Having X ~ Uniform(0,1), Y ~ Uniform(1,3) independent what's the pdf of Z = X/Y.
This means I can write the PDFs as follows $$f_X(x) = 1$$ for $ x \in \left(0,1\right)$ and 0 otherwise
$$f_Y(y) = \frac{1}{2}$$ for $x \in \left(1,3\right)$ and 0 otherwise.
Using the following formula for division of independent random variables:
$$f_Z(u) = \int_0^{\infty}f_Y(y)f_X(yu)dy$$
$f_x(uy)$ is non-zero when $uy \in \left(0,1\right)$ so $y \in \left(0,\frac{1}{u}\right)$. The maximum $u$ is then $\frac{1}{3}$. What would be the right domain of integration in the formula above? Should the resulting PDF be then parameterized by $u$?
Best Answer
Actually, probability density functions for $X$ and $Y$ include indicator function (using Iverson bracket notation) $$ f_X(x) = [ 0 < x < 1] \qquad f_Y(y) = \frac{1}{2} [ 1 < y < 3] $$ Now: $$\begin{eqnarray} f_Z(z) &=& \int_0^\infty y f_Y(y) f_X(z \, y) \mathrm{d} y = \frac{1}{2} \int_1^3 y [ 0 < z \, y < 1 ] \mathrm{d} y \\ &=& [0<z<1] \int_1^3 \frac{y}{2} [z \, y < 1 ] \mathrm{d} y \\ &=& [0<z<1] \frac{1}{4} \left( \min\left(3^2,\frac{1}{z^2}\right) - 1 \right)\\ &=& [0<z<1] \min\left(2,\frac{1}{4}\left(\frac{1}{z^2}-1\right)\right) \end{eqnarray} $$