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$X, X_1, X_2,\ldots $ are real random variables with $\mathbb{P}(X_n\leq x)\to \mathbb{P}(X\leq x)$ whenever $\mathbb{P}(X=x)=0$.

Why does $X_n\stackrel{L}{\to} X$? At the least, where would I begin?

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2 Answers 2

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A sequence $X_1,X_2,\ldots$ of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable $X$ if $$ \lim_{n\to\infty}F_n(x)=F(x) $$ for every number $x\in\mathbb R$ at which $F$ is continuous, where $F_n(x)=\mathbb P(X_n\le x)$ and $F(x)=\mathbb P(X\le x)$.

Thus, we need to show that $F(x)$ is continuous at $x$ if and only if $\mathbb P(X=x)=0$. $F$ is continuous from the right, so we need to investigate continuity from the left. Since $$ \mathbb P(X=x)=F(x)-\lim_{y\uparrow x}F(y), $$ we have that $\mathbb P(X=x)=0$ if and only if $F(x)=\lim_{y\uparrow x}F(y)$.

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  • $\begingroup$ @StefanHansen I added the definition of convergence in distribution. It may help to see the relation. $\endgroup$
    – Cm7F7Bb
    Mar 6, 2014 at 9:39
  • $\begingroup$ Nevermind my comment, I took it for granted that the definition of convergence in distribution was ${\rm E}[f(X_n)]\to{\rm E}[f(X)]$ for continuous, bounded $f$. This doesn't seem to be the definition the OP is working with. Sorry about the fuzz. $\endgroup$ Mar 6, 2014 at 9:43
  • $\begingroup$ @StefanHansen No problem. These two definitions are equivalent. $\endgroup$
    – Cm7F7Bb
    Mar 6, 2014 at 9:45
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Start with the definition of what is the convergence. There is also a theorem (http://fr.wikipedia.org/wiki/Convergence_en_loi, can't find the english version) that says that you need to prove that $P(X_n \in A) -> P(X \in A)$ for any A an open

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  • $\begingroup$ Are you talking about the portmanteau theorem? $\endgroup$
    – Kashif
    Mar 6, 2014 at 9:27
  • $\begingroup$ @Glassjawed yes $\endgroup$
    – Thomas
    Mar 6, 2014 at 12:11

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