Let $X_1,X_2,\dots,$ be i.i.d. random variables with $\mathbb{E}[X_1]=0,\operatorname{Var}(X_1)<\infty$. Assume that $\sum_{k=1}^{n}a_{nk}^2=\frac{1}{n}$, let $S_n=\sum_{k=1}^{n}a_{nk}X_k$, prove that
$$S_n\rightarrow0~~a.s.$$
It seems that this looks like the triangle array of some law of large numbers, we wish to use Borel-Cantelli to deal with this problem, but it seems hard since all $a_{nk}$ are arbitrary. How should we deal with the $a_{nk}$s here in this problem?
Best Answer
Note that the usual law of large numbers could look as a particular case of this result, namely, by taking $a_{n,i}=1/n$ for each $1\leqslant i\leqslant n$. But the SLLN only requires $X_1$ to be integrable.
The difficulty here is that we cannot use maximal inequality because the weights may not depend on $n$ nicely.
But we can use a truncation argument: for a fixed $R$, let $X_{i,\leqslant R}:=X_i\mathbf{1}_{\{\lvert X_i\rvert\leqslant R\}}$ and $X_{i,\gt R}:=X_i\mathbf{1}_{\{\lvert X_i\rvert\gt R\}}$. Then $$ \tag{1} S_n=\sum_{i=1}^na_{n,i}\left(X_{i,\leqslant R}-\mathbb E\left[X_{i,\leqslant R}\right]\right)+\sum_{i=1}^n a_{n,i}\mathbb E\left[X_{i,\leqslant R}\right]+ \sum_{i=1}^na_{n,i}X_{i,\gt R}.$$ Observe that since $X_i$ is centered and has the same distribution as $X_1$, $$ \sum_{i=1}^n a_{n,i}\mathbb E\left[X_{i,\leqslant R}\right]=-\sum_{i=1}^n a_{n,i}\mathbb E\left[X_{i,\gt R}\right]=-\sum_{i=1}^na_{n,i}\mathbb E\left[X_{1,\gt R}\right] $$ hence by the Cauchy-Schwarz inequality, $$\tag{2}\left\lvert \sum_{i=1}^n a_{n,i}\mathbb E\left[X_{i,\leqslant R}\right]\right\rvert\leqslant \mathbb E\left[\lvert X_{1}\rvert \mathbf{1}_{\{\lvert X_1\rvert>R\}}\right].$$
An application of Cauchy-Schwarz inequality gives $$\tag{3} \left\lvert \sum_{i=1}^na_{n,i}X_{i,\gt R}\right\rvert\leqslant \sqrt{\sum_{i=1}^na_{n,i}^2\sum_{j=1}^nX_{j,>R}^2}=\sqrt{\frac 1n\sum_{j=1}^nX_{j,>R}^2}. $$ The combination of (1), (2) and (3) with the strong law of large numbers applied to the i.i.d. integrable sequence $\left(X_{j,>R}^2\right)_{j\geqslant 1}$ give the almost sure inequality $$\tag{4} \limsup_{n\to\infty}\left\lvert S_n \right\rvert\leqslant \limsup_{n\to\infty}\left\lvert \sum_{i=1}^na_{n,i}\left(X_{i,\leqslant R}-\mathbb E\left[X_{i,\leqslant R}\right]\right)\right\rvert+\mathbb E\left[\lvert X_{1}\rvert \mathbf{1}_{\{\lvert X_1\rvert>R\}}\right]+\mathbb E\left[X_1^2\mathbf{1}_{\{\lvert X_1\rvert>R\}}\right] $$ By Hoeffding's inequality , the following estimate takes place: $$ \mathbb P\left(\left\lvert \sum_{i=1}^na_{n,i}\left(X_{i,\leqslant R}-\mathbb E\left[X_{i,\leqslant R}\right]\right)\right\rvert>\varepsilon\right) \leqslant 2\exp\left(-\frac{\varepsilon^2}{4\sum_{i=1}^na_{n,i}^2}\right) = 2\exp\left(-n\frac{\varepsilon^2}{4}\right). $$ By the Borel-Cantelli lemma, the convergence $$ \lim_{n\to\infty}\left\lvert \sum_{i=1}^na_{n,i}\left(X_{i,\leqslant R}-\mathbb E\left[X_{i,\leqslant R}\right]\right)\right\rvert=0 $$ takes place for each $R>0$ hence (4) gives $$ \limsup_{n\to\infty}\left\lvert S_n \right\rvert\leqslant \mathbb E\left[\lvert X_{1}\rvert \mathbf{1}_{\{\lvert X_1\rvert>R\}}\right]+\mathbb E\left[X_1^2\mathbf{1}_{\{\lvert X_1\rvert>R\}}\right]. $$ Since $R$ is arbitrary, the conclusion follows by the monotone or dominated convergence theorem.