a valid, but strange density, I checked.

Thanks again Matheagle..

I got $\displaystyle f_{Y}(x) = \frac{log(2)-log(y)}{2}$

and finally I would use this to find the condiitional pdf for X given Y by:

$\displaystyle \frac{f(x,y)}{f_{Y}(y)}$

Why did you call this density strange?
• Oct 19th 2010, 10:31 PM
matheagle
It's just weird looking to me, it's not a multiple of anything I am used to seeing.