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Suppose real value random variables satisfy $X_{n} \Rightarrow X$ (convergence in distribution) as $n\to \infty$ in the same probability space $(\Omega, \mathcal F, \mathbb P)$. It is well known that $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$ for all continuous bounded real functions $f:\mathbb R \to \mathbb R$.

[Q.] If $f$ is continuous and linear growth, i.e. $|f(x)| < K(1 + |x|)$ for some constant $K$, can you find counter-example for $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$? What additional conditions are needed to still have $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$?

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  • $\begingroup$ I think if you have convergence to a finite value for $f(x):=|x|$, that is, convergence of the absolute moments $\mathbb E |X_{n}| \to \mathbb E |X| < +\infty$, then you have it for all your $f$ . $\endgroup$ Oct 4, 2012 at 6:43

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Since $f(X_{n}) \Rightarrow f(X)$ as long as $f$ is continuous, by redefining $X_{n}:= f(X_{n})$, it is equivalent to ask following question:

[Q1.] If $X_{n}\Rightarrow X$, then what additional condition is needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$?

[Ex.] Let $X_{n}: [0,1] \mapsto \mathbb R$ be given by $X_{n} (\omega) = n I_{(0,1/n)}(\omega)$ and $\mathbb P$ be Lesbegue meaure. Then, $X_{n} \Rightarrow X:=0$, since it is indeed almost sure convergence. However, it violates $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$.

So one immediate sufficient condition needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$ is that,

$X_{n} \to X$ almost surely, and satisfies other conditions of Dominated (or Monotone) Convergence Theorem.

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$\lim_{n \rightarrow \infty}\mathbb{E}[f(X_n)] = \mathbb{E}[f(X)]$ for all continuous $f$ of linear growth if and only if $X_n \Rightarrow X$ and $\lim_{n \rightarrow \infty}\mathbb{E}[|X_n|] = \mathbb{E}[|X|]$. This is exactly convergence in (first order) Wasserstein distance.

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