Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is convergence in distribution seems to be defined differently in Wikipedia and in Kai Lai Chung's book. My view is that the one by Wikipedia is a standard definition of convergence in distribution, and the one by Chung is actually vague convergence not convergence in distribution. I wonder if the two definitions are equivalent, and why?

Following are relevant quotes from the two sources:

  1. From Wikipedia

    A sequence $\{X_1, X_2, …\}$ of random variables is said to converge in distribution to a random variable $X$ if $$ \lim_{n\to \infty} F_n(x) = F(x)$$ for every number $x ∈ \mathbb{R}$ at which $F$ is continuous. Here $F_n$ and $F$ are the cumulative distribution functions of random variables $X_n$ and $X$ correspondingly.

  2. From Kai Lai Chung's A course in probability theory, consider (sub)probability measures (s.p.m.'s or p.m.'s) on $\mathbb{R}$.

    definition of convergence "in distribution" (in dist.)- A sequence of r.v.'s $\{X_n\}$ is said to converge in distribution to d.f. $F$ iff the sequence $\{F_n\}$ of corresponding d.f.'s converges vaguely to the d.f. $F$.

    My rephrase of vague convergence of a sequence of distribution functions (d.f.'s) based on the same book is

    We say that $F_n$ converges vaguely to $F$, if their s.p.m.'s are $\mu_n$ and $\mu$, and $\mu_n$ converges to $\mu$ vaguely.

    On p85 of Chung's book, vague convergence of a sequence of s.p.m.'s is defined as

    a sequence of subprobability measures (s.p.m.'s) $\{ \mu_n, n\geq 1 \}$ is said vaguely converge to another subprobablity measure $\mu$ on $\mathbb{R}$, if there exists a dense subset $D$ of the real line $\mathbb{R}$ so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b]) \rightarrow \mu((a,b])$.

Thanks and regards!

share|cite|improve this question
Every countably infinite subset of the real line, and only those, can be discontinuity sets for weakly increasing functions, and in particular for CDFs. Next question: which dense subsets of hte line can occur in the role of what you call $D$? – Michael Hardy Feb 27 '12 at 23:13
@MichaelHardy: Thanks! (1) are your both questions addressing the definition of vague convergence of a sequence of s.p.m.'s in the last quote? (2) What is your first point trying to say? (3) To your second question, the definition in the last quote says existence of a dense subset D in R, which depends on the sequence $\{\mu_n\}$ and $\mu$. That is my understanding. – Tim Feb 28 '12 at 1:13
Tim, these definitions are equivalent. The only difference is the usual difference between convergence in distribution and vague convergence, which is that vague convergence can be to an arbitrary measure (i.e. it need not have mass 1) while convergence in distribution requires that the limiting measure assign mass one to the probability space. It takes a little analysis to show that convergence on a dense set of points is sufficient for convergence at all points of continuity, but if I recall a full proof is in Durrett. – Chris Janjigian Feb 28 '12 at 3:05
@Chris: Thanks! (1) Distribution functions can be defined for general measures on $\mathbb{R}$, not just probability measures. So why can't convergence in distribution be defined for general measures? (2) I appreciate if you can point out where in Durrett's which book the equivalence of convergence in distribution and vague convergence appears. I hope to see if the equivalence is for general measures, not just for probability measures. – Tim Feb 28 '12 at 3:14
Convergence in distribution is not equivalent to vague convergence; it is equivalent to vague convergence with the added assumption of tightness, which is what prevents mass from escaping (it is a conservation requirement). You can certainly define a notion of convergence of the distribution functions pointwise for an arbitrary measure but that is actually vague convergence if you do not have a tightness requirement. That theory is quite well developed as it is just the usual weak convergence of measures. – Chris Janjigian Feb 28 '12 at 3:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.