Let $(X_n)_{n\geq 1}$ be a sequence of random variables defined on the $d-$simplex ($d\geq 1$) : $\Sigma_d=\big\lbrace\boldsymbol{x}\in\mathbb{R}_+^d,\,\sum_{1\leq i\leq n} x_i=1\big\rbrace$. Assuming that there exists $\alpha\in\Sigma_d$ such that for $n\geq 1$, $\mathbb{E}[X_n]=\alpha$, and that the sequence of covariance matrices $\text{Cov}(X_n)$ converges to the matrix with coefficients $M_{ij}=\alpha_i(\delta_{ij}-\alpha_j)$ ($i,j\in \{1,...,d\}$ and $\delta_{ij}$ is the Kronecker symbol), does the following weak convergence holds: $$\lim\limits_{n\rightarrow\infty}\,\mathbb{P}_{X_n}\longrightarrow \sum\limits_{1\leq i\leq d} \alpha_i \delta_{e_i},$$ where $(e_i)_{1\leq i\leq d}$ is the canonical base of $\mathbb{R}^d$, and $\delta_{x}$ the Dirac measure for $x\in\mathbb{R}^d$?

The specific case for which every $X_n$ follow a Dirichlet law was solved: Weak convergence of Dirichlet distributions to a "multi-Bernoulli" distribution .

  • $\begingroup$ So, are you satisfied with the linked answer? $\endgroup$ Oct 23, 2020 at 16:40
  • $\begingroup$ Yes, thank you. I was also wondering: if for $1\leq i\leq d,\,\text{Var}(X_i)\rightarrow 0$, how can we adapt this proof to show that $X\overset{\mathbb{P}}{\rightarrow}\alpha$? $\endgroup$
    – G. Panel
    Oct 23, 2020 at 21:05
  • $\begingroup$ The extent to which probabilists don't like to talk about probability measures never ends to amaze me :) $\endgroup$
    – R W
    Oct 24, 2020 at 1:45
  • 1
    $\begingroup$ @G.Panel : The question in your comment is much easier, and the answer to it is yes, too, which follows because, if $X_{n,1}\to Y_1,\dots,X_{n,d}\to Y_d$ in probability, then $(X_{n,1},\dots,X_{n,d})\to(Y_1,\dots,Y_d)$ in probability, by the norm inequality. $\endgroup$ Oct 25, 2020 at 0:23

1 Answer 1


$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$This answer is similar to the one linked by the OP.

Indeed, let $a:=\alpha$ and $(X_{n,1},\dots,X_{n,d})$.

We have $EX_{n,1}=a_1$ and $Var\,X_{n,1}\to(1-a_1)a_1$, whence $EX_{n,1}^2\to a_1$ and $E(1-X_{n,1})X_{n,1}\to0$. So, for each $t\in(0,1)$, $$P(t\le X_{n,1}\le1-t)=E1(t\le X_{n,1}\le1-t) \le\frac{E(1-X_{n,1})X_{n,1}}{(1-t)t}\to0$$ and hence $$P(t_{n,1}\le X_{n,1}\le1-t_{n,1})\to0\tag{1}$$ for sequence $(t_{n,1})$ in $(0,1)$ converging to $0$. Passing to subsequences if needed, without loss of generality we may assume that $P(X_{n,1}>1-t_{n,1})\to p$ for some $p\in[0,1]$. Hence, by (1), $P(X_{n,1}<t_{n,1})\to1-p$, which implies $$a_1=EX_{n,1}=EX_{n,1}\,1(0\le X_{n,1}<t_{n,1})+EX_{n,1}\,1(t_{n,1}\le X_{n,1}\le1-t_{n,1}) +EX_{n,1}\,1(1\ge X_{n,1}>1-t_{n,1})\to0+0+p,$$ so that $p=a_1$ and $P(X_{n,1}>1-t_{n,1})\to a_1$.

Similarly, for each $j\in[d]:=\{1,\dots,d\}$ and some sequence $(t_{n,j})$ in $(0,1)$ converging to $0$, $$P(X_{n,j}>1-t_{n,j})\to a_j,$$ whence $$P(X_{n,j}\le 1-t_{n,j}\ \forall j\in[d])\to1-\sum_{j=1}^d a_j=0.$$

So, for any continuous function $f\colon\R^d\to\R$, $$Ef(X_n)=\sum_{j=1}^d Ef(X_n)1(X_{n,j}>1-t_{n,j}) \\ +Ef(X)1(X_{n,j}\le 1-t_{n,j}\ \forall j\in[d]) \\ \to\sum_{j=1}^d f(e_j)a_j+0,$$ where $e_j$ is the $j$th standard basis vector of $\R^d$; here we used the implications $$X_{n,j}>1-t_{n,j}\iff1\ge X_{n,j}>1-t_{n,j}\\ \implies0\le X_{n,i}<t_{n,j}<1-t_{n,i}\ \forall i\in[d]\setminus\{j\},$$ which hold for all large enough $n$; these implications imply that the events $\{X_{n,j}>1-t_{n,j}\}$ are pairwise disjoint in $j$.

Thus, your conjecture holds.


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