Lower bounding an alternating series with signs from a martingale difference sequence

Question

Let $\epsilon_n \in \{-1, 1\}$ be a martingale difference sequence, in the sense that

$$M_n := \sum_{i = 0}^n \epsilon_i$$

is a martingale.

We assume $\epsilon_0 = \pm 1$ with probability $\frac{1}{2}$ each.

Question: Is it true that there exists some absolute constant $C > 0$ such that for any deterministic, strictly positive sequence $a_n$ of real numbers with $1 \leq \sum_{n=1}^\infty a_n < \infty$, we have

$$\mathbb E \left [ \big |\sum_{n=1}^\infty \epsilon_n a_n \big | \right ] \geq C \sqrt {\sum_{n = 1}^\infty a_n}?$$

Remark: Note that the sum on the LHS converges a.s. and in $L^1$ by the martingale convergence theorem.

Answer 1

$\newcommand\ep\epsilon$No. Suppose e.g. that for some natural $N$ we have $a_1=\cdots=a_N=1/N$ and $a_i=0$ for $i>N$. Then $\sum_1^\infty a_n=1$ and $$E\Big|\sum_1^\infty\ep_n a_n\Big|\le\sqrt{E\Big(\sum_1^\infty\ep_n a_n\Big)^2}=\frac1{\sqrt N}\to0$$ as $N\to\infty$, so that the inequality in question does not hold for any universal constant $C>0$.

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Remark: Actually, your $\ep_i$'s must be independent Rademacher random variables, in view of your martingale condition and because $\ep_n\in\{-1,1\}$.