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Let $X,Y$ be independent Uniform$(0,1)$ random variables. Set $Z = \frac{\min\{X,Y\}}{\max\{X,Y\}}$. Find the distribution of $Z$.

So I need to calculate $\mathbb{P}(Z \le z)$. I wasn't sure if I should use:

$\max\{X,Y\} = \frac{1}{2}(X+Y + \lvert X -Y \rvert)$ and $\min \{X,Y\} = \frac{1}{2}(X+Y - \lvert X -Y \rvert)$.

If I do this is seems to make it more complicated than maybe it should be. I haven't done any problems with a ratio or min and max yet.

StubbornAtom
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oliverjones
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    A more general question about the inverse of this ratio for arbitrary distribution: https://math.stackexchange.com/q/3814215/16192 – gt6989b Sep 04 '20 at 15:40
  • Distribution of an inverse of this ratio https://math.stackexchange.com/questions/662601/if-x-and-y-are-uniformly-distributed-on-0-1-what-is-the-distribution-of/662630#662630 – gt6989b Sep 04 '20 at 17:14

1 Answers1

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For each $z\in (0,1)$,

\begin{align} P(Z\leqslant z)&=P(Z\leqslant z, X\geqslant Y)+P(Z\leqslant z, X<Y) \\&=P\left(\frac{Y}{X}\leqslant z, X\geqslant Y\right)+P\left(\frac{X}{Y}\leqslant z, X<Y \right) \\&=2P\left(\frac{X}{Y}\leqslant z, X<Y \right) \\&=2\int_0^1 P\left(X\leqslant yz, X<y\mid Y=y\right)\, dy \\&=2\int_0^1 P(X\leqslant yz)\, dy \\&=2\int_0^1 yz\,dy \\&=z \end{align}

Hence $Z$ is also a $U(0,1)$ variable.

StubbornAtom
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  • Can you explain this? It isn't quite clear to me why this is how the problem is completed. First I don't see why you drop one part of the sum and just takes a factor of 2. Or even why we can write it as a sum. – oliverjones Oct 14 '18 at 14:04
  • @oliverjones We can write the probability as a sum due to the total probability theorem (see the very first equation). The $2$ appears as both probabilities are the same. Both probabilities are same because $X$ and $Y$ are independent and identically distributed. – StubbornAtom Oct 14 '18 at 14:07
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    @oliverjones I deliberately left out the details when I posted. Let me know if anything is unclear. I would edit the answer accordingly. – StubbornAtom Oct 14 '18 at 14:24
  • Oh right that makes sense. So I see you are then conditioning by $Y = y$, why in the next line can you drop both $X <y$ and the conditioning by $Y = y$? This is uniform(0,1) so I understand the $P(X \le yz)= yz$. – oliverjones Oct 14 '18 at 15:13
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    @oliverjones I drop the condition in the next line because $X$ is independent of $Y$. And I don't drop $X<y$. Observe that for $0<z<1$ (the support of $Z$), $$X<y,X\leqslant yz\implies X\leqslant yz $$ – StubbornAtom Oct 14 '18 at 15:19