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Suppose that $X$ and $Y$ are two i.i.d. uniform random variables on the interval $[0,1]$

Let $Z=X/Y$, I am finding the cdf of $Z$, i.e. $ \Pr(Z\leq z) $.

Now, I came up with two ways of doing this. One produces a correct answer consistent with the pdf here: http://mathworld.wolfram.com/UniformRatioDistribution.html, the other does not. Why is the second method wrong?

First Method

$\newcommand{\rd}{\mathrm{d}} \Pr(Z\leq z) = \Pr(X/Y\leq z) = \Pr(X\leq zY) = \int^{1}_{0}\int^{\min(1,zy)}_{0} \rd x \rd y = \int^{1}_{0}\min(1,zy)\ \rd y$ $ = \left\{ \begin{array}{lr} \int^{1/z}_{0}zy\ \rd y + \int^{1}_{1/z} \rd y& : z > 1\\ \int^{1}_{0}zy\ \rd y & : z \leq 1 \end{array} \right. $ $ = \left\{ \begin{array}{lr} 1 - \frac{1}{2z} & : z > 1\\ \frac{z}{2} & : z \leq 1 \end{array} \right. $

This appears correct.

Second Method

$\Pr(X/Y\leq z) = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(zY \geq 1) + \Pr(X \leq zY\ |\ zY < 1)\Pr(zY < 1) $ by total probability

$ = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(Y \geq 1/z) + \Pr(X \leq zY\ |\ zY < 1)\Pr(Y < 1/z)$

Taking $z>1$ yields $ (1)(1-\frac{1}{z}) + (\int^{1/z}_{0}\int^{zy}_{0} \rd x \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + (\int^{1/z}_{0}zy\ \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + \frac{1}{2z^{2}}$

This is already different. Why is this wrong?

Thanks!

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Here is a hint.

Consider carefully the term $\mathbb P( X \leq z Y \mid z Y < 1 )$. In particular, for concreteness, choose $z = 2$, so that we are considering the event $\mathbb P( X \leq 2 Y \mid Y < 1/2 )$.

Now, look at this picture (which is very closely related to the above probability).

Conditional probability plot for ratio of uniforms

Now, does that conditional probability depend on our particular choice of $z$?

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  • $\begingroup$ I guess the more formal description of the picture is: Let $R = zY$ we can see that the pdf for $R$ is $1/z$ so $\Pr(X \leq R \wedge R < 1) = \int^{1}_{0}\int^{r}_{0} 1/z\ dx\ dr = \frac{1}{2z}$ But I still do not understand why setting the integral up as I did before does not work. Even if the value of z plays no real role, why does setting x's integral limit to zy and then setting y's integral limit to 1/z not rectify this? $\endgroup$ – Junier Sep 14 '11 at 2:25
  • $\begingroup$ Ah, ok I think maybe I got it. So we are going to have this contracted region {(x,y) : y < 1/z} on the unit square, then we are going to expand that very region by z so {(x,y) : y < z/z}. I.e. all the unit square again. The region where x<y is 1/2. But how do we formalize this intuition mathematically; i.e. following this contract, expand route formally? And what are some tips for avoid these types of mistakes? $\endgroup$ – Junier Sep 14 '11 at 2:58
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    $\begingroup$ @Junier Drawing a picture often helps :-). $\endgroup$ – whuber Sep 14 '11 at 4:39
  • $\begingroup$ +1 @whuber. When in doubt, draw a picture. This seems to invariably clarify problems I'm having. $\endgroup$ – Fomite Sep 14 '11 at 5:12
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    $\begingroup$ It depends on how formally you want to formalize this mathematically. First note that $\int_0^{1/z} \int_{0}^{zy} \mathrm{d}x \mathrm{d}y$ is the joint probability $\mathbb P(0 \leq X \leq zY, 0 \leq Y \leq 1/z)$, not the conditional probability you were trying to calculate. This is a pretty common mistake. Note that dividing what you have by $\mathbb P(0 \leq Y \leq 1/z) = 1/z$ recovers the correct answer. (This is just Bayes' rule.) $\endgroup$ – cardinal Sep 14 '11 at 9:06

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