MLE for Poisson distribution is undefined with all-zero observations

Question

Suppose $x_1, x_2, ..., x_n$ are observed values of IID random variables $X_1, X2, ..., X_n$ from Poisson distribution, i.e., $P(x)=e^{-\lambda} \frac{\lambda^x}{x!}$.

The MLE for $\lambda$ is only defined (as the average of observed values) when there is at least one value of $x_i$ greater than zero; otherwise the log likelihood function is $l(\lambda)=-\lambda n$ and the MLE is undefined ($?$).

So, I am confused when computing the bias of this estimator, since the case of all-zero valued observations actually has non-zero probability, how to account this when finding the average of $\hat{\lambda}$.

Any answer is appreciated.

Answer 1

The likelihood function of the Poisson given observations $x_1, x_2, \ldots, x_n$ is

$$ l(\lambda; x) = \prod_i e^{-\lambda}\frac{\lambda^{x_i}}{x_i!} = \frac{e^{-n\lambda}}{x_1!x_2!\cdots x_n!}\lambda^{x_1 + x_2 + \cdots + x_n}$$

If $x_1 = x_2 = \cdots = x_n = 0$ then this becomes

$$ l(\lambda; x) = e^{- n \lambda} $$

Which is maximized when $\lambda = 0$.

So the MLE does exist in this case, it is $\lambda = 0$.

Answer 2

I still keep the opinion that MLE is undefined when all the observations are zero. However, this does not affect on the expectation or variance of this MLE, since the average of all-zero observations is zero, which does not have any effect on these computation, whether MLE is defined or undefined at this point.