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?
 A: I guess that when you write $X_1$ is the random number chosen at $(0,1)$, you mean that the law of $X_1$ is the uniform distribution on $(0,1)$, and that when you say that $X_2$ is the random number chosen at $(1-X_1,1)$, you mean that the law of $X_2$, conditionally to $X_1$, is the uniform law on $(1-X_1,1)$, and so on. Hence, the conditional expectancy of $X_{n+1}$ knowing $X_n$ is given by $$\mathbb{E}[X_{n+1}|X_n] = 1 - \frac{X_n}{2}.$$
Hence, we have $$\mathbb{E}[X_{n+1}] = \mathbb{E}\left[\mathbb{E}[X_{n+1} |X_n] \right] = 1 - \frac{\mathbb{E}[X_n]}{2}.$$
One can then check by induction that $\mathbb{E}[X_n]$ is given for $n \geq 1$ by $$\mathbb{E}[X_n] = \frac{2}{3} + \frac{1}{3}\left(- \frac{1}{2}\right)^{n}.$$
A: 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 $E[X_n]\to 2/3$.
