Empirical distribution of vector

Question

Currently, I am reading about the empirical distribution and doubting a statement, which is my own intuition. Let's say we have a vector $\mathbf{x} \in \mathbb{R}^n$ converges in distribution to the empirical distribution $\mu$ as $n$ tends to infinity. Can we conclude that $$\frac{1}{n}\sum_{i=1}^{n}x_i = \mathbb{E}[X],\text{ almost surely},$$ where $X$ is a random variable and $X \sim \mu$? If it holds, then in what condition of function $f$ we have $$\frac{1}{n}\sum_{i=1}^{n}f(x_i) = \mathbb{E}[f(X)] ?$$

Answer 1

This is my understanding (or "intuition"): In terms of the Dirac delta function $\delta(x)$ we define the probability density $\mu(x)$ by $$\mu(x)=\frac{1}{n}\mathbb{E}\left[\sum_{i=1}^n\delta(x-x_i)\right].$$ No large-$n$ limit here. Then, by construction, $$\mathbb{E}\left[n^{-1}\sum_{i=1}^n f(x_i)\right]=\int f(x)\mu(x)\,dx\equiv \mathbb{E}[f(X)].$$ Now for large $n$ you may want to remove the expectation value from the left-hand-side of the equation, based on the usual argument that sample-to-sample fluctuations decay as $1/\sqrt n_{\rm eff}$, where $n_{\rm eff}$ is the number of $x_i$'s that contribute to $f(x)$. If this function is sharply peaked on the scale of the mean separation of the $x_i$'s, then $n_{\rm eff}$ is of order unity and this argument will fail.