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Given a sequence $X_i$ of independant variables we denote $S_n$ as $\sum_{i=1}^{n}X_i$.

We know that $S_n$ converges in distribtion (weak* convergence).

Does that imply that $S_n$ converges almost surely?

When $\forall_{i}\ \mathbb{E}X_i = 0$, then, I guess, the sum sequence is a martingale and the convergence comes from definition.

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1 Answer 1

up vote 2 down vote accepted

Under the assumption of independence and weak convergence, we actually have almost everywhere convergence.

  • We have to show that there is convergence in probability; once it will be done, we will conclude with this thread, which is a weaker version of Levy's Equivalence Theorem that we are trying to prove here.

  • Let $\mu_{m,n}$ the law of $\sum_{j=n+1}^mX_j$, that is, the probability measure associated to the random variable $\sum_{j=n+1}^mX_j$. By the assumption, the set $\{\mu_{m,n},m\geqslant n+1\}$ is tight. Assume that $\{S_n\}$ is not Cauchy in probability. Then we can find $\delta>0$ and construct by induction a sequence $\{(m(k),n(k)\}$ of integers such that $m(k)\geqslant n(k)+1$ and $n(k),m(k)$ increase, and $$\forall k,\quad E[\min\{1,|S_{n(k)}-S_{m(k)}|\}]>\delta.$$

  • We have, using independence, that $\widehat{\mu_m}(t)= \widehat{\mu_n}(t)\widehat{\mu_{m,n}}(t)$, where for a probability measure $\nu$, $\widehat\nu (t):=\int_{\Bbb R}e^{itx}d\nu(x)$, the Fourier transform or the characteristic function of $\nu$. So $\widehat{\mu_{m(k),n(k)}}\to 1$ and $\mu_{m(k),n(k)}\to \delta_0$.
  • By the second point, $\mu_{m(k),n(k)}(\{x,|x|\geqslant\delta\})>\frac 1{1+\delta}$ for all $k$. Portmanteau theorem yields a contradiction with third point.

  • It works for random variables with values in separable Banach spaces. Actually, this sketch of proof is taken from an exercise in A. Araújo and E. Giné's book The Central Limit Theorem for Real and Banach Valued Random Variables. The third step is a bit more technical as we have to deal with characteristic functionals, which generalize characteristic functions.

  • Note that we didn't use assumption about the existence of the expectation or moments of any order.

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Hi Davide. I think making a clear statement at the beginning about what you're actually proving would help make this post more transparent. Cheers. :-) –  cardinal Jan 2 '13 at 15:31
    
(+1) From me. :-) –  cardinal Jan 2 '13 at 15:35
    
How do you define $\hat \mu$ ? What does "Let $\mu_{m,n}$ the law of $\sum^m_{j=n+1}Xj$" mean? @DavideGiraudo –  Golob Jan 2 '13 at 21:21
    
@Golob I've added this. Do you know what the law of a real valued random variable is? –  Davide Giraudo Jan 2 '13 at 21:25
1  
ah, the distribution. Yes, I do. Thanks alot. –  Golob Jan 3 '13 at 1:16

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