If X1 X2 are independent Uniform variates on (0,1), Find the distribution of Z=X1/X2.
I tried using the CDF method where P(X1<=zX2) is equal to z/2 when z is in(0,1). However, I am unable to find the CDF when z is greater than 1
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$\begingroup$ $P(X_1\leq z X_2)=1/2$ is not correct $\forall z\in (0,1)$ $\endgroup$– gunesCommented May 12, 2020 at 14:53
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$\begingroup$ Sorry, it’s z/2 $\endgroup$– user284824Commented May 12, 2020 at 14:54
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$\begingroup$ stats.stackexchange.com/q/15522/119261, stats.stackexchange.com/q/185683/119261 $\endgroup$– StubbornAtomCommented May 12, 2020 at 15:33
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2 Answers
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We have \begin{align*} \mathbb P( Z \leq z ) &= \mathbb P(X_1 \leq zX_2) \\ &= \int_0^1 \mathbb P( X_1 \leq zx_2) dx_2 \end{align*}
When $z \leq 1$ then $\forall x_2 \in (0,1)$, $ \mathbb P( X_1 \leq zx_2)= zx_2$ and $\mathbb P( Z \leq z ) = \frac{z}{2}$.
When $z\geq 1$ you have to split the integral into two parts: a first one where $zx_2 \leq 1$ for all $x_2$ and a second one where $zx_2 \geq 1$.
Since $zx_2 \leq 1 \iff x_2 \leq \frac{1}{z}$, we have for $z\geq 1$,
\begin{align*}
\mathbb P( Z \leq z ) &=\int_0^{z^{-1}} \mathbb P( X_1 \leq zx_2) dx_2 + \int_{z^{-1}}^1 \mathbb P( X_1 \leq zx_2) dx_2 \\
&=\int_0^{z^{-1}} zx_2 dx_2 + \int_{z^{-1}}^1 dx_2 \\
&= 1 - \frac{1}{2z}
\end{align*}
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The support is the unit square. When $z>1$, the line $X_1=zX_2$ crosses the unit square where $x_1=1,x_2=1/z$. The area over the line (intersecting the unit square) is $1-{1\over {2z}}$. This area equals to the probability you ask since the PDF is uniform and equals to $1$.
