This question might have been asked here before, but I have not managed to found an answer to my question. So for $k = 1,2,\dots$, let $X_i \sim \mathrm{Poi}(\lambda)$ be i.i.d. r.v.s and $Y = \frac{1}{K}\sum_{k=1}^{K}X_k$.
- Is the theoretical expected value of $Y$ equal to $\lambda$?
- Supposing $\mathrm{E}(Y) = \lambda$, does it follow that $\mathrm{Var}(Y) = \lambda$?
- If all $X_k$ would still be independent, but now $X_k \sim \mathrm{Poi}(\lambda_k)$ it is true that $\sum_{k=1}^{K}X_k \sim \mathrm{Poi}(\sum_{k=1}^{K}\lambda_k)$. But is now $Y \sim \mathrm{Poi}(\frac{1}{K}\sum_{k=1}^{K}\lambda_k)$? How would the change in the intensity of some $X_k$ affect the theoretical expected value and variance of $Y$?
All you need is the first part of 3. (which is a correct conclusion), and the fact that the mean and variance of Poisson-type RV are both equal to its only parameter. But you have to be more careful when dividing your sum by $K$; this divides the mean of the sum by $K$, but its variance by $K^2$. Also, the resulting Y has then non-integer values and cannot thus be Poisson.