Finding the variance of the estimator for the maximum likelihood for the Poisson distribution

Question

If $K_1, \dots, K_n$ are i.i.d. Poisson distributions with parameter $\beta$ I have worked out that the maximum likelihood estimate is $$\hat\beta (k_1, \dots, k_n) = \frac{1}{n} \sum_{i=1}^n k_i$$ for data $k_1, \dots, k_n$. Therefore we can define the corresponding estimator $$T = \frac{1}{n} \sum_{i=1}^n K_i .$$ My question is how would you work out the variance of this estimator?

In particular, as each $K_i$ follows a Poisson distribution with parameter $\beta$ I know, from the properties of the Poisson, that the distribution $\sum_{i=1}^n K_i$ will follow a Poisson distribution with parameter $n \beta$, but what is the distribution of $T$?

Answer 1

$T$ is distributed... as a Poisson variable scaled by $n$. Hence the variance of $T$ is $1/n^2 \times n\beta$.

Answer 2

Remember that $$ \mathbb{Var}\left(\sum_{i=1}^n a_i X_i\right) = \sum_{i=1}^n a_i^2\,\mathbb{Var}(X_i) + 2 \sum_{1\leq i<j\leq n} a_i\,a_j\,\mathbb{Cov}(X_i X_j) \, , $$ always. But, if the $X_i$'s are independent, what is the value of $\mathbb{Cov}(X_i X_j)$? That's all you need to answer the question.