I can't seem to figure out why these expectations turn out the way they do, I am currently studying about the Fisher Information. If $X_1,X_2,...,X_n $ are all iid Poission($\lambda$) , then going through the process I reach this stage;
$$\frac{1}{\lambda^2}\mathbb{E}[(\sum\limits_{i=1}^n X_i-n\lambda)^2]$$ Ultimately I am told as the distribution of $\sum\limits_{i=1}^n X_i$ is Po($\lambda$) that this is equal to $$=\frac{n}{\lambda}$$
My question is when you get an expectation looking like the above how do you go about finding it, regardless of the distribution?
For this specific problem, $\text{E}(X_i) = \lambda$ and $\text{Var}(X_i)=\lambda$ for $i=1,\ldots,n$. This is can be calculated from the Poisson distribution. Furthermore, the sum of $n$ independent Poisson$(\lambda)$ random variables is $$ Z = \sum_{i=1}^n X_i \sim \text{Poisson}(n\lambda). $$ Hence, $\text{E}\left(Z\right) = \text{Var}\left(Z\right) = n\lambda$. Therefore, $$ \frac{1}{\lambda^2}\text{E}\left(\sum_{i=1}^nX_i - n\lambda\right)^2 = \frac{1}{\lambda^2}\text{E}\left(Z - \text{E}Z\right)^2 = \frac{1}{\lambda^2}\text{var}\left(Z\right) = \frac{n\lambda}{\lambda^2} = \frac{n}{\lambda}, $$ so the expectation above is the variance of $Z$. So, if you see an expression like $$ \text{E}\left( \sum_{i=1}^n X_i - n\text{E}X \right)^2, $$ what you have is the variance of the sum.