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My question is quite simple, but I was unable to find an answer by googling, since you can't exactly google syntax. What does the $\in \cdot$ mean in: $$\lim_{n\to\inf}||P(S_n\in\cdot)-P(S_n+k\in\cdot)||_{tv}=0$$ This is from coupling lectures in probability theory.

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closed as too localized by Wadim Zudilin, José Figueroa-O'Farrill, Yemon Choi, Andrés E. Caicedo, Martin Brandenburg Jan 9 '11 at 9:59

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I think this question would belong better on (and in any case you should really explain your notation, since people do not necessarily have access to the same notes or textbook that you do) – Yemon Choi Jan 9 '11 at 3:20
Hmm, Jimmie likes to answer Jimmie's questions... Some kind of pathology. – Wadim Zudilin Jan 9 '11 at 4:16
up vote 1 down vote accepted

If $S_n$ is a real-valued random variable, then I would read $P(S_n \in \cdot)$ as denoting the probability measure $P \circ S_n^{-1}$ on $\mathbb{R}$, i.e. the set function $B \mapsto P(S_n \in B)$ where $B \in \mathcal{B}\_{\mathbb{R}}$, the Borel $\sigma$-algebra on $\mathbb{R}$. The quantity in norm bars is then the total variation distance between these two measures, which means the given expression can be rewritten $$ \lim_{n \to \infty} \sup_{B \in \mathcal{B}_\mathbb{R}} |P(S_n \in B) - P(S_n + k \in B)| $$

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I think the answer is that the $\cdot$ induces an anonymous function, e.g. the above is equal to:

$$\lim_{n\to\inf}||f||_{tv}=0$$ with f a function on the appropriate domain having $$f(x) := P(S_n=x)-P(S_n+k=x)$$

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