an estimation of the expected value of a Poisson process Let $N$ be a Poisson process with intensity $\lambda$, I want to prove that for any $c>0$,
$$\limsup_{t\rightarrow\infty}P(\sup_{0\leq s\leq t}(N_s-\lambda s)\geq c\sqrt{\lambda t})\leq \frac{1}{c\sqrt{2\pi}}$$
By Doob's inequality to the martingale $\{N_t-\lambda t\}_{t\geq 0}$, we get
$$P(\sup_{0\leq s\leq t}(N_s-\lambda s)\geq c\sqrt{\lambda t})\leq\frac{E[(N_t-\lambda t)^+]}{c\sqrt{\lambda t}}$$
So it is enough to show that
$$\limsup_{t\rightarrow\infty}\frac{E[(N_t-\lambda t)^+]}{\sqrt{t}}\leq\sqrt{\frac{\lambda}{2\pi}}$$
But the bound $E[(N_t-\lambda t)^2]$ is a little big for this estimation, could some give a more delicate estimation of $E[(N_t-\lambda t)^+]$? Thanks a lot!
 A: I think that I have found a solution:
Firstly I prove that for any Poisson random variable $N$ of paramter $\lambda$, we have 
$$E[(N-\lambda)^+]=E[(N-\lambda)^-]=e^{-\lambda}\frac{\lambda^{[\lambda]+1}}{[\lambda]}$$
where $[\lambda]$ the largest integer less than or equal to $\lambda$.
Since $(N-\lambda)^+-(N-\lambda)^-=N-\lambda$, then
$$E[(N-\lambda)^+]=E[(N-\lambda)^-]=\sum_{k=0}^{[\lambda]}e^{-\lambda}\frac{\lambda^k}{k!}(\lambda-k)=e^{-\lambda}\sum_{k=0}^{[\lambda]}\frac{\lambda^{k+1}}{k!}-e^{-\lambda}\sum_{k=0}^{[\lambda]-1}\frac{\lambda^{k+1}}{k!}=e^{-\lambda}\frac{\lambda^{[\lambda]+1}}{[\lambda]!}$$
Secondly by Stirling's formulae we get 
\begin{eqnarray*}
&&\lim_{t\rightarrow\infty}\frac{E[(N_t-\lambda t)^+]}{\sqrt{\lambda t}}\\
&=&\lim_{t\rightarrow\infty}\frac{e^{-\lambda t}(\lambda t)^{[\lambda t]+1}}{\sqrt{\lambda t}[\lambda t]!}\\
&=&\lim_{t\rightarrow\infty}\frac{e^{-\lambda t}\sqrt{\lambda t}(\lambda t)^{[\lambda t]}}{[\lambda t]!}\\
&=&\lim_{t\rightarrow\infty}\frac{e^{-\lambda t}\sqrt{\lambda t}(\frac{\lambda t}{[\lambda t]})^{[\lambda t]}}{[\lambda t]!}\times \frac{e^{[\lambda t]}[\lambda t]!}{\sqrt{2\pi[\lambda t]}}\\
&=&\lim_{t\rightarrow\infty}\frac{e^{[\lambda t]-\lambda t}(\frac{\lambda t}{[\lambda t]})^{[\lambda t]}}{\sqrt{2\pi}}\\
&=&\lim_{t\rightarrow\infty}\frac{1}{\sqrt{2\pi}}e^{[\lambda t]-\lambda t}e^{[\lambda t]\log(\frac{\lambda t}{[\lambda t]})}\\
&=&\lim_{t\rightarrow\infty}\frac{1}{\sqrt{2\pi}}e^{[\lambda t]-\lambda t}e^{\lambda t-[\lambda t]+o(\frac{\lambda t-[\lambda t]}{[\lambda t]})}\\
&=&\frac{1}{\sqrt{2\pi}}
\end{eqnarray*}
which terminates the proof
