Transformation of two independent uniform random variables Suppose $X,Y \sim \text{Uniform} \left(0,1 \right)$ are independent. Then I need to find the PDF for $W=X/Y$.
By the CDF technique this is seen to be :
$$F_W( w)=\int_{0}^1 \int_{0}^{wy} \mathrm{dxdy}=w/2$$
And therefore 
$f_W (w)= 1/2 $, a uniform distribution on $(0,2)$
My question is, assuming that I have not made a mistake anywhere, what is the intuition behind this result? I would think that since $Y$ can be close to zero the quotient would be unbounded. Instead it seems that it can only take values up to $2$.
Thank you.
 A: What you want is
\begin{align}
F_W(w)&=\Pr[W\leq w]\\
&=\Pr[X/Y\leq w]\\
&=\Pr[X\leq wY]&\text{(because $Y\geq0$)}\\
&=\Pr[(X,Y)\in\{(x,y)\in(0,1)\times(0,1):x\leq wy\}].
\end{align}
Let us denote $A(w)=\{(x,y)\in(0,1)\times(0,1):x\leq wy\}.$
Then,
since $X$ and $Y$ are independent,
we see that
\begin{align*}
F_W(w)=\int_{A(w)}~d(x,y).
\end{align*}
As hinted in Michael's answer,
the way we parametrize $A(w)$ will depend on $w$ itself:
1. If $w\geq 1$,
then $x$ can take any value in $(0,1)$,
and for any fixed $x$,
$y$ has to be such that $x/w\leq y\leq1$,
thus
\begin{align*}
F_W(w)=\int_0^1\int_{x/w}^{1}~dydx=\int_0^1(1-x/w)~dx=1-1/(2w).
\end{align*}
This makes sense from a geometrical point of view,
because in this case,
$A(w)$ is the square $(0,1)\times(0,1)$ to which you remove the triangle $T$ with corners $(0,0)$, $(1,0)$, and $(1,1/w)$; and $T$ has an area of $1/2\times 1\times 1/w$.
2. If $0<w\leq 1$,
then $y$ can take any value from $0$ to $1$,
but $x$ is restricted to $0\leq x\leq wy$
\begin{align*}
F_W(w)=\int_0^1\int_{0}^{wy}~dxdy=w/2.
\end{align*}
A: If the ratio is less than 1, then I agree the PDF is 1/2.
If the ratio is above 1, then I think the PDF is $1/2w^2$.
The set of values $(x,y)$ with $w<y/x<w+dw$ is a triangle with vertices at $(0,0),(1/w,1)$ and $(1/(w+dw),1)$.  Its area is roughly $dw/(2w^2)$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#88f}{\large%
\int_{0}^{1}\int_{0}^{1}\delta\pars{w - {x \over y}}\dd x\,\dd y}
=\Theta\pars{w}\int_{0}^{1}\int_{0}^{1}{\delta\pars{x - wy} \over \verts{1/y}}
\,\dd x\,\dd y
\\[3mm]&=\Theta\pars{w}\int_{0}^{1}y\,\Theta\pars{1 - wy}\,\dd y
=\Theta\pars{w}\int_{0}^{1}y\,\Theta\pars{{1 \over w} - y}\,\dd y
\\[3mm]&=\Theta\pars{w}\Theta\pars{1 - {1 \over w}}\int_{0}^{1/w}y\,\dd y
+\Theta\pars{w}\Theta\pars{{1 \over w} - 1}\int_{0}^{1}y\,\dd y
\\[5mm]&=\Theta\pars{w}\bracks{{\Theta\pars{w - 1} \over 2w^{2}}
+{\Theta\pars{1 - w} \over 2}}
=\color{#88f}{\large\left\lbrace\begin{array}{ccl}
0 & \color{#000}{\mbox{if}} & w < 0
\\[3mm]
\half & \color{#000}{\mbox{if}} & 0 < w \leq 1
\\[3mm]
{1 \over 2w^{2}} & \color{#000}{\mbox{if}} & w > 1
\end{array}\right.}
\end{align}



