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Let $U$ be uniform on $(0,\ 1)$. Show that $\min(U,\ 1-U)$ is uniform on $(0,\ 1/2)$ and that $\max(U,\ 1-U)$ is uniform on $(1/2,\ 1)$.

I'm not sure how to approach... the only hint i have is that a Random Variable U is Uniform on (0,1) if for all 0<=u<=1, P[U<=u]=u the only thing that I have seen is that the function min(U,1-U)=U for all 0<=U<=1/2 and that is is quite like the behave of a uniform variable on (0,1/2) but not sure is that's proof enough

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  • $\begingroup$ There are many ways to solve this problem. To keep it focused, could you tell us what steps have you taken toward trying to answer this question yourself? $\endgroup$
    – whuber ♦
    Commented Jan 18, 2015 at 18:31
  • $\begingroup$ I'm not shure how to approach... the only hint i have is that a Random Variable U is Uniform on (0,1) if for all 0<=u<=1, P[U<=u]=u $\endgroup$
    – Mor_H
    Commented Jan 18, 2015 at 18:45
  • $\begingroup$ the only thing that I have seen is that the function min(U,1-U)=U for all 0<=U<=1/2 and that is is quite like the behave of a uniform variable on (0,1/2) but not sure is that's proof enough $\endgroup$
    – Mor_H
    Commented Jan 18, 2015 at 18:53

1 Answer 1

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If $X = max(U, 1-U)$ then it's distribution function is

$F_X = Pr(X \leq x)$ $ = Pr\{ max(U, 1-U) \leq x \}$

You should note that the maximum between two numbers is less than x iff both numbers are below x, so

$F_X = Pr\{ max(U, 1-U) \leq x \} = Pr\{U \leq x , (1 -U ) \leq x\} = Pr\{(1-x) \leq U \leq x \}$

This event has zero probability if x is below $\frac{1}{2}$ (in that case you would be asking for the probability of an impossible event, since $0.5 \leq 1- x$ and $x \leq 0.5$). In terms of distributions, then, you would have:

$F_X = Pr\{(1-x) \leq U \leq x \} = [F_U(x) - F_U(1-x)].I_{(x \geq 0.5)}$

Taking derivatives (since they exist except in a finite number of points), you can find $X$ density:

$f_X = [f_U(x) +f_U(1-x)].I_{(x \geq 0.5)} = [I_{(0,1)}(x)+I_{(0,1)}(1-x)].I_{(x \geq 0.5)}$

Since $I_{(0,1)}(x)=I_{(0,1)}(1-x)$, you finally get:

$f_X = 2.I_{(0,1)}(x).I_{(x \geq 0.5)}=2.I_{(0.5,1)}(x)$

So the maximum is distributed uniformly between $0.5$ and $1$.

For the minimum, you have that $min(U,1-U) = 1 - max(U,1-U)$ you have just proven that the maximum is uniform between $.5$ and $1$, so this equality means that the minimum is uniform between $0$ and $.5$ (being a linear function of a uniform r.v.)

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    $\begingroup$ +1 There was no need to take derivatives. By starting with the minimum you can use your technique to deduce that $\Pr(0\le \min(U,1-U)\le x)=\Pr(0\le U\le x)+\Pr(1\le U\ge 1-x)=2x$ for $0\le x\le 1/2$. Since the probability of the interval $[0,x]$ is thereby seen to be proportional to its length $x=x-0$, you may immediately conclude the distribution is uniform. Then the result for the maximum follows, as you remark at the end. Since you're new here, please now take a minute to review our guidelines for answering self-study questions. $\endgroup$
    – whuber ♦
    Commented Jan 18, 2015 at 22:40
  • $\begingroup$ thanks whuber, I already reviewed the guidelines... also thank you for giving me other option to solve it, but I still think that the conclution is more obvious using derivatives. but I have a question what do you refer to this:"Since the probability of the interval [0,x] is thereby seen to be proportional to its length x=x−0, you may immediately conclude the distribution is uniform"? what has to do, that the probability is proportional to the length, to being a uniform distribution? $\endgroup$
    – Mor_H
    Commented Jan 19, 2015 at 0:55
  • $\begingroup$ Mor_H, That's the definition of uniform that you gave. (My previous comment was addressed to Cristián Antuña. Normally self-study questions are not answered in full unless they are old.) $\endgroup$
    – whuber ♦
    Commented Jan 19, 2015 at 1:41
  • $\begingroup$ Oh, I'm sorry @Mor_H, didn't know that. Still getting to know the rules. Won't happen again. $\endgroup$
    – Cristián Antuña
    Commented Jan 19, 2015 at 1:53

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