Suppose $\{X_1,...X_n\}:\Omega \to \mathbb{R}^p$ be i.i.d. random vectors with common probability law/measure $p$, i.e. $Prob(X_i^{-1}(E))=p(E) \forall E \subset \mathbb{R}^p $ Borel measurable.

Consider the random Dirac measures $\delta_{X_i}$, and their average, which is a random probability measure on $\mathbb{R}^p$, defined by $\frac{1}{n}\sum_{i=1}^{n} \delta_{X_i}$. I'd like to know if $\frac{1}{n}\sum_{i=1}^{n} \delta_{X_i}$ weakly converges to the deterministic measure $p$,

i.e. for every continuous, bounded function $f: \mathbb{R}^p \to \mathbb{R}$, must we have:

$$\frac{1}{n}\sum_{i=1}^{n} {f(X_i)} \to \int_{\mathbb{R}^p } f(x)dp(x)$$ as a convergence in probability of a sequence of random variables $\frac{1}{n}\sum_{i=1}^{n} {f(X_i)}$?

ON a related note, I'd also like to know if the following is true or not:

If a sequence of random measures converges to a deterministic measure in probability, is it equivalent to have the same convergence almost surely? This question is motivated by the fact a when a sequence of random variables converge to a sonstant in probability, the convergence is a.s.

P.S. I understand that this question might be elemenrtary to many of you, so some references would be greatly apreciated!

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    $\begingroup$ If I'm not mistaken, this is exactly the Glivenko-Cantelli theorem. $\endgroup$ Jan 24, 2020 at 15:35
  • $\begingroup$ @NateEldredge Honestly I'm not familiar with the many theorems in probability, but I just looked up Glivenko-Cantelli theorem, and you're right, it's what I was looking for. I can add "reference request" as a topic and modify the question, of you're free to vote to close it. BTW: what's the answer to the second question which concerns the equivalence of convergence in probability and a.s. convergence of a sequence of random measures? $\endgroup$ Jan 24, 2020 at 15:39
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    $\begingroup$ I guess Glivenko-Cantelli shows something a little stronger: that the cdfs actually converge uniformly (for weak convergence, you only need pointwise convergence at points of continuity). This only makes any difference if the distribution $p$ has a discrete part (i.e. atoms). It also gives a.s. convergence. $\endgroup$ Jan 24, 2020 at 17:58
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    $\begingroup$ By the way, you've probably found this by now, but the usual term for $\frac{1}{n} \sum^n \delta_{X_i}$ is "empirical distribution" or "empirical measure". $\endgroup$ Jan 24, 2020 at 19:09
  • $\begingroup$ @NateEldredge sorry for the late reply in comments. Yes I agree that the the Glivenko-Cantelli theorem proves something stronger that weak convergence of measures, and yes I've found that it's called the "empirical distribution". But looking at all these "law of large number type" theorems of random measures, I wonder if there's a lecture note/blog post that'd nicely combine and compare them. So the kind of note/post I'm looking for is something that'll first state the WLLN/SLLN for r.v., and then would do the same for random measures, i.e. measure-valued random variables (contd.) $\endgroup$ Jan 25, 2020 at 9:39

1 Answer 1


For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ of iid random vectors. Then, by the strong law of large numbers (SLLN) , $$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$ almost surely (a.s.), which, as desired, implies the weak law of large numbers (WLLN) , that is, the convergence in probability (which is always implied by the a.s. convergence).

Concerning your second question, it is of course not true that "when a sequence of random variables converge[s] to a [constant] in probability, the convergence is a.s." -- That would be the same as to say that the convergence in probability always implies the a.s. convergence, because $Y_n\to Y$ a.s. (respectively, in probability) if and only if $Y_n-Y$ converges to the constant $0$ a.s. (respectively, in probability).


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