Let $X$, $Y$ and $Z$ independent uniform (0,1) random variables:
Find $P(X/Y \leq t)$ and $P(XY \leq t)$
I started with: $P(X/Y \leq t)$ by using Jacobian:
So, $Z= X/Y$, $W = Y$ , my Jacobian gave me: $w$
$F_{Z,W}(z,w) = w$ .
From now on I started to have some problems to define the interval of $F_{Z}(z)$.
Z as I can see goes from zero to infinity right?
The answer in Casella is: $F_{Z} = 1/2 (t)$ , for $t>1$ and $1/2 - (1-t)$, for $t \leq 1$
Any help guys?
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $\mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t \leq 0$, then clearly it is $0$. If $t \in (0, 1)$, then $yt \in (0, 1)$ for all $y \in (0, 1)$, hence $$P(X/Y \leq t) = \int_0^1 P(X/y \leq t) f_Y(y) dy = \int_0^1 P(X \leq yt) dy = \int_0^1 yt dy = \frac{1}{2}t.$$ Similar argument shows that for any $t > 1$, \begin{align} P(X / Y \leq t) & = \int_0^1 P(X/y \leq t) f_Y(y)dy = \int_0^1 P(X \leq yt) dy \\ & = \int_0^{t^{-1}} yt dy + \int_{t^{-1}}^1 dy = \frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - \frac{1}{2}t^{-1}. \end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.