# independence

• Jan 27th 2010, 08:40 PM
helpwithassgn
independence
I hope this is in the right place.

Given the joint pdf
f (u, v) = u + v , u > 0, v < 1

determine if U and V are independent

i tried to find f(u) by integrating the joint pdf from 0 to infinity... but my answer comes up to be infinity?

for v, i integrated the joint pdf from negative infinity to 1

is this the only way to find independence? f(u, v) = f(u) * f(v)?
• Jan 27th 2010, 09:11 PM
matheagle
BY inspection the RVs are dependent.
AND you need to find the correct region where the density is nonzero.
• Jan 27th 2010, 09:14 PM
helpwithassgn
Quote:

Originally Posted by matheagle
BY inspection the RVs are dependent.
AND you need to find the correct region where the density is nonzero.

Seems like they are dependent as well. But I still need to find the marginal pdf's for the other parts of the question.

What is the correct region? I was only given u > 0 and v < 1
so I thought u goes from 0 to infinity and v goes from negative infinity to 1
• Jan 27th 2010, 09:18 PM
matheagle
Quote:

Originally Posted by helpwithassgn
Seems like they are dependent as well. But I still need to find the marginal pdf's for the other parts of the question.

What is the correct region? I was only given u > 0 and v < 1
so I thought u goes from 0 to infinity and v goes from negative infinity to 1

THAT is not a valid region.
A density cannot be negative.
After you find the correct region, maybe 0<v<1 and 0<u<1.
The rvs are DEPENDENT since the joint density cannot factor into two functions, one of u and one of v.
• Jan 27th 2010, 09:26 PM
helpwithassgn
Quote:

Originally Posted by matheagle
THAT is not a valid region.
A density cannot be negative.
After you find the correct region, maybe 0<v<1 and 0<u<1.
The rvs are DEPENDENT since the joint density cannot factor into two functions, once of u and one of v.

Why can't a density be negative? If you look at a normal distribution, the pdf goes from negative infinity to positive infinity.
• Jan 27th 2010, 09:43 PM
matheagle
you are confusing the region and the FUNCTION

\$\displaystyle f(x,y)\ge 0\$ and u+v can be negative in YOUR region
SO, why don't you find the correct region.
ONCE again, I'll bet that 0<u,v<1 is the correct region.
• Jan 27th 2010, 10:04 PM
helpwithassgn
Quote:

Originally Posted by matheagle
you are confusing the region and the FUNCTION

\$\displaystyle f(x,y)\ge 0\$ and u+v can be negative in YOUR region
SO, why don't you find the correct region.
ONCE again, I'll bet that 0<u,v<1 is the correct region.

I get what you're saying now.
I think the correct region for v is 0< v < 1
but can't u be 0 < u < infinity?
• Jan 27th 2010, 10:15 PM
matheagle
Quote:

Originally Posted by helpwithassgn
I get what you're saying now.
I think the correct region for v is 0< v < 1
but can't u be 0 < u < infinity?

because the double integral must equal ONE.
These facts come from the definition of a density.
• Jan 27th 2010, 10:29 PM
helpwithassgn
Quote:

Originally Posted by matheagle
because the double integral must equal ONE.
These facts come from the defintion of a density.

Thanks, I think I should be able to solve it now. I'll be back tomorrow if I can't solve it so let's hope I get it! :)