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Let's assume we have sequence $(X_n)$ of r.v. on a probability space $(\Omega,\mathcal{F},P)$ and we denote by $\mu_n$ the distribution of $X_n$. Now we assume that the sequence of distributions is tight, therefore there is a subsequence such that

$$ \mu_{n_k} \overset{d}{\longrightarrow}\mu$$

for $k \to \infty$. Thus there's a r.v. $X$ with distribution $\mu$. This should follow by a lemma of Lebesgue-Stieltjes. The question is, what can I say about $X$. It isn't a r.v. on $(\Omega,\mathcal{F},P)$, right? From the proof of the lemma of Lebesgue-Stieltjes it's a r.v. on $((0,1),\mathcal{B}(0,1),\lambda)$ where $\lambda$ is the Lebesgue measure. If the originally sequence has some properties, for example boundedness can we deduce that $ E(|X|)<\infty$ ? What about other properties? My problem is, I do not see a connection between the $(X_n)$ and $X$.

Thank you for your help.


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As a concrete example, using Chris's idea, Fatou's lemma shows that $E(|X|)\leq \liminf_n E(|X_n|)$. – Byron Schmuland Feb 10 '12 at 16:41
@ Byron Schmuland: I know the equivalent definition of weak convergence described in Chris' answer. But I do not see why in this case $ Y_n = X_n$. My situation is different: If we look at $(|X_n|)$ we can use Fatou as you mentioned and we denote $ |X|:=\lim_n |X_n|\in [0,\infty]$. But I don't see why $X$ should be a r.v. and why $ \mu$ should be the distribution of $X$. From the definition of tightness we know that there exists $(Y_n)$ as Chris said, but the sequence $(Y_n)$ is different to the sequence $(X_n)$, isn't it? – user20869 Feb 10 '12 at 17:49
$Y_n$ is different from $X_n$, but they have the same distribution and hence the same expected absolute mean: $E(|Y_n|)=E(|X_n|)$. – Byron Schmuland Feb 10 '12 at 17:57
Ah....and it's also true that $E(|X|)=E(|Y|)$ and then everything is clear. Thanks a lot! – user20869 Feb 10 '12 at 18:15
up vote 1 down vote accepted

In general, there is not necessarily any connection between $X_n$ and $X$. In fact, convergence in distribution is defined in a way that there is no reason to require that the $X_n$ sequence be defined on a single probability space.

There is, however, a relationship that you can force if you want to find out some information about the target random variable. Given the sequence of probability laws $\mu_n$ corresponding to $X_n$ we can define new random variables $Y_n$ on $([0,1],\mathcal{B}, \lambda$) where $\lambda$ is Lebesgue so that $Y_n \to Y$ a.s. and where the law of $Y$ is given by $\mu$. Using this representation you can often deduce some information about the random variables like what you were looking for.

You may want to take a look at Rick Durrett's book Probability Theory and Examples.

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