The expected value of the sum of Poisson distribution sample series with upper bound on the sampled value

Solution 1:

One might note that $$ x\mathbb{P}(X=x)=\lambda \mathbb{P}(X=x-1) $$ and therefore,

$$ \mathbb{E}\mathbb{1}_{X< \lambda}X=\sum_{i=1}^{\lceil \lambda-1\rceil} i\mathbb{P}(X=i)=\lambda \sum_{j=0}^{\lceil\lambda-2\rceil} \mathbb{P}(X=j)=\lambda\mathbb{P}(X\leq \lambda-2), $$ which, in turn, gives \begin{align} \mathbb{E}C(X)&=\mathbb{E}\mathbb{1}_{X< \lambda}X+\mathbb{E}1_{X\geq\lambda}\lambda\\ &=\lambda \mathbb{P}(X\leq \lambda-2)+\lambda \mathbb{P}(X\geq \lambda)\\ &=\lambda(1-\mathbb{P}(X\in (\lambda-2,\lambda))) \end{align} Now, there's exactly one natural number in $(\lambda-2,\lambda)$, call it $\lambda^*$ and thus, $$ \mathbb{E}C(X)=\lambda\left(1-e^{-\lambda}\frac{\lambda^{\lambda^*}}{\lambda^*!}\right) $$