We say $U_1$, $U_2$ and $ U_3$ are independent uniform random numbers. We are arrange them, like $U_{(1)}$, $U_{(2)}$ and $U_{(3)}$ from small to big.

I have to determine what the pdf, cdf are. How can I find them if I don't know the interval where the numbers are chosen from? Can somebody explain me?

  • $\begingroup$ uniform likely means standard uniform, you are choosing from $(0,1)$. But I would think it would not matter too much. I would try to perhaps start with 2 variables and note the first is the min and the second is the max. And then upgrade to 3 variable case $\endgroup$ – gt6989b Sep 30 '18 at 14:01

First do it for $V_1,V_2,V_3$ with standard uniform distribution (so defined on $[0,1]$).

Then let $a,b$ be real numbers with $a<b$ and let $U_i:=a+(b-a)V_i$.

Then $U_i$ will have uniform distribution over $[a,b]$ and you can easily find that: $$U_{(i)}=a+(b-a)V_{(i)}\text{ for }i=1,2,3$$

These equalities will enable you to find CDF and PDF of the $U_{(i)}$.

  • $\begingroup$ But then it looks like that every pdf is $\frac{1}{b-a}$ just like the uniform distribution itself. For every $U_{(i)}$ $\endgroup$ – Hans Sep 30 '18 at 14:26
  • $\begingroup$ What I thought, was I choose $U_{(i)}$, for example $U_{(2)}$, then the density function for $U_{(2)}$ = $\frac{1}{U_{(1)} - a}$ and for $U_{(3)}$ = $\frac{1}{U_{(2)} - a}$, and then integrate for the cdf $\endgroup$ – Hans Sep 30 '18 at 14:30
  • $\begingroup$ If $V_1,V_2,V_3$ have standard uniform distribution and are independent then $P(V_{(3)}\leq x)=P(V_1\leq x,V_2\leq x,V_3\leq x)=P(V_1\leq x)P(V_2\leq x)P(V_3\leq x)=x^3$ if $x\in[0,1]$. This gives you the CDF of $V_{(3)}$ and on base of $U_{(3)}=a+(b-a)V_{(3)}$ you can find the CDF of $U_{(3)}$. $\endgroup$ – drhab Sep 30 '18 at 16:10

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