# Question on Convergence in distribution

Belyaev and Sjöstedt-de Luna introduced the notion of weakly approaching sequences of distributions, generalizing the weak convergence without imposing the limiting distribution.

Definition. Two sequences of random variables $$\{Y_n\}$$ and $$\{X_n\}$$ are said to have weakly approaching distributon laws, $$\{\mathcal{L}(Y_n)\}$$ and $$\{\mathcal{L}(X_n)\}$$, if for any bounded continuous function $$f$$, $$E(f(Y_n))-E(f(X_n))\to 0$$ as $$n\to\infty$$, and we write $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n), \ n\to\infty$$.

I know that $$Y_n$$ converges in distribution/weakly to $$X$$, denoted by $$Y_n\overset{d}{\to}X$$, if for any bounded continuous function $$f$$, $$E(f(Y_n))-E(f(X))\to 0$$ as $$n\to\infty$$, by portmanteau Lemma.

My question is: when $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$$ will imply $$Y_n\overset{d}{\to}X$$?

I believe that $$X_n\overset{d}{\to}X$$ is sufficient. But I cannot argue why.

My attempt

Suppose that $$X_n\to X$$ in distribution. Then the portmanteau Lemma (see Lemma 2.2 of Van der Vaart's Asymptotic Statistics) gives $$\mathcal{L}(X_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X)$$. Therefore $$E(f(Y_n))-E(f(X))=E(f(Y_n))-E(f(X_n))+E(f(X_n))-E(f(X))\to 0$$ for any bounded continuous $$f$$, by hypothesis.

This shows that if $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$$ and $$X_n\overset{d}{\to}X$$, then $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X)$$ . By portmanteau Lemma again, $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X)$$ implies $$Y_n\overset{d}{\to}X$$.

• Just an comment on nomenclature: it's the 'portmanteau lemma', from the word 'portmanteau', indicating that it has a lot of things stuffed in together. It's not named after a Monsieur Portmanteau Commented May 21, 2020 at 6:40
• @ThomasLumley thanks Commented May 21, 2020 at 16:58

Theorem: Suppose that $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$$. Then for any random variable $$L$$ we have: $$X_n\overset{d}{\to}L \quad \iff \quad Y_n\overset{d}{\to}L.$$
Proof: Let $$f$$ be an arbitrary bounded continuous function, so we have the limit: $$\lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) = 0.$$ ($$\implies$$) Since $$X_n\overset{d}{\to}L$$ we have $$\lim_{n \to \infty} \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) = 0$$ which gives: \begin{aligned} \text{Limit} &\equiv \lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(L)) \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) + \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) \Big] \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) \Big] + \lim_{n \to \infty} \Big[ \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) \Big] \\[6pt] &= 0+0 = 0. \\[6pt] \end{aligned} The proof of the reverse implication ($$\impliedby$$) is identical. $$\blacksquare$$
Indeed, if $$E(f(Y_n))-E(f(X_n))\to 0$$ and $$E(f(X_n))-E(f(X))\to 0$$, elementary facts about sequences would tell you $$E(f(Y_n))-E(f(X))\to 0,$$ i.e. (since $$f$$ is arbitrary bounded and continuous function) $$Y_n\overset{d}{\to}X.$$
(Necessity also holds as well as sufficiency. Suppose $$\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$$, then $$Y_n$$ converges in distribution if and only $$X_n$$ does also.)