Counterexamples and Kolmogorov zero-one law (on the convergence of $\sum_{k=1}^\infty (\xi_k - \mathbf{E}(\xi_k))$)

Question

The following is a formulation of Kolmogorov zero-one rule:

Let $\{\xi_n\}_{n=1}^\infty$ be a sequence of random variables on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$ such that $\xi_k \in L^1(\Omega)$ for all $k \in \mathbb{N}$.

Theorem 1 Suppose

1. $\xi_1, \xi_2, ...$ are independent;

2. $\sum_{k=1}^\infty \mathbf{Var}(\xi_k) < \infty$

Then $\sum_{k=1}^\infty (\xi_k - \mathbf{E}(\xi_k))$ converges almost everywhere.

Theorem 2 Suppose

1. $\xi_1, \xi_2, ...$ are independent;

2. $\sum_{k=1}^\infty \mathbf{Var}(\xi_k) = \infty$;

3. there exists $C>0$ such that $|\xi_k|\le C$ for all $k \in \mathbb{N}$.

Then $\sum_{k=1}^\infty (\xi_k - \mathbf{E}(\xi_k))$ diverges almost everywhere.

• What counterexamples show that Theorem 1 does not hold if either assumption 1. or assumption 2. is removed?

• What counterexamples show that Theorem 2 does not hold if either assumption 1. or assumption 2. or assumption 3. is removed?

Answer 1

Theorem 1 does not hold without assumption 1: Let $\xi$ be such that $\mathbb{P}(\xi=\pm 1)=1/2$. If we set $$\xi_k := \frac{1}{k} \xi,$$ then $\sum_k \text{var}(\xi_k)< \infty$, but $$\sum_{k \geq 1} (\xi_k-\mathbb{E}\xi_k) = \sum_{k \geq 1} \left( \frac{1}{k} \xi \right)$$ fails to converge with probability 1.

Theorem 1 does not without assumption 2: Let $\xi_k$, $k \geq 3$, be independent random variables such that $$\mathbb{P}(\xi_k = -k^2) = \frac{1}{k^2} \qquad \mathbb{P}(\xi_k = k^2/(k^2-1)) = 1- \frac{1}{k^2}.$$ A straight-forward computation shows that $\mathbb{E}\xi_k=0$ for all $k \geq 3$. Since $$\sum_{k \geq 3} \mathbb{P}(\xi_k=-k^2) < \infty$$ it follows from the Borel-Cantelli lemma that $\sharp\{k \geq 3; \xi_k(\omega)=-k^2\}< \infty$ with probability 1 and therefore $\xi_k(\omega) = \frac{k^2}{k^2-1}$ for $k \geq k_0(\omega)$ sufficiently large with probability $1$; in particular $\sum_{k \geq 1} \xi_k$ does not converge.

Theorem 2 does not hold without assumption 1: Let $\xi \in L^2$ be such that $\mathbb{P}(\xi=\pm 1)=\frac{1}{2}$. If wet set

$$\xi_k := \frac{(-1)^k}{\sqrt{k}} \xi$$

then $\mathbb{E}(\xi_k)=0$, $\sum_k \text{var}(\xi_k) = \infty$ and $|\xi_k| \leq 1$, but $\sum_k (\xi_k-\mathbb{E}\xi_k)$ converges with probability 1.

Theorem 2 does not hold without assumption 2: This is obvious from Theorem 1; just choose any sequence of independent and bounded random variables such that $\sum_k \text{var}(\xi_k)<\infty$.

Theorem 2 does not hold without assumption 3: Let $(\xi_k)_{k \geq 3}$ be a sequence of independent random variables such that $$\mathbb{P}(\xi_k = \pm k^2) = \frac{1}{k^2} \qquad \mathbb{P}(\xi_k=0) = 1- \frac{2}{k^2}.$$ Then $\mathbb{E}(\xi_k)=0$ and $\text{var}(\xi_k)= 2$ for all $k$; in particular $\sum_k \text{var}(\xi_k)=\infty$. Since $\sum_k \mathbb{P}(\xi_k \neq 0)< \infty$, it follows from the Borel Cantelli lemma that $\xi_k(\omega)=0$ for $k \geq k_0(\omega)$ sufficiently large with probability $1$, and therefore $\sum_k (\xi_k-\mathbb{E}(\xi_k)) = \sum_k \xi_k$ converges with probability $1$.