# convergence in distribution from convergence of norm

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.

• How is convergence in distribution defined on the normed space $S$? – angryavian May 20 at 15:57
• @angryavian , I updated the question – LrM May 20 at 16:12
• What have you tried ? – Gabriel Romon May 20 at 16:29
• yes... I apologise... this is Theorem 3.1 from Billingsley. – LrM May 20 at 16:40

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?