Weak convergence of Dirichlet distributions to a "multi-Bernoulli" distribution

Question

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 $\mathbb{R}^n$)?

Answer 1

$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$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 $X_j>1-t\iff1>X_j>1-t\implies0<X_i<t<1-t\ \forall i\in[n]\setminus\{j\}$.

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 $s=1$.