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If I have 2 uniform random numbers $r_1$ and $r_2$ in $[0,1)$, what is the distribution of $\min(r_1, r_2)$?

I have a problem I'm trying to get my arms around, and getting some context on this will help.

(BTW, my next step is to look at the minimum of a set of uniform randoms, $r_1$ to $r_n$, but this is my starting point.)

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I don't know if it helps, but $$\min(r_1,r_2)=\frac12\bigl(r_1+r_2-|r_1-r_2|\bigr).$$ –  Cameron Buie Apr 4 '13 at 19:35
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$P(\min(r_1,r_2)\geq x)=P(r_1\geq x,r_2\geq x)=P(r_1\geq x)P(r_2\geq x)$ if we assume that $r_1$ and $r_2$ are independent. –  Stefan Hansen Apr 4 '13 at 19:37
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1 Answer

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If $r_1, r_2$ $-$ independent random variables with uniform distribution on $[0,1)$, then it is easy to use geometric interpretation of probability.

Universal sample space $\Omega$ for pair $(r_1,r_2)$ is the square $[0,1) \times [0,1)$.

Cumulative distribution function for random variable $r = \min \{r_1,r_2\}$ is

$F(r_0) = P(r\leqslant r_0) = \displaystyle\frac{{\rm mes}\;A(r_0)}{{\rm mes}\;\Omega}$,
where $A(r_0)\subset \Omega$ is the subset of $\Omega$, where $\min\{r_1,r_2\}\leqslant r_0$;
other words, complement of $A(r_0)$ is $\overline {A(r_0)} = \{(r_1,r_2)\in\Omega: r_1>r_0, r_2>r_0\}$ .

In fact, $\overline {A(r_0)}$ is the square: $(r_0,1)\times (r_0,1)$.

So, ${\rm mes}\:A(r_0) = {\rm mes}\:\Omega - {\rm mes}\:\overline {A(r_0)} = 1 - (1-r_0)^2 = 2r_0 - r_0^2$.

Hence $F(r_0) = \displaystyle\frac{2r_0-r_0^2}{1} = 2r_0 - r_0^2$, if $r_0 \in (0,1)$.

Finally, cumulative distribution function is

$F(r) = \;\;\;\begin{array}{lll} 0, \;\;{\rm if}\; r<0; \\ 2r-r^2, {\rm if}\; r\in [0,1); \\ 1, \;\;{\rm if}\; r\geqslant 1. \end{array}$

Probability density function is $f(r) = F'(r)$:

$f(r)= 2-2r$, if $r\in [0,1)$; and is zero-function outside.


(In the case of $(r_1, \ldots, r_n)$ we'll consider $\Omega$ as $n$-dimensional cube. ${\rm mes} A(r_0) = 1 - (1-r_0)^n$. )

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Wonderful! Thank you very much! –  Rob Y Apr 5 '13 at 15:21
    
BTW I was going down the n-cube path last night, in a much less formal way, but tripped over the math at the end (been too long). But your solution completes what I couldn't! So that's nice. FYI the application is one of making selections from a huge set. I want to figure out what will be the first item to grab, so I'll treat the space as a 60-million-dimension cube and use this to pick the first value, and then continue that way until I've made the number of selections I need. (I think I mixed up the analogy, but you've given me the piece I need to make it work :) ) –  Rob Y Apr 5 '13 at 16:06
    
In case you're interested, here's the application of your work. Thank you again! looselyconnected.wordpress.com/2013/04/09/… –  Rob Y Apr 9 '13 at 21:50
    
@RobY: Wow! Great. Thank you too. –  Oleg567 Apr 10 '13 at 3:00
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