Joint Density problem again: What am I doing wrong?

[hwill205]

Hello,

I would gladly appreciate any and all help with this joint density problem a practice problem for an exam. Please excuse my lack of use of the proper symbols, I don't know how to express them online unfortunately :

Joint Density of two random variables, U and V is:

f(u,v)= 1/u for 0<v<u, and 0<u<1

1. Find marginal densities of U and V.

For marginal density of u, I did the integral of 1/u with respect to v from 0 to u. I got the integral to be v/u. And then when you plug in 0 and u, I got 1. This doesn't look right to me at all.

For the marginal density of v, I did the the integral of 1/u with respect to u and I got ln(u). I then plugged in 0 and 1 and I got 0. For the ln u, when I plugged in the 0, I just did the limit as u approaches 0 and that is zero. So it should be 0-0, which is zero.

Does this even look remotely correct? I don't believe it is.

2. Find E(u) and E(v)

For E(u), I just did the double integral (from 0 to 1 and 0 to u) of 1, since u*(1/u) equals 1. I got 1/2.

For E(v), I did the double integral (from 0 to 1 and 0 to u) of v/u and I got 1/8.

Again, these answers don't look right. Any and all help is GREATLY appreciated.

 

[statdad]

To find the density of $$u$$ you need to integrate the joint density with respect to $$v$$ over its range: since you know $$0 < v < u < 1$$, you know the integrals limits.

Similarly, to find the density of $$v$$ you integrate the joint density over the range of $$u$$. Try for the densities again.

 

[hwill205]

Ok I tried it again and this is what I get

marginal density of u= Integral (from 0 to u) of 1/u dv. For that I get v/u and I plug in the 0 and u. I am still getting 1.

marginal density of v= Integral (from v to 1) of 1/u du. This is ln (u). Plugging in v and 1, I got - ln(v) for my answer.


These still don't seem right.

 

[lanedance]

so for 0<u<1
$$p(u)du = (\int p(u,v) dv)du = (\int_0^u 1/u dv)du = (v/u)_0^u du = du$$
so the marginal density of u is constant between 0 and 1

note its is correctly normalised and intuitively makes sense. p(u,v) is only defined on the traingle bounded by u=0, u=v and u=1, with amplitude 1/u

The marginal density is the probailty of u, reagrdless of v. It will be proportional to the volume of a vertical sliver for the triangle, the length along the v axis will be u (as the line u=v), the length along the u axis is du, and the height is 1/u, given p(u)du = (u)(1/u)du = du

 

[statdad]

Ok I tried it again and this is what I get

marginal density of u= Integral (from 0 to u) of 1/u dv. For that I get v/u and I plug in the 0 and u. I am still getting 1.

marginal density of v= Integral (from v to 1) of 1/u du. This is ln (u). Plugging in v and 1, I got - ln(v) for my answer.


These still don't seem right.

Yes, the marginal density of u is $$g(u) = 1, \quad 0 < u < 1$$ (makes more sense when you write out the interval). And, the marginal density of v is $$h(v) = -\ln v, \quad 0 < 1$$.