Let $N$ be a Poisson process with parameter $\lambda$, that is, for $a>b\geq0$, there is $$P[N(a,b)=k]=\frac{((ab)\lambda)^k}{k!}e^{(ab)\lambda}.$$ Now denote $N_t=N[0,t)$ and define $$ M_t=N_t\lambda t\qquad 0\leq t<\infty. $$ we can easily show that $M$ is a Martingale. Now the question is, prove that for any $c>0$, there is \begin{align*} \limsup_{t\rightarrow\infty}&P\left[\sup_{s\leq t}M_s\geq c\sqrt{\lambda t}\right]\leq\frac{1}{c\sqrt{2\pi}}\\ &E\left[\sup_{s\leq u\leq t}\left(\frac{M_u}{u}\right)^2\right]\leq\frac{4t\lambda}{s^2} \end{align*} The main difficulty I meet is how to describe the $sup$ of random variables and how to make use of the fact that $M$ is a martingale to evaluate the probability.

$\begingroup$ Crossposted: math.stackexchange.com/questions/2714345/…. $\endgroup$– jeqMar 30, 2018 at 15:13

$\begingroup$ @jeq I’m sorry but I didn’t get answer there, and I don’t know where is the best place to post this question. $\endgroup$– aurora_borealisMar 30, 2018 at 15:23
1 Answer
$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$
If $(S_t)_{t\ge0}$ is a nonnegative submartingale, then the following Dood inequalities hold: \begin{equation} P(\sup_{0\le u\le t} S_u\ge x)\le ES_u/x \tag{1} \end{equation} for $x>0$ and \begin{equation} \\sup_{0\le u\le t} S_u\_p\le\frac p{p1}\,\S_t\_p \tag{2} \end{equation} for $p>1$, where $t>0$ and $\X\_p:=(EX^p)^{1/p}$.
Using (1) with $S_u=\max(0,M_u)$, we have \begin{multline*} P(\sup_{s\le t}M_s\ge c\sqrt{\la t}) =P(\sup_{s\le t}\max(0,M_s)\ge c\sqrt{\la t}) \\ \le\frac{E\max(0,M_t)}{c\sqrt{\la t}} \underset{t\to\infty}\longrightarrow\frac{E\max(0,Z)}{c}=\frac1{c\sqrt{2\pi}} \end{multline*} by the central limit theorem, where $Z\sim N(0,1)$.
Using (2) with $S_u=M_u^2$, for $0<s<t$ we have \begin{equation} E\sup_{s\le u\le t}\left(\frac{M_u}{u}\right)^2 \le \frac1{s^2}\,E\sup_{s\le u\le t}M_u^2 \le \frac1{s^2}\,\Big(\frac 2{21}\Big)^2\,EM_t^2 =\frac{4t\la}{s^2}. \end{equation}