Standard Uniform Distribution

Question

I am trying to show that a random variable $X_2$ has a standard uniform distribution. I have:

$\alpha \subset(0,1), X_1 \sim U[0,1],$ and

$X_2 = \begin{cases} X_1 &\mbox{if } X_1< \alpha \\ 1+\alpha-X_1 & \mbox{if } X_1 \ge \alpha. \end{cases} $

I tried splitting things into two cases and plotting, but I ended up with a V rather than a standard normal. I think I am missing something obvious. Sorry for the simple question!

Answer 1

for $0 \leq t \leq 1$,

\begin{align} P(X_2 < t) &= P(X_2 < t, X_1 < \alpha) + P(X_2 < t, X_1 \geq \alpha)\\ & = P(X_1 < t, X_1 < \alpha) + P( 1+ \alpha - X_1 < t, X_1 \geq \alpha) \\ & = \min(t, \alpha) + 1 - \min(1,\max(\alpha, 1 + \alpha -t))\\ & = \min(t, \alpha) + 1 - \min(1, 1 + \alpha -t)\\ & = \min(t, \alpha) - \min(0, \alpha -t)\\ & = \begin{cases} t, t < \alpha \\ t, \alpha \leq t \end{cases} \end{align}

Answer 2

Hint: Your distribution is clearly uniform if $X_i < \alpha$. So you just need to show that the distribution is uniform when $X_1 \geq \alpha$. For this use the reflection principle, showing there is a measure-preserving bijection between $X_1$ and $1 + \alpha - X_1$ when $X_1 \geq \alpha$.