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I am reading a very complex paper consider a set of random variables $\left\{ X_{i}\right\} _{i=1}^{\infty }$ whose common distribution $F_{X}$ belong to the domain of attraction of an $\alpha -stable$ law $\gamma $ with the following characteristic function \begin{equation*} \exp \int \left( e^{itx}-1-itx\right) d\mu \left( c_{1},c_{2};\alpha \right) \left( x\right) \end{equation*} where $d\mu \left( c_{1},c_{2};\alpha \right)(x)$ is defined as $c_{1}x^{-1-\alpha }dx $ if $x>0 $ and $ c_{2}\left\vert x\right\vert ^{-1-\alpha }dx$ if $x<0$

In this paper they used

\begin{equation*} \frac{\sum_{i=1}^{n}X_{i}-n\mathbb{E}\left( X_{1}\right) }{a_{n}}\underset{w}% {\rightarrow }\gamma \end{equation*}

is $\gamma$ a random variable or a measure? if it was a measure I can not understand the notation in the above formula? If it was a measure may we say that the characteristic function of the \bigskip $\alpha -stable$ random variable $\gamma $ is as follow \bigskip

\begin{equation*} \phi _{\gamma }\left( t\right) =\exp \int \left( e^{itx}-1-itx\right) d\mu \left( c_{1},c_{2};\alpha \right) \left( x\right) \end{equation*}

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  • $\begingroup$ You don't understand what it means that a sequence of random variables converges weakly to a measure? $\endgroup$
    – Stefan Hansen
    Commented Sep 2, 2014 at 11:18
  • $\begingroup$ It is not clear to me any reference you suggest? I am confused with notation. Sometimes they use convergence in distribution and other times weak convergence. I understand that convergence weakly and convergence in distribution are used exchangeable. But most books uses convergence in distribution when they associate random variable. While on the other hand they associate weak convergence with their measure. Can you give any further guidance what exactly is meant by random variable converges weakly to a measure?! $\endgroup$
    – Anonymous
    Commented Sep 2, 2014 at 11:25
  • $\begingroup$ Try the definition of weak convergence. Random variables converges weakly if their distribution converges weakly. This is also stated in the wiki-article. $\endgroup$
    – Stefan Hansen
    Commented Sep 2, 2014 at 11:28
  • $\begingroup$ Are you suggesting the following? \begin{equation*} \lim_{n\rightarrow \infty }\mathbb{P}\left( \frac{\sum_{i=1}^{n}X_{i}-n% \mathbb{E}\left( X_{1}\right) }{a_{n}}\leq x\right) =\gamma (-\infty ,x] \end{equation*} $\endgroup$
    – Anonymous
    Commented Sep 2, 2014 at 11:32
  • $\begingroup$ In all continuity points of the function $x\mapsto \gamma((-\infty,x])$ at least, i.e. it holds for all $x$ with $\gamma(\{x\})=0$. This is equivalent to saying that $\lim_{n\to\infty}{\rm E}[f(Y_n)]=\int f\,\mathrm d\gamma$ for all bounded, continuous $f$, where $Y_n$ is your sequence of random variables. The latter is the usual definition of weak convergence / convergence in distribution. $\endgroup$
    – Stefan Hansen
    Commented Sep 2, 2014 at 11:39

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