Finding joint distribution (problem with limits of integration)

Question

I don't know how to advance in the following problem:

Let $X$, $Y$ and $Z$ independent random variables equally distributed with uniform distribution over $[0,1]$.

Find the joint pdf of $W:=XY$ and $V:=Z^2$.

I tried to answer this problem by declaring a new random variable $U:= Y$ (to my opinion necessary to get the transformation).

Then:
$w=xy,$
$v=z^2,$
$u=y.$

We can see that dividing the first equation by the second one:
$x=\dfrac{w}{u},$
$y=u,$
$z=\sqrt{v}.$

Consider the transformation $h(x(w,v,u),y(w,v,u),z(w,v,u))=\left(\dfrac{w}{u},u,\sqrt{v}\right)$ gives us

$$f_{WVU}(w,v,u)=|\boldsymbol{J(h)}|f_{XYZ}(h(x,t,z))=\frac{1}{2u\sqrt{v}}.$$

To find the pdf of $W,V$: $$f_{WV}(w,v)=\int_u\frac{1}{2u\sqrt{v}}du.$$ However I don't know what limits to use. Any help is appreciated.

See e.g. http://math.stackexchange.com/questions/375967/probability-density-function-of-a-product-of-uniform-random-variables — Clement C., Nov 04 '14 at 02:01
@ClementC. Yes, my problem was just with the limits of integration. Thanks. — Vladimir Vargas, Nov 04 '14 at 02:04

score 1 · Accepted Answer · answered Nov 04 '14 at 01:54

I don't know what limits to use.

Note that $x=w/u$, $y=u$, $z=\sqrt{v}$ with $0\leqslant x,y,z\leqslant1$ hence the domain of integration is $$0\leqslant w/u,u,\sqrt{v}\leqslant1,$$ or, equivalently, $$0\leqslant w\leqslant u\leqslant1,\qquad0\leqslant v\leqslant1.$$

Find the joint pdf of $W:=XY$ and $V:=Z^2$.

This can be simplified by noting that $W$ and $V$ are independent hence their marginal densities suffice to solve the question.

score 0 · Answer 2 · answered Nov 04 '14 at 02:03

0

Notice that:

$$f_{WVU}(w,v,u)=|\boldsymbol{J(h)}|f_{XYZ}(h(x,t,z))=|\boldsymbol{J(h)}|f_X\left(\dfrac{w}{u}\right)\chi_{[0,1]}(w)f_Y(u)\chi_{[w,1]}(u)f_Z(\sqrt{v})\chi_{[0,1]}(v).$$

answered Nov 04 '14 at 02:03

Vladimir Vargas

4,655

Finding joint distribution (problem with limits of integration)

2 Answers2