How Do You Calculate the Density of a Random Vector with Given Conditions?

[Krizalid1]

I'm pretty rusted with this stuff. How about a hand?

Let \(\displaystyle (U,V)\) be a random vector such that \(\displaystyle f_V(u)=\dfrac{3}{v^3}I_{[3,\infty)}(v)\) and \(\displaystyle U/V=v\) has uniform distribution over the interval \(\displaystyle ]0,3v[.\) Find:

a) A density for the random vector \(\displaystyle (U,V)\).

b) A density for the random variable \(\displaystyle U.\)

c) A density for \(\displaystyle V/U=u.\)

Thanks!

 

[Valkarie]

a) The density of the random vector (U,V) is given by the product of the densities of U and V, f_U(u) and f_V(v), respectively. Since U/V=v is uniformly distributed over the interval ]0,3v[, we have that f_U(u) = 3/(3v^2) for 0 < v < 3 and 0 otherwise. Thus, the density for (U,V) is f_U(u)f_V(v) = (3/(3v^2)) * (3/v^3)I_{[3,\infty)}(v) = 1/v^5 I_{[3,\infty)}(v).b) The density of the random variable U is given by integrating the density of the random vector (U,V) over all possible values of V: f_U(u) = \int_{3}^{\infty} 1/v^5 dv = -1/v^4\big|_{3}^{\infty} = -1/3^4.c) The density of V/U=u is given by the change of variables formula: f_{V/U}(u) = f_V(v) * |\frac{\partial u}{\partial v}| = f_V(v) * |\frac{1}{u}| = \frac{3}{u^4}I_{[3,\infty)}(v).