If you assume that counts in sample units would be distributed according to a Poisson distribution, but the data that you have are observations of only presence (count would be 1 or more) or absence (count would be 0) in sample units, is there a way to estimate the mean, $\lambda$, of the Poisson distribution from the proportion of sample units with a count of zero?

I know that for a Poisson distribution the probability of a zero is $$P_{x = 0} = e^{-\lambda}$$ and the probability of a count of 1 or more is $$P_{x \geq 1} = 1 - P_{x = 0} = 1 - e^{-\lambda}$$

Also, I know that the maximum likelihood estimator is $$\hat{\lambda} = \bar{X}$$

So for the binary data, can you use the negative log of the proportion of zeros as an estimate of $\lambda$? Since $$log(P_{x = 0}) = log(e^{-\lambda})$$ $$log(P_{x = 0}) = -\lambda$$ $$\lambda = -log(P_{x = 0})$$

  • $\begingroup$ Do you know these data come from a Poisson distribution? Any overdispersion or zero-inflation would invalidate this method. $\endgroup$ Commented Nov 17, 2019 at 7:47
  • $\begingroup$ I am just trying to understand for a Poisson first before moving on to other distributions. $\endgroup$
    – Jdub
    Commented Nov 17, 2019 at 16:38
  • $\begingroup$ You're going to have to a hard time using this for any distribution with more than 1 parameter. The reason this might work for data which you do know are Poisson distributed is because mean $=$ variance. $\endgroup$ Commented Nov 18, 2019 at 2:54

1 Answer 1


If the underlying distribution has a Poisson distribution with parameter $\lambda$ and from a sample size of $n$ there are $n_0$ zeros, then the maximum likelihood estimator of $\lambda$ is $-\log(n_0/n)$.

So this is what you've posted if $P_{x=0}$ is the same thing as $n_0/n$.

An estimate of the variance is $1/n_0-1/n$.


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