TITLE: Asymptotic behavior of $X_n$ in a Dirichlet vector $(X_1, ..., X_n)$
QUESTION [5 upvotes]: Let $(\alpha_k)$ be a sequence of positive numbers and let $(Y_k)$ be a sequence of independent random variables $Y_k \sim \text{Gamma}(\alpha_k,1)$. Set $X_n=\dfrac{Y_n}{\sum_{i=1}^nY_i}$.
(edit) The title is not appropriate: $(X_1, ..., X_n)$ is not a Dirichlet vector, because $X_k=\dfrac{Y_k}{\sum_{i=1}^kY_i} \neq \dfrac{Y_k}{\sum_{i=1}^nY_i}$. Sorry for the confusion.

First question:


Is it possible that these two conditions simultaneously occur:

$\sum X_n < +\infty$ almost surely?
$\sum \mathbb{E}[X_n] = \infty$?


(edit) Answer to question 1: No. One has $\sum_{i=1}^nY_i =\dfrac{1}{\prod_{i=2}^n(1-X_i)}$, hence $\sum X_n < +\infty$ $\iff$ $\sum Y_n < \infty$ $\iff$ $\sum \alpha_n < \infty$ (because $\sum Y_n \sim \text{Gamma}(\sum \alpha_n,1))$. Now, $E[X_n]=\dfrac{\alpha_n}{\sum_{i=1}^n\alpha_i}$, and by a similar reasoning, $\sum E[X_n] < \infty$ $\iff$ $\sum \alpha_n < \infty$.

Second question:


Is it possible that these three conditions simultaneously occur:

$\sum X_n = +\infty$ almost surely?
$\sum X_n^2 < \infty$ almost surely?
$\sum(\mathbb{E}[X_n])^2 = \infty$?

REPLY [2 votes]: The answer to the second question is no too. 
Set $D_k^n=\dfrac{Y_k}{\sum_{i=1}^nY_i}$. The neutrality property of Dirichlet vectors says that $D_n^n=:X_n$ is independent of $(D_1^{n-1}, \ldots, D_{n-1}^{n-1})$. But $X_k=\dfrac{D_k^{n-1}}{D_1^{n-1}+\ldots+D_k^{n-1}}$ for every $k=1, \ldots, n-1$, therefore $(X_n)$ is a sequence of bounded independent random variables.
Then, by a variant of Borel-Cantelli's lemma, $\sum X_n^2 < \infty$ $\iff$ $\sum E(X_n^2)<\infty$. But $E(X_n^2)\geq E(X_n)^2$, hence the second and third bullets are opposite.