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I know that $(\gamma_i)$ is a square summable sequence and that $$ L_n :=\sum_{i=1}^n \gamma_i (W_i^2-1) \quad \quad \forall n\in \mathbb{N} $$ where $(W_i)$ is an i.i.d. sequence of std. normal distributed rvs.

I know that $(L_n)$ converges almost surely (Khinchin-Kolmogorov) towards $\sum_{i=1}^\infty \gamma_i (W_i^2-1)\in \mathcal{L}^2$.

One can derive that $$ Var \left( \sum_{i=1}^\infty \gamma_i (W_i^2-1) \right) = \sum_{i=1}^\infty \gamma_i^2 Var (W_i^2-1) = 2 \sum_{i=1}^\infty \gamma_i^2 < \infty. $$ and we especially know that it has mean.

Problem how do i derive the expectation of the limit?

I know that
$$ E\left( \sum_{i=1}^\infty \gamma_i (W_i^2-1) \right) = \sum_{i=1}^\infty E\gamma_i (W_i^2-1) =0 $$ if $\sum_{i=1}^\infty E|\gamma_i (W_i^2-1)|=\sum_{i=1}^\infty |\gamma_i| \, E| (W_i^2-1)|=c\sum_{i=1}^\infty |\gamma_i| \, <\infty$, but i don't have the absolute convergence, only square summability.

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    You know that $$R_n=\sum_{i=n}^\infty \gamma_i (W_i^2-1)$$ converges to $0$ in $L^2$ and you are asking if $$E(|R_n|)\to0$$ Well, the inequality $$E(|R_n|)^2\leqslant E(R_n^2)\to0$$ answers this this question, no? – 2017-02-08
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    Well in order to use Fubini's theorem i need that $$ E\left( \sum_{i=1}^\infty |\gamma_i (W_i^2-1)| \right) < \infty $$ and not that $E|R_n| \to 0$ right? – 2017-02-08
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    Please reread my comment. No Fubini here, only the remark that if $L=L_n+R_{n+1}$ with $E(L_n)=0$ for every $n$ and $E(|R_n|)\to0$ then $E(L)=0$. Yes, as soon as you turn to sums $$\sum|\gamma_i(W_i^2-1)|$$ you are doomed... but $E(|R_n|)$ does not use these, only sums$$\left|\sum\gamma_i(W_i^2-1)\right|$$ – 2017-02-08
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    Thanks Did, please paste your comment as an answer. – 2017-02-08

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