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?

Thanks in advance!

  • $\begingroup$ 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) $\sum_{i\geq 1}\operatorname{Var}(Y_i)/i^2<\infty$. $\endgroup$
    – charlus
    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 $n\geq J$, $$\frac{c_1X_1+...+c_nX_n}{n}=\frac{c_1X_1+...+c_{J-1}X_{J-1}}{n}+\frac{Y_J+...+Y_n}{n}\longrightarrow 0$$

  • 1
    $\begingroup$ Thanks! Thanks brilliant. $\endgroup$
    – Zorualyh
    Jan 4 at 13:19

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