I am trying to solve the following problem and would very much appreciate some help.
Let $x_1,\dots,x_n$ identically distributed continuous random variables that take values in $(0,1)$ and satisfy the equality $x_1+\dots+x_n=1$ always (not just in expectation). What is the value of $E[x_i^2]$?
In principle, I do not want to pose restrictions on the distribution, although I am mostly interested in the uniform case.
Some initial ideas are that $E(x_i)=1/n$ by symmetry and for the uniform case that:
$\begin{align} E(x_i^2 \mid x_1+\dots+x_n=1)&=\frac{1}{n} E\left(\sum\limits_{i=1}^n x_i^2 \mid x_1+\dots+x_n=1\right)=\\ &=\frac{1}{n}\int\limits_0^1 \int\limits_0^{1-x_1}\dots\int\limits_0^{1-\sum\limits_1^{n-2}} \sum\limits_1^{n-1}x_i^2+\left(1-\sum\limits_1^{n-1}x_i\right)^2 dx_{n-1}\dots dx_2 dx_1 \end{align}$
But I seem to have serious trouble solving this integral.
Any help is more than welcome.
Well, there is a Dirichlet distribution with similar properties, but it is parametrised by $\vec{\alpha}$. In simplest case all $\alpha_i$ = 1, $i=1...n$, $\alpha_0 = n$. Mean and Variance are well-known. $E[X_i] = \frac{1}{n}$, $Var[X_i]=\frac{n-1}{n^2 (n+1)}$. Making $E[X^2_i]$ as sum of Variance and Mean squared, one can arrive to the answer $E[X^2_i]=\frac{2}{n(n+1)}$.
Reference https://en.wikipedia.org/wiki/Dirichlet_distribution
More complex case where there is non-trivial $\vec{\alpha}$ could be easily constructed