$X$ and $Y$ are independent standard uniform random variables. What is the density of $Z = X/Y$?
So far I have:
$$f_X(x) = f_Y(y) = 1\text{ if }0 \le x,y \le 1$$
$$\begin{align}
f_Z(z) & = \int_{-\infty}^\infty f_X(zx)f_Y(x)|x| dx \\
& = \int_{-\infty}^\infty f_X(zx)f_Y(x)x dx \\
& = \int_{-\infty}^\infty f_X(zx)x dx \\
\end{align}$$
I think that $f_X(zx)$ in the integrand can be replaced with $1$ but how would I change the bounds of the integral?
Best Answer
$\begin{align} {f}_{Z} ({z} )d{z}&=\int_{-\infty}^{\infty} {f}_{{X},{Y}} ( {x},{y} )d{x}d{y} \\& = \int_{-\infty}^{\infty} {f}_{X} ({x} ) {f}_{Y} ({y} )d{x}d{y} \tag*{by independence} \\& = \int_{-\infty}^{\infty} {f}_{X} ({y}{z} ) {f}_{Y} ({y} ){y}d{z}d{y} \tag*{${x}={y}{z}$, $d{x}=({z}+d{z}){y}-{y}{z}$} \\ \Rightarrow {f}_{Z} ({z} )&=\int_{-\infty}^{\infty} {f}_{X} ({y}{z} ) {f}_{Y} ({y} ){y}d{y}\\ &=\int_{-\infty}^{\infty} \textbf{1}_{({y}{z}\in({0},{1}))} \textbf{1}_{({y}\in({0},{1}))}{y}d{y}\\ \end{align} $ Case 1 $${f}_{Z} ({z} )\int_{0}^{1}{y}d{y}=\frac{1}{2} $$ Case 2 $${f}_{Z} ({z} )\int_{0}^{\frac{1}{z}}{y}d{y}=\frac{1}{2{z}^2} $$ $${f}_{Z} ({z} )= \left\{\begin{array}{ll} \frac{1}{2}&{z}\in({0},{1})\\ \frac{1}{{2}{z}^2}&{z}\ge{1}\\ {0}&\textrm{else}\end{array}\right. \tag*{$\color{red}{\blacksquare}$}$$ I think I did that right. Maybe later I'll add the pictures...maybe...