# Weak convergence of Dirichlet distributions to a “multi-Bernoulli” distribution

For a positive vector $$\alpha\in\mathbb{R}^n$$ ($$n\geq 1$$), denote by $$\text{Dir}(\alpha)$$ the Dirichlet distribution with parameter $$\alpha$$. In terms of weak convergence, is it true that, if $$\sum\limits_{i=1}^n\alpha_i=1$$, then $$\lim\limits_{\varepsilon\rightarrow 0^+}\text{Dir}(\varepsilon\alpha)\longrightarrow \sum\limits_{i=1}^n \alpha_i \delta_{\lbrace e_i\rbrace}$$ (where $$(e_i)_{1\leq i\leq n}$$ is the canonical base of $$\mathbb{R}^n$$)?

$$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$$For any $$a=(a_1,\dots,a_n)\in(0,\infty)^n$$ and any real $$t\in(0,1/2)$$, let $$X=(X_1,\dots,X_n)$$ have the Dirichlet distribution with parameter $$ta$$. Then $$X_1$$ has the beta distribution with parameters $$ta_1$$ and $$tb_1$$, where $$b_1:=s-a_1$$ and $$s:=a_1+\dots+a_n.$$
Let $$t\downarrow0$$. Then $$\Ga(t)=\Ga(1+t)/t\sim1/t$$ and hence
$$P(X_1>1-t)=\frac{\Ga(ts)}{\Ga(ta_1)\Ga(tb_1)}\,J \sim\frac{ta_1b_1}s\,J,$$ where $$J:=\int_{1-t}^1 x^{ta_1-1}(1-x)^{tb_1-1}\,dx \sim\int_{1-t}^1 (1-x)^{tb_1-1}\,dx=\frac{t^{tb_1}}{tb_1}\sim\frac1{tb_1},$$ so that $$P(X_1>1-t)\to\dfrac{a_1}s$$. Similarly, for each $$j\in[n]:=\{1,\dots,n\}$$, $$P(X_j>1-t)\to\dfrac{a_j}s.$$ Hence, $$P(X_j\le 1-t\ \forall j\in[n])\to1-\sum_{j=1}^n\dfrac{a_j}s=0.$$
So, for any continuous function $$f\colon\R^n\to\R$$, $$Ef(X)=\sum_{j=1}^n Ef(X)1(X_j>1-t)+Ef(X)1(X_j\le 1-t\ \forall j\in[n]) \to\sum_{j=1}^n f(e_j)\dfrac{a_j}s+0,$$ where $$e_j$$ is the $$j$$th standard basis vector of $$\R^n$$; here we used the implications $$X_j>1-t\iff1>X_j>1-t\implies0.
Thus, the Dirichlet distribution with parameter $$ta$$ converges to $$\sum\limits_{j=1}^n \dfrac{a_j}s \delta_{\{e_i\}}$$ as $$t\downarrow0$$. That is, your conjecture holds iff $$s=1$$.