= Solution
A <maximum-likelihood estimator> maximizes the <likelihood> as a function of the parameter for the observed data. Under the usual regularity conditions, a scalar <maximum-likelihood estimator> satisfies $\sqrt n(\hat\theta-\theta)\Rightarrow N(0,I_1(\theta)^{-1})$, where $I_1$ is the <Fisher information> in one observation.
For a <Poisson distribution> with $\theta>0$, the <log-likelihood> is $\ell(\theta)=\sum_jY_j\log\theta-n\theta-\sum_j\log(Y_j!)$. Solving $\ell'(\theta)=0$ gives \b[$\boxed{\hat\theta=\bar Y}$.] Its <expectation> is $\theta$ and its <variance> is $\theta/n$. If all observations are zero, the maximum over $\theta\geq0$ occurs at zero; over $\theta>0$ this is a boundary supremum. The one-observation <score function> is $Y/\theta-1$, so $I_1(\theta)=\operatorname{Var}(Y)/\theta^2=1/\theta$ and the asymptotic variance of $\sqrt n(\bar Y-\theta)$ is $\theta$.
For $n\geq2$, let $S=(n-1)^{-1}\sum_j(Y_j-\bar Y)^2$. The identity
$$
\sum_j(Y_j-\bar Y)^2=\sum_j(Y_j-\theta)^2-n(\bar Y-\theta)^2
$$
gives $\mathbb E S=(n\theta-\theta)/(n-1)=\theta$, so $S$ is another <unbiased estimator>. However, the total <score function> has <variance> $n/\theta$. For any regular <unbiased estimator> $T$, differentiating $\mathbb E_\theta T=\theta$ gives $\operatorname{Cov}(T,\ell')=1$, so the <Cauchy-Schwarz inequality> yields $\operatorname{Var}(T)\geq\theta/n$. The <sample mean> attains this <Cramér-Rao bound> exactly. \b[There is no variance advantage in preferring $S$ to $\bar Y$.]
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