# Weak convergence to a "multi-Bernoulli" distribution

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 .

• So, are you satisfied with the linked answer? Oct 23, 2020 at 16:40
• 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 $X\overset{\mathbb{P}}{\rightarrow}\alpha$? Oct 23, 2020 at 21:05
• The extent to which probabilists don't like to talk about probability measures never ends to amaze me :)
– R W
Oct 24, 2020 at 1:45
• @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. Oct 25, 2020 at 0:23

$$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$$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), $$P(X_{n,1}, which implies $$a_1=EX_{n,1}=EX_{n,1}\,1(0\le 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} which hold for all large enough $$n$$; these implications imply that the events $$\{X_{n,j}>1-t_{n,j}\}$$ are pairwise disjoint in $$j$$.