Transformation of two independent uniform random variables

Question

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.

Answer 1

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*}

Answer 2

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 $dw/(2w^2)$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}