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In this post: Exercise regarding Poisson processes and the uniform distribution
It is noted that U and 1-U are identically distributed for the r.v. U which is uniformly distributed on (0,1). If a definition for two random variables being identically distributed is:
Random variables X and Y are identically distributed if $ F_{X}(x) = F_{Y}(x) $ for all x.

Then let $ X = U$ and $ Y = 1 - U $.
So $ F_{U}(u) = F_{1-U}(u) = Pr[U \le u ] = Pr[1-U \le u] $
I don't understand the last equality (if it is even correct). Isn't it that $1-U$ will be the complement of $U$ so the CDFs will be inverted?

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Let $U$ be uniformly distributed on $[0,1]$. Then, for $0\le u \le 1$, we have $F_U(u)=u$.

Now let $V=1-U$. We find $F_V(v)$ for $0\le v\le 1$. We have $$F_V(v)=\Pr(1-U\le v)=\Pr(U\ge 1-v)=1-\Pr(U\le 1-v)=1-(1-v)=v.$$

To complete the calculation, we need to show that $F_V(v)=0$ if $v\lt 0$ and that $F_V(v)=1$ if $v\gt 1$. This is straightforward. For example, if $v\lt 0$, then $\Pr(V\le v)=\Pr(1-U\le v)=\Pr(U\ge 1-v)$. Since $1-v\gt 1$, this probability is $0$.

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  • $\begingroup$ Very clear, thank you very much. $\endgroup$
    – rhl
    Mar 26 '13 at 3:42
  • $\begingroup$ What if we have Cauchy distribution here (instead of uniform distribution)? $\endgroup$
    – Wen
    Aug 21 '16 at 4:37
  • $\begingroup$ @Wen: There are many versions of the Cauchy. If $W$ is standard Cauchy, then $W$ and $-W$ have the same distribution. $\endgroup$ Aug 21 '16 at 5:19
  • $\begingroup$ @AndréNicolas Thanks for the reply. I meant standard Cauchy and I think $W$ and $\frac{1}{W}$ have the same distribution. I tried use similar steps in your post but there is a problem since $W$ here can be either positive or negative. Like we can't say $P(\frac{1}{W} \leq t)=P(W \geq \frac{1}{t})$ $\endgroup$
    – Wen
    Aug 21 '16 at 5:36
  • $\begingroup$ @Wen: One can use similar calculation. If $t$ is positive, we have $1/W\le t$ if $W$ is negative or if $W\ge 1/t$. $\endgroup$ Aug 21 '16 at 16:02

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