Distribution of $1-U$

Question

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 $1-U$.

Answer 1

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.

Answer 2

@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.