= Solution
Write $S=\sum_iX_i$. The <likelihood function> for the <Poisson distribution> is proportional to $e^{-n\theta}\theta^S$. For $S>0$, differentiation of the <log-likelihood> gives $-n+S/\theta=0$, and its second derivative is negative. For $S=0$, the likelihood decreases with $\theta$ and has its maximum at the boundary zero. Hence
$$
\boxed{\widehat\theta=\overline X=\frac Sn.}
$$
The estimator is \b[unbiased], since $\mathbb E\widehat\theta=n^{-1}\sum_i\mathbb EX_i=\theta$. Its variance is $\theta/n$. If the parameter space excludes zero, an all-zero sample gives a supremum as $\theta\downarrow0$ rather than an attained maximum.
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