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I'm currently probability from two different sources: the classic text by Billingsley and the course notes of an instructor at my university. I've run into a terminology conflict that I was hoping someone might clear up.

Let $F_1, F_2, ...$ be a sequence of distribution functions. According to Billingsley, $F_n$ is said to converge weakly to $F$ if (i) $F_n (x) \to F(x)$ for $x$ such that $F$ is continuous at $x$ and (ii) $F$ is also a distribution function.

Conversely, my notes say $F_n \to F$ weakly requires (i) but not (ii). If both (i) and (ii) are satisfied then we say $F_n \to F$ completely.

So, which is the more standard use? If I subscribe to Billingsley's definition, does complete convergence then have some different meaning [e.g. by strengthening (i)]?

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In my past experiences, I have used both completely and weakly to mean Billingsley's definition. But I understand where the professor is coming from, as it's nice to be able to distinguish between them.

So I thought I would check to see if this is true by looking briefly through google. And what did I find? There are some places, like wikipedia, that agree with my experiences. But there are many little papers, such as here, that agree with your professor. In fact, the only common thing I saw is that people always defined which one they used carefully, so that there was no confusion.

How like mathematicians to be exactly so! In terms of utility, I am now inclined to use the professor's definition. It's easier to be exact. But to answer your question, neither.

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  • $\begingroup$ Well said. Amazingly (or perhaps not so amazingly if the use isn't standard) the paper you pulled up was authored by my professor. I like my professor's definition as well, but it makes studying both materials at the same time a little frustrating since the fact that they conflict alters the statements of several theorems, as well as their proofs. $\endgroup$ – guy Jul 4 '11 at 21:10

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