# Estimate mean of Poisson from binary data

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 $$log(P_{x = 0}) = log(e^{-\lambda})$$ $$log(P_{x = 0}) = -\lambda$$ $$\lambda = -log(P_{x = 0})$$

• Do you know these data come from a Poisson distribution? Any overdispersion or zero-inflation would invalidate this method. – Frans Rodenburg Nov 17 '19 at 7:47
• I am just trying to understand for a Poisson first before moving on to other distributions. – Jdub Nov 17 '19 at 16:38
• 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. – Frans Rodenburg Nov 18 '19 at 2:54

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 $$1/n_0-1/n$$.