absolute value of poisson random variable's deviation from a constant Let $q\in \{0,1,2,\dotsc\}$ be a fixed constant and $N$ be a Poisson random variable with parameter $\lambda$. Let $X=|N-q|$. Determine $\mathbb{E}[|N-q|]$, if possible. 
So, we start out, $\mathbb{E}[X]=\sum_{n=1}^\infty P[X\geq n]=\sum_{n=1}^\infty (1-P[X< n]).$ Now, $P[X<n]=P[|N-q|<n]=P[q-n<N<q+n]=F_N(q+n-1)-F_N(q-n+1)$, where $F_N$ is the cumulative distribution function of $N$. But here I am having trouble simplifying this last expression. The cdf can be written in terms of the (incomplete) gamma function but I'm having trouble using that too. Any suggestions would be appreciated. 
 A: $$|N - q| = \begin{cases}
N-q, & N \geq q \\
q - N, & N < q\text{.}
\end{cases}$$
Hence,
$$\begin{align}
\mathbb{E}[|N-q|] &= \sum_{k=0}^{q-1}(q-k)\cdot \dfrac{e^{-\lambda}\lambda^{k}}{k!} + \sum_{k=q}^{\infty}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!} \\
&= \sum_{k=0}^{\infty}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!}+\sum_{k=0}^{q-1}(q-k)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!}-\sum_{k=0}^{q-1}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!} \\
&= \mathbb{E}[N-q]-2\sum_{k=0}^{q-1}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!} \\
&= \lambda-q-2\sum_{k=0}^{q-1}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!}
\end{align}$$
I am not sure if this can be simplified further.
Edited: WolframAlpha says that 
$$\sum_{k=0}^{q-1}(k-q)\cdot\dfrac{e^{-\lambda}\lambda^{k}}{k!} = \dfrac{\lambda\left[\Gamma(q+1, \lambda)-e^{-\lambda}\lambda^{q}\right]}{\Gamma(q+1)} - \dfrac{\Gamma(q+1, \lambda)}{\Gamma(q)}$$
where $\Gamma(\cdot, \cdot)$ is the incomplete Gamma function.
A: Hint:
$|N-q|=(N-q)1_{N\geq q}+ (q-N)1_{N\leq q}$
