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
$$|N - q| = \begin{cases} N-q, & N \geq q \\ q - N, & N < q\text{.} \end{cases}$$ 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.
Hint:
$|N-q|=(N-q)1_{N\geq q}+ (q-N)1_{N\leq q}$