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I have a problem that I have trouble to think of the proof. It is obvious that the answer will be Uniform Distribution intuition wise, but how do I prove it?

Question: Let $U$ be a continuous uniform $(0,1)$ random variable. Find the distribution of $1-U$.

lydias
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  • First of all, do you understand what "find the distribution" means? –  Mar 02 '17 at 22:31
  • Secondly, what is a continuous uniform $(0,1)$ random variable? –  Mar 02 '17 at 22:31
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    What do you mean by "visualize the solution"? Graph of the distribution function? –  Mar 02 '17 at 22:32
  • @Jack Hi Jack, I do understand 'find the distribution' means. From my understanding, if the question asks to "find the distribution", then the answer will be Uniform Distribution, but this seems to be too easy intuition wise. I must have missed some details. I should have asked for the proof of the question not just for clarification. Thanks for the clarifications from BruceEt and Ethan Boker. – lydias Mar 03 '17 at 16:35

2 Answers2

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First, the CDF of $U$ is $F_y(t) = t$, for $0 \le t \le 1.$ Now let $V = 1-U$. We seek the CDF of $V:$

$$F_V(t) = P(V \le t) = P(1 - U \le t) = P(U \ge 1 - t)\\ = 1 - P(U \le (1 - t)) = 1 - (1-t) = t.$$

So $V$ has the same CDF as $U,$ and hence the same distribution.

BruceET
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@BruceET 's answer is correct. Here's another that addresses your intuition, with less algebra.

In everyday language, $U$ is choosing a random point uniformly between $0$ and $1$. You can specify your point by randomly (uniformly) choosing its distance $x$ from $0$, or equally well by its distance from $1$. That's $1-x$. So you could choose $1-x$ instead of $x$ and have all the same probabilities. That means anything you could say about $U$ you can say just as well about $1-U$. They have the same distribution.

Ethan Bolker
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