If $X \sim U(0,1)$ and
$\begin{align} \\[10pt] Y = {} & \begin{cases} \frac 3 2 - X & \text{if } \frac 1 2 < X< 1 \\[10pt] \frac 3 4 - X & \text{if } \frac 1 4 < X < \frac 1 2 \\[10pt] \frac 3 8 - X & \text{if } \frac 1 8 < X < \frac 1 4 \\[10pt] \text{and so on.} \end{cases} \end{align}$
How can we say that $Y$ has the same distribution of $X$? Should I calculate $Y$ for every interval in which $X$ is contained, for example:
$Y= \frac 3 2 - X$ if $\frac 1 2 <X< 1$ means that in the interval $(\frac 1 2,1)$, $Y \sim U(\frac 1 2,1)$ or for
$Y=\frac 3 4 - X$ if $\frac 1 4 <X< \frac 1 2$ means that in the interval $(\frac 1 4,\frac 1 2)$ $Y \sim U(\frac 1 2,\frac 5 2)$?
But if this is correct I don't see how $X \sim Y$.
I've also tried to draw a graph of the two random variables in the square$[0,1]x[0,1]$ but the situation is still unclear.
If X lies between 1/2 and 1 then Y= 3/2- X lies between 3/2- 1= 1/2 and 3/2- 1= 1/2 the same as X.
If X lies between 1/4 and 1/2 then Y= 3/4- X lies between 3/4- 1/2= 1/4 and 3/4- 1/4= 1/2, the same as X.
If X lies between 1/8 and 1/4, then Y= 3/8- X lies between 3/8- 1/4]= 1/8 and 3/8- 1/8= 1/4, the same as X.
Yes, Y has exactly the same distribution as X.