Obtain $\lim _{t \rightarrow \infty} \mathbb{E}\{|N(t)-\lambda t|\}$ for a Poisson process $\boldsymbol{N}(\boldsymbol{t})$ with rate $\lambda$. I want to Obtain $$\lim _{t \rightarrow \infty} \mathbb{E}\{|N(t)-\lambda t|\}$$ for a Poisson process $\boldsymbol{N}(\boldsymbol{t})$ with rate $\lambda$.
I tried to go with upper-bounds and lower-bounds and involving a convergence in distribution but with no success
Any suggestions??
 A: If $X$ is a Poisson distribution with parameter $\lambda$, then we have
$$
\mathbf{E}|X-\lambda|=\sum_{k=0}^{\infty}|k-\lambda|\frac{e^{-\lambda}\lambda^k}{k!}=\sum_{k>\lambda} (k-\lambda)\frac{e^{-\lambda}\lambda^k}{k!}+\sum_{k\leq \lambda}(\lambda-k)\frac{e^{-\lambda}\lambda^k}{k!}.
$$
Since $\mathbf{E}(X-\lambda)=0$, we have
$$
\sum_{k>\lambda} (k-\lambda)\frac{e^{-\lambda}\lambda^k}{k!}= -\sum_{k\leq \lambda} (k-\lambda)\frac{e^{-\lambda}\lambda^k}{k!}. 
$$
Thus,
$$
\mathbf{E}|X-\lambda|=2\sum_{k\leq \lambda} (\lambda-k)\frac{e^{-\lambda}\lambda^k}{k!}=2\sum_{k\leq \lambda}\left( \frac{e^{-\lambda}\lambda^{k+1}}{k!}-\frac{e^{-\lambda}\lambda^k}{(k-1)!}\right)=\frac{2e^{-\lambda}\lambda^{\lfloor \lambda\rfloor+1}}{\lfloor \lambda\rfloor !}. 
$$
Since $N(t)$ is Poisson distributed with parameter $\lambda t$, we have
$$\mathbf{E}|N(t)-\lambda t| = \frac{2e^{-\lambda t}(\lambda t)^{\lfloor \lambda t\rfloor+1}}{\lfloor \lambda t \rfloor !}.$$
We use this expression and Stirling's formula to conclude that the limit is $\infty$. To see this,
$$
\frac{2e^{-\lambda t}(\lambda t)^{\lfloor \lambda t\rfloor+1}}{\lfloor \lambda t \rfloor !}\sim 2e^{-\lambda t}(\lambda t)^{\lfloor \lambda t\rfloor+1} \left(\frac e{\lfloor\lambda t\rfloor}\right)^{\lfloor \lambda t \rfloor }\frac1{\sqrt{2\pi\lfloor \lambda t \rfloor}}
$$
as $t\rightarrow\infty$.
