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I know the "usual" CLT for iid, square integrable sequence of random variables $(X_i)$. In the theorem $$\operatorname{Var}\Big( \sum_{i = 1}^n X_i\Big) = n^2 \operatorname{Var}(X_1) \to \infty$$ when $n \to \infty$.

If the random variables $(X_i)$ are non longer equally distributed, but still independent, one can use Linderberg CLT, but one requirement is that $$s_n^2 =\operatorname{Var}\Big( \sum_{i = 1}^n X_i\Big) = \sum_{i = 1}^n \operatorname{Var}(X_i) \to \infty$$ when $n \to \infty$.

I wonder what can be said, for independent random variables $(X_i)$, if $$\operatorname{Var}\Big( \sum_{i = 1}^n X_i\Big) \to c \in] 0,\infty[$$

Do we have the convergence of $$\sum_{i =1}^n X_i - \mathbb{E}[X_i]$$ in distribution to some random variable? Is there a theory for that, such general, problem?

It would be amazing if the answer is yes, but if not, is there a theory of the less general problem? : $(B_i)$ are independent Bernoulli random variables with parameter $(p_i)$ ($p_i \in [0,1]$) such that $0 < \sum_{i = 1}^{\infty} p_i < \infty$. Thus $\sum_{i =1}^{\infty} p_i(1-p_i) < \infty$. What can be said about the limit in distribution (if it exists) of $$S_n = \sum_{i =1}^n B_i $$

I know that for fixed $n$, $S_n$ is a poisson binomial distribution. Thanks to Le Cam theorem (https://en.m.wikipedia.org/wiki/Le_Cam%27s_theorem) we can have an approximation of $S_n$ with Poisson random variable, but I don't see how I could help for this problem.

Is there a way to put hypothesis on $p_i$ ti be able to say something about the limit in law of $S_n$?

Thanks,

Edit : I think I can prove that $S_n$ cannot converge in distribution to a law $\mathcal{L}$ such that if $X+Y$ follow the law $\mathcal{L}$ then $X$ and $Y$ does. And if $X, Y$ are $\mathcal{L}$, then $X+Y$ is.

By contradiction : if $S_n = \sum_{i = 1}^n B_i \to Z$ where $Z$ follow the law $\mathcal{L}$, then we have also $S_n - B_1 \to Z'$ where $Z'$ follow $\mathcal{L} $. Thus $B_1 + S_n - B_1 \to Z$ implies, by independence, $B_1 + Z'= Z$ (in distribution), so $B_1$ follows $\mathcal{L}$. Contradiction. Consequently the limit can't be gaussian, Cauchy, Levy. Is my argument right?

K.defaoite
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jvc
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    For the case where the variance is bounded (first question), check "martingales bounded in L^2", the limit random variable may be basically anything. The point is that the first terms may have strong influence on the limit, which is typically precluded in the "standard" limit theorems. – Olivier Mar 30 '23 at 19:17
  • Thanks! I should have though about martingale. And, concerning the Bernoulli case. Do you think that there is enough structure to say something? – jvc Mar 30 '23 at 19:35

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By independence, the assumption $\operatorname{Var}\Big( \sum_{i = 1}^n X_i\Big) \to c \in] 0,\infty[$ implies the convergence of the series $\sum_{i=1}^\infty\operatorname{Var}(X_i)$, which in turn gives the converge in $\mathbb L^2$ (hence in probability) of the sequence $\left(\sum_{i=1}^nX_i\right)_{n\geqslant 1}$. Then a result guarantees the almost sure convergence.

Davide Giraudo
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