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Assume we have two sequences of random elements $X_{n}$ and $Y_{n}$ taking values in some normes space $S$, defined on the same probability space. Next, assume we know that $$ X_n \overset{d}{\to} X $$ and $$ ||X_n-Y_{n}|| \overset{a.s.}{\to} 0. $$

Is the following correct $$ Y_n \overset{d}{\to} X? $$

Remark. Convergence in distribution for random elements is

1)$\mathbb{E}[f(X_{n})] \to \mathbb{E}[f(X)]$ for all bounded uniformly continuous functions.

2)$\limsup_{n}\mathbb{P}[X_{n}\in F]\leq \limsup_{n}\mathbb{P}[X\in F]$ for all closed $F$.

3)$\liminf_{n}\mathbb{P}[X_{n}\in G]\geq \limsup_{n}\mathbb{P}[X\in G]$ for all closed $F$.

4)$\mathbb{P}[X_{n}\in A] \to \mathbb{P}[X\in A]$ for all $X$-continuity sets A.

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  • $\begingroup$ How is convergence in distribution defined on the normed space $S$? $\endgroup$ – angryavian May 20 at 15:57
  • $\begingroup$ @angryavian , I updated the question $\endgroup$ – LrM May 20 at 16:12
  • $\begingroup$ What have you tried ? $\endgroup$ – Gabriel Romon May 20 at 16:29
  • $\begingroup$ yes... I apologise... this is Theorem 3.1 from Billingsley. $\endgroup$ – LrM May 20 at 16:40
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Let $A_{n,\epsilon} = \{\|X_n - Y_n\| < \epsilon\}$. Let $f$ be a bounded uniformly continuous function. Let $\sup_{x \in S} |f(x)| \le B$. We have $$|E[f(Y_n)] - E[f(X)]| \le \big|E[f(Y_n) \mathbf{1}_{A_{n,\epsilon}}] - E[f(X)]\big| + \big|E[f(Y_n) \mathbf{1}_{A_{n,\epsilon}^c}]\big|.$$

The second term is bounded by $B\cdot P(A_{n,\epsilon}^c)$ which converges to zero.

For the first term, note that $$E[f(Y_n) \mathbf{1}_{A_{n, \epsilon}}] = E[f(X_n) \mathbf{1}_{A_{n, \epsilon}}] + E[(f(Y_n) - f(X_n)) \mathbf{1}_{A_{n, \epsilon}}]$$ so $$\big|E[f(Y_n) \mathbf{1}_{A_{n,\epsilon}}] - E[f(X)]\big| \le \big|E[f(X_n) \mathbf{1}_{A_{n, \epsilon}}] - E[f(X)]\big| + \epsilon.$$ Can you take it from here?

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