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I have the following problem:

Suppose that a point $X_1$ is chosen from a uniform distribution on the interval $[0, 1]$, and that after the value $X_1=x_1$ is observed, a point $X_2$ is chosen from a uniform distribution on $[x_1, 1]$. The variables $X_3, X_4,\ldots$ are generated in the same way. In general, for $j=1, 2,\ldots$, after $X_j=x_j$ is observed, $X_{j+1}$ s chosen from a uniform distribution on $[x_j, 1]$. Calculate $E[X_n]$.

So, I get the the probability for $X_n$ to be equal to some value $a\in [x_{n-1}, 1]$ is $\dfrac{1}{1-x_n}$, given that $X_1=x_1, X_2=x_2,\ldots X_{n-1}=x_{n-1}$. But I don't really know how to continue to get the expected value from here. Any help would be really appreciated.

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    $\begingroup$ $\dfrac{1}{1-x_n}$ is a density rather than a probability $\endgroup$
    – Henry
    Jan 9 '20 at 8:57
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Hint : \begin{align*} \mathbb E[X_n] &= \mathbb E[\mathbb E[X_n|X_{n-1}]]\\ &= \mathbb E\left[ \frac{1+X_{n-1}}{2} \right]\\ &= \frac{1}{2} + \frac{\mathbb E[X_{n-1}]}{2} \end{align*}

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