# uniform random numbers and pdf and cdf

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?

• 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 – 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 $$a and let $$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)}$$.
• But then it looks like that every pdf is $\frac{1}{b-a}$ just like the uniform distribution itself. For every $U_{(i)}$ – Hans Sep 30 '18 at 14:26
• 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 – Hans Sep 30 '18 at 14:30
• 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 $U_{(3)}$. – drhab Sep 30 '18 at 16:10