0
$\begingroup$

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 .

$\endgroup$
4
  • $\begingroup$ So, are you satisfied with the linked answer? $\endgroup$ – Iosif Pinelis Oct 23 '20 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 '20 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 '20 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. $\endgroup$ – Iosif Pinelis Oct 25 '20 at 0:23
2
$\begingroup$

$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.