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 $Y_n\overset{d}{\to}X$.

Thanks in advance!

  • 2
    $\begingroup$ 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 $\endgroup$ Commented May 21, 2020 at 6:40
  • 1
    $\begingroup$ @ThomasLumley thanks $\endgroup$ Commented May 21, 2020 at 16:58

2 Answers 2


This one is a simple case of combining two limits into one. In the present case the condition is necessary and sufficient, so you have a stronger result than the one you are positing.

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.)


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