# Proof of a corollary for the Strong Law of Large Number

Here is Theorem 1.13 from Mathematical Statistics Jun Shao.

$$X_i$$ are i.i.d. A necessary and sufficient condition for the existence of constant $$c$$ for which \begin{align} \frac{1}{n}\sum_{i=1}^n X_i \longrightarrow c\quad a.s. \end{align} is that $$E|X_1|<\infty$$ in which case $$c=EX_1$$. And \begin{align} \frac{1}{n}\sum_{i=1}^nc_i(X_i-EX_1)\longrightarrow 0 \quad a.s. \end{align} for every bounded real $$\{c_i\}$$

The first part is just the SLLN. The proof of the second part is marked as an exercise. However, I am having trouble to see this. My attempt is the following: To prove the result, it is sufficient to prove $$P(|c_i(X_i-EX_1)|\geq \epsilon \ i.o.)=0$$ for every $$\epsilon$$. And this is equivalent to prove $$P(|X_i-EX_1|\geq \epsilon \ i.o.) = 0$$ since $$c_i$$ is bounded. At this step, I try to use Borel-Cantelli, but it seems does not work. Does anyone have any idea for this?

• First, we can assume that $X_i$ has 0 mean by replacing it by $X_i-\mathbb{E}X_i$. Then I think that you can simply adapt the proof of the LLN, that is: let $Y_i=c_iX_i\mathbf{1}_{\lvert c_iX_i\rvert\leq i}$ and show that (1) $\lim\mathbb{E}Y_n=0$, (2) a.s. there exists $J\geq 1$ such that for all $i\geq J$, $Y_i=c_iX_i$ (3) $\sum_{i\geq 1}\operatorname{Var}(Y_i)/i^2<\infty$. Jan 4 at 11:35
Expanding a bit on my comment: replace $$X_i$$ by $$X_i-\mathbb{E}X_i$$. Let $$X$$ be distributed as the $$(X_i)_{i\geq 1}$$ and $$Y_i=c_iX_i\mathbf{1}_{\lvert c_iX_i\rvert\leq i}$$. By dominating convergence, $$\lim\mathbb{E}Y_i=0$$. Denoting $$M=\sup_{i\geq 1}\lvert c_i\rvert$$, $$\sum_{i=1}^\infty\mathbb{P}(c_iX_i\neq Y_i)=\sum_{i=1}^\infty\mathbb{P}(\lvert c_iX\rvert>i)\leq\sum_{i=1}^n\mathbb{P}(M\lvert X\rvert>i)\leq\int_0^\infty\mathbb{P}(M\lvert X\rvert>t)dt=\mathbb{E}M\lvert X\rvert<\infty$$ So the Borel-Cantelli lemma applies and a.s., there exists $$J\geq 1$$ such that for all $$i\geq J$$ the equality $$c_iX_i=Y_i$$ holds. Finally, $$\sum_{i=1}^\infty\frac{\operatorname{Var}(Y_i)}{i^2}\leq\sum_{i=1}^n\frac{\mathbb{E}Y_i^2}{i^2}=\mathbb{E}\left[\lvert X\rvert^2\sum_{i=1}^\infty\frac{\mathbf{1}_{\lvert c_iX\rvert>i}}{i^2}\right]\leq c\mathbb{E}\lvert X\rvert<\infty$$ for some constant $$c\geq 0$$. We can now apply the following lemma with $$Z_n=Y_n-\mathbb{E}Y_n$$.
Let $$(Z_n)_{n\geq 1}$$ be a sequence of independent r.v. with $$\mathbb{E}Z_n=0$$ and $$\sum_{n\geq 1}\operatorname{Var}(Z_n)/n^2<\infty$$. Denoting $$S_n=Z_1+...+Z_n$$, the sequence $$(S_n/n)_{n\geq 1}$$ converges a.s. to $$0$$.
Since $$(\mathbb{E}Y_i)_{i\geq 1}$$ converges to $$0$$ $$\frac{Y_1+...+Y_n}{n}=\frac{Z_1+...+Z_n}{n}+\frac{\mathbb{E}Y_1+...+\mathbb{E}Y_n}{n}$$ converges to $$0$$ as well a.s. Finally, a.s., for all $$n\geq J$$, $$\frac{c_1X_1+...+c_nX_n}{n}=\frac{c_1X_1+...+c_{J-1}X_{J-1}}{n}+\frac{Y_J+...+Y_n}{n}\longrightarrow 0$$