Problem similar to Kolmogorov's inequality using martingale.

Question

Suppose that $X_k$ is a sequence of independent random variables with mean zero and variance $1$. Let $S_k=X_1+\cdots+X_k$ and let $$ h(\lambda)=\limsup_{n \rightarrow \infty}P\left(\max_{1\leq k\leq n}|S_k|\geq \lambda n^{1/2}\right).$$ Do we have that for each $p>0$ there exists a $C_p$ such that: $$ h(\lambda)\leq C_p\lambda^{-p},\quad\lambda>0? $$

Answer 1

The following proof shows the claim for $p \leq 2$:

First of all, since $\mathbb{P}(\ldots) \leq 1$, it suffices to prove the claim for $\lambda \geq 1$. Moreover, since $\lambda^{-2} \leq \lambda^{-p}$ for all $p \leq 2$ and $\lambda \geq 1$, we can restrict ourselves to the case $p=2$.

It follows from Etemadi's inequality that

$$\mathbb{P} \left( \sup_{1 \leq k \leq n} |S_k| \geq \lambda n^{1/2} \right) \leq \sup_{1 \leq k \leq n} \mathbb{P}\left( |S_k| \geq \lambda n^{1/2} \right).$$

By Markov's inequality and the independence of the random variables, we conclude

$$\mathbb{P} \left( \sup_{1 \leq k \leq n} |S_k| \geq \lambda n^{1/2} \right) \leq \frac{1}{n \lambda^2} \sup_{k \leq n} \mathbb{E}(S_k^2) = \frac{1}{n \lambda^2} \sup_{k \leq n} k = \frac{1}{\lambda^2}.$$

This finishes the proof.

Remark Instead of Etemadi's inequality, we can also use Doob's maximal inequality.

Answer 2

The constant $C_2$ exists: this can be seen from Doob's inequality.

For $p\gt 2$, if the constant $C_p$ exists then $$\tag{*}\sup_{\lambda\gt 0}\lambda^p\mathbb P(|X_1|\gt \lambda)<\infty.$$ It turns out that $(*)$ is also a sufficient condition, even when $(S_n)_{n\geqslant 1}$ is a martingale difference sequence. This can be seen from Hall and Heyde, which gives that for each $t$ and each $n$ $$\mathbb P\left(\max_{1\leqslant j\leqslant n}|S_j|\gt \beta t\right) \leqslant \frac{\delta^2}{(\beta-\delta-1)^ 2} \mathbb P\left(\max_{1\leqslant j\leqslant n}|S_j|\gt t\right)+\mathbb P\left(\sum_{i=1}^n\mathbb E[X_i^2\mid\mathcal F_{i-1}]\gt \delta t\right)$$ (in the proof of Theorem 2.11, page 28). Here $\beta\gt 1$ and $0\lt \delta\lt \beta-1$ are independent on $n$ and $t$. We choose these parameters to make $$\frac{\delta^2}{(\beta-\delta-1)^ 2}\beta^p\lt 1.$$

Answer 3

This question was posed by me on the final exam in Statistics 530 at the University of Pennsylvania. The problem was handed out only very shortly before it was posted here. I love the Stack Exchange, and I am sad to see that someone has abused it in this way. J. Michael Steele.