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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.

  1. 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, $\sum E[X_n] < \infty$ $\iff$ $\sum \alpha_n < \infty$.

  1. Second question:

Is it possible that these three conditions simultaneously occur:

  • $\sum X_n = +\infty$ almost surely?

  • $\sum X_n^2 < \infty$ almost surely?

  • $\sum(\mathbb{E}[X_n])^2 = \infty$?

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  • $\begingroup$ I've played around a bit with powers and logarithms to see what happens, even trying to localize them on sparse sets to allow relatively large values of $\alpha_n$ (something in the spirit of $\alpha_n = \frac{1}{n} \mathbb{1}_{\text{n is a square}}$) and each time, the sum in your third bullet converges. Whatever I try to do, the worst decay I obtain for $\mathbb{E}(X_n)$ is $\mathcal{O}(\frac{1}{n})$. $\endgroup$
    – Hachino
    Feb 20, 2015 at 9:53
  • $\begingroup$ Addendum : by "decay", I mean that, for instance with $\alpha_n = \frac{1}{n^{\frac{1}{4}}} \mathbb{1}_{\text{n is a fourth power}}$, then $\mathbb{E}(X_n)$ is of the order of $\frac{1}{n^{\frac{1}{4}} \ln{n}} \mathbb{1}_{\text{n is a fourth power}}$, whose $\ell^2$ norm is finite. $\endgroup$
    – Hachino
    Feb 20, 2015 at 10:03
  • $\begingroup$ Hello @Hachino. For numerical investigations, I think it is clearer to set $p_n=E(X_n)$ and then $\alpha_n$ is given by $\alpha_n=\frac{p_n}{\prod_{k=2}^n (1-p_k)}$. $\endgroup$ Feb 20, 2015 at 10:14
  • $\begingroup$ I've also found this Wikipedia page claiming that your $X_n$ is some Beta random variable, perhaps this could help you ? $\endgroup$
    – Hachino
    Feb 20, 2015 at 10:38
  • $\begingroup$ @Hachino I know that. My difficulty is the second bullet. $\endgroup$ Feb 20, 2015 at 10:41

1 Answer 1

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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.

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