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I know how to find the distribution of X/Y when they are independent uniform(0,1) by drawing the integration area.

Correct answer is:

$$ P(X/Y \leqslant t) = \\ \frac{1}{2} t, t\leqslant1\\1-\frac{1}{2t},t>1 $$

But when I use bivariate transformation $U=\frac{X}{Y}$ and $V=Y$ I find the following problem.

$$ f_{U,V}(u,v) = v \text{ where }0\leqslant v \leqslant 1, 0\leqslant u \leqslant \frac{1}{v} $$

$$ f_U(u) = \int_{0}^{1} f_{U,V}(u,v)dv=\int_{0}^{1} vdv=\frac{1}{2} $$

which is the pdf when $t\leqslant1$. How to get the other part?

Thanks!

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