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$X$ 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?

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Hint: Rather than use formulas, sketch the $x$-$y$ plane and the region on which the joint density is nonzero. Then, for fixed $a$, $0 \leq a \leq 1$, find the region consisting of all points $(x,y)$ such that $\frac{x}{y} \leq a$. Then $Z = X/Y \leq a$ exactly when the random point $(X,Y)$ lies in this region. What is $P\{Z \leq a\}$? (Hint: it is related to the area of the region which you should be able to find without evaluating an integral; simple mensuration will suffice). Repeat for fixed $a$, $1 < a < \infty$. Differentiate $F_Z(a)$ to get $f_Z(a)$. –  Dilip Sarwate Oct 28 '12 at 2:04
    
@Dilip Sarwate I want to understand how to use this formula to find a density. Can you provide some hints on how to solve it using the formula? –  woaini Oct 28 '12 at 2:53
    
I have no idea how your formula $$f_Z(z) = \int_{-\infty}^\infty f_X(zx)f_Y(x)|x| dx$$ was obtained. I dislike the use of such mystical magical general formulas, which you presumably found in your textbook or were "taught" in class and so is in your class notes, because students are never able to actually use the formula in any given specific case. It is far better to learn a general method such as the one outlined in my previous comment rather than rely on a canned formula. Maybe someone else can help you. –  Dilip Sarwate Oct 28 '12 at 16:52
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