Solved – MLE for Poisson distribution is undefined with all-zero observations

maximum likelihoodpoisson distributionunbiased-estimator

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.

Best Answer

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 $\lambda = 0$.

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