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Wikipedia states that the definition of convergence in law only requires that the cumulative distribution functions of the sequence of variables converges pointwise to the CDF of the limit variable at points where the CDF of the limit variable is continuous.

If I were to define convergence in law, I think I would define it as:

$X_n$ converges to $X$ in law if for every event (measurable set of the image space) $E$, $\text P(X_n\in E)\to\text P(X\in E)$

The definition on Wikipedia seems to almost verify the above, but if $x$ is any point of discontuinity, we might not have $\text P(X_n<x)\to\text P(X<x)$. Why not just require convergence everywhere?

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  • $\begingroup$ Wikipedia itself gives an answer to your question right after the definition. $\endgroup$ – Loreno Heer Nov 5 '14 at 20:13
  • $\begingroup$ @sanjab Perhaps I should have mentioned that I don't really understand the point of the example. Given that $F_n(0)=0$ for all $0$ but $F(0)=1$, I would find it more natural to just conclude that there isn't convergence in law. I don't see how the example is supposed to motivate changing the definition. $\endgroup$ – Jack M Nov 5 '14 at 20:27
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$X_n$ converges to $X$ in law if for every event (measurable set of the image space) $E$, $\text P(X_n\in E)\to\text P(X\in E)$

is too strong a definition for weak convergence/convergence in distribution.

For instance, if $X_n$ is such that $P(X_n=1/n)=1$, and $X$ is such that $P(X=0)=1$, then it is intuitively clear that $X_n \rightarrow X$ (and it does converge according to the accepted characterization of convergence: $Ef(X_n) \rightarrow Ef(X)$ for all bounded, continuous functions f), but your definition above would lead to a wrong conclusion - take E to be the set $(0,\infty)$.

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  • $\begingroup$ I guess the trouble is that this is basically a subjective question... to me it's not intuitively clear that $X_n\to X$, since as you say, their laws disagree for certain events. $\endgroup$ – Jack M Nov 5 '14 at 20:38
  • $\begingroup$ I think it all boils down to which definition leads to a useful theory. I'm by no means an expert. But since this is classical stuff, I'm guessing probabilists think that the relaxed definition is more rich. $\endgroup$ – elexhobby Nov 5 '14 at 20:42

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