Supposing I have a Poisson distributed random variable $X \sim \text{Poiss}(\lambda)$ with a parameter $\lambda$ that could take integer values only. Let $x$ be a single observation of a random variable. I could now get the estimate $\hat{\lambda}$ of the parameter by maximizing the likelihood function: $$ \hat{\lambda} = \arg\max_{\lambda \in \mathbb{N}}{ \log{p(x|\lambda)} }, \tag{1} $$ where $p(x|\lambda)$ is the Poisson PDF. To get the estimator variance $\text{Var}(\hat{\lambda})$ I first calculate the estimator distribution as $$ f(\hat{\lambda}|\lambda_0)=\sum_{x\in\mathbb{N}}{p(x|\lambda_0)\mathbf{1}_{\{\log{p(x|\hat{\lambda})} \geq \log{p(x|\lambda)}, \forall\lambda\in\mathbb{N}\}}}. \tag{2} $$ Here $\lambda_0\in\mathbb{N}$ is the true parameter value and $\mathbf{1}_{\{\cdot\}}$ is equal to 1 if condition in brackets holds and 0 otherwise. Since I computed (2) I can easily get $\text{Var}(\hat{\lambda})$. But things get worse when I try to estimate the parameter using a sample $x_1, x_2, \dots, x_n$ of the size $n$. In this case the estimator is $$ \hat{\lambda} = \arg\max_{\lambda \in \mathbb{N}}{ \sum_{j=1}^{n}{\log{p(x_j|\lambda)}} }. \tag{3} $$ Now the estimator PDF is $$ f(\hat{\lambda}|\lambda_0)=\sum_{x_1,\dots,x_n\in\mathbb{N}}{p(x_1|\lambda_0)\cdot\ldots\cdot p(x_n|\lambda_0)\mathbf{1}_{\{\sum_{j=1}^{n}{\log{p(x_j|\hat{\lambda})}} \geq \sum_{j=1}^{n}{\log{p(x_j|\lambda)}}, \forall\lambda\in\mathbb{N}\}}}. \tag{4} $$ There is no way to compute (4) for large $n$ (about $10^4$ in my case). So my questions are:

  1. Can I somehow simplify (4) to break the sum into a product?
  2. Are there some other ways to estimate the variance of the estimator (3)?

Things that I've tried

First of all I've tried to estimate the variance using Cramer-Rao bound by taking the finite difference derivative: $$ \text{Var}(\hat{\lambda}) = \left(n\mathbb{E}\left[(\log{p(x|\lambda_0+1)}-\log{p(x|\lambda_0))^2}\right]\right)^{-1}. \tag{5} $$ As expected, that didn't work quite well: the variance in simulation was lower than this value.

Then I came across a more general Hammersley–Chapman–Robbins bound. One can find that for my case the lower bound is (e.g. see [Dahiya R.C., Commun. Stat. - Theory Methods, 15(3), 709 (1986)]) $$ \text{Var}(\hat{\lambda}) = \left(e^{n/\lambda_0}-1\right)^{-1}. \tag{6} $$ However, this bound turned out to be too tight and was unreachable for any estimator I could find due to exponential decrease of the bound with $n$ (the same problem stated in Dahiya's article).


1 Answer 1


If you need the exact variance you're in trouble, I think. If not, progress can be made.

To start with, the monotone likelihood ratio property of exponential families means that $\hat\lambda$ is either the integer below the mean $\bar X$ or the integer above the mean. The variance of $\bar X$ is $\lambda/n$.

Consider some ranges of ($n$,$\lambda$):

  • If $\sqrt{\lambda/n}\gg 1$ then $\mathrm{var}[\hat\lambda]\approx \lambda/n$, since the reduction in variance caused by making $\hat\lambda$ an integer is small compared to the standard error of the mean.

  • For slightly smaller $\lambda/n$, $\mathrm{var}[\hat\lambda]$ is a bit smaller than $\lambda/n$. We can write $$\mathrm{var}[\hat\lambda] \approx \mathrm{var}[\bar X] - \mathrm{var}[\bar X-\hat\lambda]$$ and approximate the second term by the variance of a $U[-0.5, 0.5]$ $$\mathrm{var}[\hat\lambda] \approx \mathrm{var}[\bar X] -1/12$$ (this is the opposite of the Sheppard correction for rounding)

  • The difficult case: $\lambda/n$ not far from 1. I think this needs actual calculation, but only over the relatively small number of $\hat\lambda$ values with non-negligible probability.

  • At the far extreme, if $\sqrt{\lambda/n}\ll 1$, $\mathrm{var}[\hat\lambda]\approx 0$ (and even $n\mathrm{var}[\hat\lambda]\approx 0$), since $\hat\lambda=\lambda$ with very high probability. For example, if $\sqrt{\lambda/n}=1/10$, different integers are ten standard deviations apart, so it will be very unlikely for the closest integer to $\bar X$ not to be the true $\lambda$.

  • $\begingroup$ Can I interpret the last option as follows: for a continuous-valued parameter MLE asymptotically gives a normally distributed estimator, which I can then approximate to a discrete distribution by turning the PDF into a histogram with bin centers located at the discrete values? $\endgroup$
    – Krivoi
    Commented Jun 13, 2020 at 2:37
  • $\begingroup$ Yes, but the condition is stronger than just 'continuous-valued'. For some other type of distribution it might be possible for the discrete MLE to not just be a rounded continuous MLE, but for Poisson it will be. $\endgroup$ Commented Jun 13, 2020 at 2:43
  • $\begingroup$ Can I get some examples of a discrete parameter MLE? Is it like the one from this question? $\endgroup$
    – Krivoi
    Commented Jun 13, 2020 at 8:59
  • $\begingroup$ Yes, balls in urns are a classic example where the parameter is discrete. Another two are from ecology: estimating the population of a type of animal or the number of distinct species of animal. Often, though, you're in the first situation in the answer, where the discreteness can be ignored $\endgroup$ Commented Jun 14, 2020 at 1:48

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