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.

  • 1
    $\begingroup$ what do you mean mle is undefined? $\endgroup$
    – Deep North
    Sep 26, 2017 at 3:18
  • $\begingroup$ I mean when all observed values are zero, you simply can't estimate $\lambda$ as the average, which is zero, since we always have $\lambda>0$ $\endgroup$
    – Tri Nguyen
    Sep 26, 2017 at 3:21

2 Answers 2


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 $\lambda = 0$.

  • 1
    $\begingroup$ Does it violate the assumption that $\lambda$ is always greater than zeros? $\endgroup$
    – Tri Nguyen
    Sep 26, 2017 at 3:28
  • 3
    $\begingroup$ I suppose that depends on who you are communicating with, and their level of fussiness about such things. It seems to me that a poisson distributed variable of rate $\lambda = 0$ is just a random variable that always gives zero. $\endgroup$ Sep 26, 2017 at 3:30
  • $\begingroup$ Thank you for your answer, if it is supposed that $\lambda>=0$, then we simply have no mystery anymore :). $\endgroup$
    – Tri Nguyen
    Sep 26, 2017 at 3:36
  • $\begingroup$ Why zero and not $-\infty$? $\endgroup$
    – Frank
    Aug 11, 2019 at 17:51

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.


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