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Let $S$ be a metric space equipped with a distance function $d$, and let $X_n,Y_n$ be sequences of random variables having values from $S$. Suppose that $X_n$ converges in distribution to some random variable $X$, and suppose as well that $d(X_n,Y_n)$ convergence to 0 in probability.

Does it follow that $Y_n$ convergence in distribution to $X$?

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2 Answers 2

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One can use portmanteau theorem. Fix a closed set $F$. We have to show that $\limsup_{n\to +\infty}\mu(Y_n\in F)\leqslant \mu(X\in F)$.

To this aim, define $F^\varepsilon:=\{x\in S, d(x,x')\leqslant \varepsilon\mbox{ for some }x'\in F\}$. Then $$\mu(Y_n\in F)\leqslant \mu(X_n\in F^\varepsilon,d(X_n,Y_n)\leqslant \varepsilon)+\mu(d(X_n,Y_n)\gt\varepsilon)\leqslant \\ \leqslant\mu(X_n\in F^\varepsilon)+\mu(d(X_n,Y_n)\gt\varepsilon).$$ Since $F^\varepsilon$ is closed, it follows that by the assumption that $d(X_,Y_n)\to 0$ in probability that $$\limsup_{n\to +\infty}\mu(Y_n\in F)\leqslant\mu(X\in F^\varepsilon).$$ Since $F^\varepsilon\downarrow F$, the proof is complete.


An alternative way is to show that the convergence $\mathbb E[f(Y_n)]\to \mathbb E[f(X)]$ holds for any $f\colon S\to\mathbb R$ uniformly continuous and bounded.

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  • $\begingroup$ Regarding your last remark: is it any $f$ uniformly continuous and bounded, or just any $f$ conitnuous and bounded? (cf. the issue with my answer below, in the comments) $\endgroup$
    – Clement C.
    May 11, 2014 at 19:27
  • $\begingroup$ It's (much) easier to prove for uniformly continuous function. I don't see how to prove directly for $f$ continuous and bounded but not necessarily uniformly continuous because it seems harder to exploit the convergence in probability of $d(X_n,Y_n)$ to $0$. $\endgroup$ May 11, 2014 at 19:29
  • $\begingroup$ It is easier, yes: but the characterization of convergence in distribution is with continuous bounded functions, not uniformly continuous; isn't it? $\endgroup$
    – Clement C.
    May 11, 2014 at 19:40
  • $\begingroup$ It's the definition, but we can prove that this is equivalent to prove the convergence $E[f(X_n)]\to E[f(X)]$ for any $f$ uniformly continuous and bounded. Sometimes, it's included in the statement of the portmanteau theorem. $\endgroup$ May 11, 2014 at 19:54
  • $\begingroup$ Ah, great. Thanks! $\endgroup$
    – Clement C.
    May 11, 2014 at 19:56
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Here is an attempt (let me know if you feel something is wrong or missing):

Fix any continuous and bounded function $f\colon S\to\mathbb{R}$; the goal is to prove that $\mathbb{E} f(Y_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X)$.

From the assumptions, we know that $$ \mathbb{E} f(X_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X) \tag{$\dagger$} $$ and by continuity of $f$ that $f(X_n) - f(Y_n)$ converges to $0$ in probability $(\ddagger)$; writing $$ \lvert \mathbb{E} f(Y_n) - \mathbb{E} f(X) \rvert \leq \lvert \mathbb{E}[ f(Y_n) - f(X_n)] \rvert + \lvert \mathbb{E} f(X_n) - \mathbb{E} f(X) \rvert $$ The first term goes to $0$ with $(\ddagger)$, the second by $(\dagger)$; hence $\mathbb{E} f(Y_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X)$ for any bounded continuous function $f$, i.e. $Y_n\xrightarrow[n\to\infty]{\mathcal{D}} X$.

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  • $\begingroup$ If I understand correctly, from $d(X_n,Y_n) \rightarrow_p 0$ you went to $f(X_n) - f(Y_n) \rightarrow_p 0$. Can you please elaborate a bit on that point? My confusion comes from the fact that $d(\cdot;\cdot)$ has domain $S \times S$ while $f$ has domain $S$. $\endgroup$ May 11, 2014 at 18:48
  • $\begingroup$ Reading back, I'm also getting uncertain about this part. Certainly, if $f$ were Lipschitz (or just uniformly continuous), this would be true, as then (for the Lipschitz case, say $c$-Lipschitz) $$ \mathbb{P}\{\lvert f(X_n)-f(Y_n)\rvert > \varepsilon\} \leq \mathbb{P}\left\{d(X_n,Y_n) > \frac{\varepsilon}{c}\right\} \xrightarrow[n\to\infty]{}0 $$ For $f$ only continuous and bounded, however, I do not see such a simple argument... $\endgroup$
    – Clement C.
    May 11, 2014 at 19:25
  • $\begingroup$ (following Davide Giraudo's comments above, uniform continuity is good enough). $\endgroup$
    – Clement C.
    May 11, 2014 at 19:57

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