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The definition of weak convergence of random variables is as follows:

Let $C_b(\mathbb{R}^d)$ be a functional space containing continuous and bounded functions from $\mathbb{R}^d$ to $\mathbb{R}$. Given a sequence of random variables ($X_n$) and $X$ on a probability space $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d), \mathbb{P})$. $(X_n)$ converges weakly to $X$, denoted by $X_n \xrightarrow{(d)} X$, if $\forall \phi \in C_b(\mathbb{R}^d)$, $\mathbb{E}[\phi(X_n)] \rightarrow \mathbb{E}[\phi(X)].$

I would like to show that continuous and bounded are two necessary conditions for test function $\phi$. Given $X_n \xrightarrow{(d)} X$, find a test function $\phi$ which is not continuous (or unbounded) such that $\mathbb{E}[\phi(X_n)] \nrightarrow \mathbb{E}[\phi(X)].$

I've come up with an example for one case. Consider $X_n \sim Unif(\{ \frac{k}{2^n} | 1 \leq k \leq 2^n \})$. By Riemann sum approximation, $X_n \xrightarrow{(d)} X \sim Unif([0,1])$. Let $\phi(x)= 1*\unicode{x1D7D9}_{\{\mathbb{Q} \cap [0,1]\}}(x) + 0*\unicode{x1D7D9}_{\{(\mathbb{Q} \cap [0,1])^c\}}(x)$ be a bounded but discontinuous function from $\mathbb{R}$ to $\mathbb{R}$. After calculation by the definition of expectation, we have $\mathbb{E}[\phi(X_n)] \rightarrow 1$ as $n \rightarrow \infty$ but $\mathbb{E}[\phi(X)] = 0$.

However, I have no idea about constructing a continuous but unbounded $\phi$ such that $\mathbb{E}[\phi(X_n)] \nrightarrow \mathbb{E}[\phi(X)]$ given $X_n \xrightarrow{(d)} X$.

Are there any examples for this case? Thanks!

jcm22
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  • You can find a counterexample with $\phi(x) = x$ (in which case you are simply looking at the convergence of the means), see this answer https://math.stackexchange.com/a/153294/491675 – Michh Nov 30 '23 at 18:45
  • You need to violate dominated convergence (a weaker condition is uniform integrability). Otherwise $X_n \Rightarrow X$ still implies $E[\phi(X_n)]\to E[\phi(X)]$ (by Skorokhod representation) – Andrew Nov 30 '23 at 18:52
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    @Michh Thanks, this can be an example. Given $X_n$ with probability measure defined as $\mathbb{P}(X_n = 2^n) = 1/n$ and $\mathbb{P}(X_n = 0) = 1- 1/n$, the limiting distribution of $X_n$ is a point mass at 0. Consider $\phi(x) = x$, we have $\mathbb{E}[\phi(X_n)] = (1/n)2^n+(1−1/n)0=(1/n)*2^n \rightarrow \infty$ but $\mathbb{E}[\phi(X)] = 0$. – jcm22 Dec 01 '23 at 02:30

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