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"Random variables $X$,$Y$ are independent and both have uniform distribution on interval $[0,1]$. The issue is to delineate the probability denisty function of random variables $(X,\frac{Y}{X})$ and $(\frac Y X)$.

I attempted to do it as follows:

Let $(U,V)=G(X,Y)$, where $G(x,y)=(x,\frac y x)$

then

$\begin{cases} x=u & \text{} \\ y=uv, & \text{} \\ \end{cases}$

and then after some steps

$g_{(u,v)}(u,v)= g_{(x,y)}(G^{-1}(u,v)) \times detDG^{-1}$

where $detDG^{-1}$ =det$ \left[ \begin{array}{cc|c} 1&0\\ v&u \end{array} \right]=u$

therefore I got:

$g_{(u,v)}(u,v)=u\times {indicator}_{[0,1]}(u)\times {indicator}_{[0,\infty]}(v)dudv$

however if we integrate the obtained probability density function we will get $\infty$ instead of $1$, so I must have made some mistake/s. Can anyone help me by improving my reasoning or at least showing when I am making the mistakes?

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  • $\begingroup$ Wasn't this asked very recently? Anyway, the problem is that the (u,v)-domain is not [0,1]x[0,oo). If u=x and v=y/x then uv< $__$... $\endgroup$ – Did May 7 '14 at 21:00
  • $\begingroup$ I don't know if i did properly understand your comment, however I tried chaning the domain to $[0,1] \times [0,2]$, so that when I integrate this probability density function i would get 1 as it should be, but I believe that's not correct, since $\frac{Y}{X}$ can equal to any non-negative numer so to numbers greater than 2 as well. $\endgroup$ – user141766 May 7 '14 at 22:35
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If $u=x$ and $v=y/x$ and if the $(x,y)$-domain is $0<x<1$ and $0<y<1$, then $x=u$ and $y=uv$ hence the $(u,v)$-domain $D$ is defined by the inequalities $$0<u<1,\qquad0<uv<1,$$ or, equivalently, $$D=\{(u,v)\mid0<u<1,\,0<v<1/u\}. $$ Note that the integral of $u$ on $D$ is $$ \int_0^1u\int_0^{1/u}\mathrm dv\,\mathrm du=\int_0^1u\frac1u\,\mathrm du=1, $$ as it should be for the function $g$ defined by $$ g(u,v)=u\,\mathbf 1_D(u,v), $$ to be a legitimate PDF.

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