Let $\mu$ be some probability distribution on $\Bbb R$ with a continuous density function $f$. Let $X = (x_i)$ be a finite subset of $\Bbb R$ and define a discrete distribution $\mu_X$ that gives to $x_i$ probability of $$\frac{f(x_i)}{\sum_j f(x_j)}$$ Are there any conditions on the choice of $X$ to make sure that $\mu_X$ weakly converges to $\mu$ if I add more and more points to $X$?

  • $\begingroup$ $\sum_j f(x_j)$ might be zero. Also, which weakly convergence do you mean? In case of the weakly $L^p$ convergences this question makes no sense, because $\mu_x$ and $\mu$ have a different domain of definition. $\endgroup$ – Adam Feb 15 '16 at 18:15
  • $\begingroup$ @Adam: we can assume $f$ is positive everywhere. By weak convergence I mean $\int fd\mu_n \to \int f d\mu$ for all bounded continuous $f$. $\endgroup$ – Ulysses Feb 16 '16 at 16:11

If you allow that the $(x_i)$ are random variables distributed according to $\mu$ and independent, then it follows for every $g \in L^1(\mu)$ that

$$ \frac{1}{N} \sum_{i=1}^N g(x_i) \to \int g \mu $$ converges in probability. For details, see Law of large numbers.

Edit: Maybe if you take more and more rational points from $\mathbb{Q}$ you might get convergence for bounded, continuous functions.

Edit2: How is this related to your question?

You want to know if $\int g d\mu_n \to \int g d \mu $ for all bounded continues g. In your case we have

$$ \int g d\mu_n = \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_{j=1}^n f(x_j)}.$$

So your question is what restriction on the sequence $(x_i)$ is necessary so that for every continuous bounded function g we get

$$ \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_{j=1}^n f(x_j)} \to \int g \mu .$$

In the case where $\mu$ is simply the Lebesgue Measue, we have $f=1$ and thus $$\int g d\mu_n = \frac{1}{N} \sum_{i=1}^N g(x_i) .$$

In the case where $\mu$ is arbitrary, notice that if you pick the points $(x_i)$ all over $\mathbb{R}$, and the points $(y_i)$ distributed according to $\mu$, then you have $$ \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_j f(x_j)} \approx \frac{1}{n} \sum_{i=1}^n g(y_i).$$

  • $\begingroup$ Thanks, but how is that relevant for my question? There I ask about the convergence of distributions, whereas you talk about the convergence of reals. $\endgroup$ – Ulysses Feb 24 '16 at 13:54
  • $\begingroup$ @Ulysses I added a short section to explain the relation to your question. $\endgroup$ – Adam Feb 24 '16 at 15:17

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