Expected value of sequence of rv I'm stuck with this problem.
Let $X_n$ be a sequence of random variables with $X_1$ the random number chosen at $(0,1)$. $X_2$ be the random number chosen at $(1-X_1,1)$ and so on. This is,  $X_n$ is the random number chosen at $(1 - X_{n-1}, 1)$. What is $E[X_n]$?
I have used conditional expectation in my first approach and order statistics in the second, but I'm not sure about my interpretation of the problem. Any hint?

First of all compute the conditional expectation:
$$\begin{aligned}E[X_n|X_{n-1}]&=\frac{1}{1-(1-X_{n-1})}\int_{(X_{n-1}-1,1)}xdx=\\
&=\frac{1}{X_{n-1}}\frac{1}{2}(1-(X_{n-1}-1)^2)=\\
&=\frac{1-X_{n-1}^2-1+2X_{n-1}}{2X_{n-1}}=\\
&=\frac{2-X_{n-1}}{2}
\end{aligned}$$
We can recast the problem as a sequential problem:
$$E[X_n]=\frac{2-E[X_{n-1}]}{2}\implies (a_{n})_{n \in \mathbb{N}}:a_{n+1}=\frac{2-a_{n}}{2}=1-\frac{a_n}{2}$$
The pattern becomes clear (recall $a_1=0.5$)
$$\begin{align}a_1&=a_1 \\
a_2&=1-\frac{a_1}{2} \\
a_3&=1-\frac{1}{2}+\frac{a_1}{4} \\
a_4&=1-\frac{1}{2}+\frac{1}{4}-\frac{a_1}{8} \\
a_5&=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{a_1}{16}\\
&\mathrel{\vdots} \\
a_n&=\sum_{k=0}^{n-2}\frac{(-1)^{k}}{2^{k}}+\frac{a_1(-1)^{n-1}}{2^{n-1}}, \quad n\geq 2\\&=\frac{1-\bigl(-\frac{1}{2}\bigr)^{n-1}}{\frac{3}{2}}+\frac{1}{2}\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}+\Bigl(\frac{1}{2}-\frac{2}{3}\Bigr)\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}-\frac{1}{6}\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}+\frac{1}{3}\Bigl(-\frac{1}{2}\Bigr)^{n}\end{align}$$
In the limit $E[X_n]\to 2/3$.
