absolute value of poisson random variable's deviation from a constant

Question

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.

Answer 1

$$|N - q| = \begin{cases} N-q, & N \geq q \\ q - N, & N < q\text{.} \end{cases}$$ 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.

Answer 2

Hint:

$|N-q|=(N-q)1_{N\geq q}+ (q-N)1_{N\leq q}$