I need your expertise in solving the following problem:
Let $Y_{1},Y_{2},...$ be a sequence of gaussian independent random variables with zero mean and unit variance. Let $X_{i}=\exp \left( a_{i}Y_{i}\right) $ , where we suppose that $\sum \limits_{i\geq 1}a_{i}Y_{i}\rightarrow Y$ a.s.
Is that $E\left[ \prod \limits_{i=1}^{\infty }X_{i}\right] <\infty $?
The assumption $\displaystyle\sum_{i\geq 1}a_{i}Y_{i}\rightarrow Y$ a.s. is crucial, here, because (together with $EY_i=0$) it's equivalent with $\displaystyle\sum_{i\geq 1}a^2_{i}<\infty$ (cf. a.s. convergence of sum of normal random variables). Since $\displaystyle Ee^{a_iY_i}=e^{a^2_{i}/2}$, we have (using the independence of the $X_i$ in the first step) $$E\left[ \prod_{i=1}^{\infty }X_{i}\right]\le\prod_{i=1}^{\infty }EX_{i}=\exp{\left(\frac12\sum_{i\geq 1}a^2_{i}\right)} <\infty.$$ Because of $EY_i=0$, assumptions about $\displaystyle\sum_{i\geq 1}a_{i}$ are not necessary.
EDIT: The strict reasoning would be $$E\left[ \prod_{i=1}^nX_{i}\right]=\prod_{i=1}^nEX_{i}\le\exp{\left(\frac12\sum_{i\geq 1}a^2_{i}\right)} <\infty,$$ so $$E\left[ \prod_{i=1}^{\infty }X_{i}\right]\le\liminf_{n\to\infty} \prod_{i=1}^nEX_{i}\le\exp{\left(\frac12\sum_{i\geq 1}a^2_{i}\right)} <\infty$$ due to Fatou's lemma, but that's fairly obvious.
REMARK: It's actually possible to prove $$E\left[ \prod_{i=1}^{\infty }X_{i}\right]=\prod_{i=1}^{\infty }EX_{i},$$ but the proof isn't exactly obvious. Let $$Z_n=\prod_{i=1}^nX_{i}=\exp{\left(\sum^n_{i= 1}a_{i}Y_i\right)}.$$ Denoting by $\mathcal{F}_n$ the $\sigma$-algebra generated by $X_1,\ldots,X_n$, we see that $$E(Z_{n+1}|\mathcal{F}_n)=Z_n\,EX_{n+1}=Z_n\,e^{a^2_{n+1}/2}\ge Z_n,$$ i.e. $Z_n$ is a non-negative submartingale. Moreover, it's quadratically integrable, since $$EZ^2_n=\exp{\left(2\sum^n_{i=1}a^2_{i}\right)}\le\exp{\left(2\sum_{i\geq 1}a^2_{i}\right)}$$ If we set $\displaystyle S_n=\sup_{k\le n}Z_k$ and $S_\infty=\sup Z_k$, we have due to Doob's martingale inequality $$ES_n\le(ES^2_n)^{1/2}\le 2(EZ^2_n)^{1/2}\le2\exp{\left(\sum_{i\geq 1}a^2_{i}\right)}<\infty$$ and (using Fatou's lemma) $ES\infty\le2\exp{\left(\sum_{i\geq 1}a^2_{i}\right)}<\infty$, so we can use the dominated convergence theorem, and thus $$E\left(\lim_{n\to\infty}Z_n\right)=\lim_{n\to\infty}EZ_n.$$