I am trying to give a theoretical background for a sample process.
I am sampling values from a Poisson process with a fixed $\lambda$. I am trying to calculate the expected value of $n$ samples from Poisson distribution when considering only the maximal value between the random value and $\lambda$.
Consider the following function:
$$ C(x)= \begin{cases} x, & x \le \lambda\\ \lambda, & x > \lambda\\ \end{cases} $$
I am trying to calculate the expected value of: $\sum_{i=0}^{n}C(X)$
My first attempt was to try and calculate $\mathbb{P}(X \le \lambda)$:
$\mathbb{P}(X \le \lambda) = e^{-\lambda}\sum_{i=0}^{\lambda}\frac{\lambda^{i}}{i!}$
But I am not sure how to proceed from here. Any suggestions?
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, $$ \mathbb{E}C(X)=\lambda\left(1-e^{-\lambda}\frac{\lambda^{\lambda^*}}{\lambda^*!}\right) $$