How to calculate a confidence level for a Poisson distribution?

Question

Would like to know how confident I can be in my $\lambda$. Anyone know of a way to set upper and lower confidence levels for a Poisson distribution?

• Observations ($n$) = 88
• Sample mean ($\lambda$) = 47.18182

what would the 95% confidence look like for this?

Answer 1

For Poisson, the mean and the variance are both $\lambda$. If you want the confidence interval around lambda, you can calculate the standard error as $\sqrt{\lambda / n}$.

The 95-percent confidence interval is $\hat{\lambda} \pm 1.96\sqrt{\hat{\lambda} / n}$.

Answer 2

Patil & Kulkarni (2012, "Comparison of Confidence Intervals for the Poisson Mean: Some New Aspects", REVSTAT - Statistical Journal) discuss 19 different ways to calculate a confidence interval for the mean of a Poisson distribution.

Answer 3

In addition to the answers that others have provided, another approach to this problem is achieved through a model based approach. The central limit theorem approach is certainly valid, and the bootstrapped estimates offer a lot of protection from small sample and mode misspecification issues.

For sheer efficiency, you can get a better confidence interval for $\lambda$ by using a regression model based approach. No need to go through derivations, but a simple calculation in R goes like this:

x <- rpois(100, 14)
exp(confint(glm(x ~ 1, family=poisson)))

This is a non-symmetric interval estimate, mind you, since the natural parameter of the poisson glm is the log relative rate! This is an advantage since there is a tendency for count data to be skewed to the right.

The above approach has a formula and it is :

$$\exp\left( \log \hat{\lambda} \pm \sqrt{\frac{1}{n\hat{\lambda} }}\right)$$

This confidence interval is "efficient" in the sense that it comes from maximum likelihood estimation on the natural parameter (log) scale for Poisson data, and provides a tighter confidence interval than the one based on the count scale while maintaining the nominal 95% coverage.

Answer 4

Given an observation from a Poisson distribution,

• the number of events counted is n.
• the mean ($\lambda$) and variance ($\sigma^2$) are equal.

Step by step,

• The estimate for the mean is $\hat \lambda = n \approx \lambda$
• Assuming the number of events is big enough ($n \gt 20$), the standard error is the standard deviation $\sigma$, which we can also estimate,

$$stderr = \sigma = \sqrt{\lambda} \approx \sqrt{n} $$

Now, the 95% confidence interval is,

$$ I = \hat \lambda \pm 1.96 \space stderr = n \pm 1.96 \space \sqrt{n}$$

[Edited] Some calculations based on the question data,

• Assuming the $\lambda$ indicated in the question has been externally checked or was given to us, i.e., it is a good piece of information not an estimation.

  I am making this assumption as the original question does not provide any context about the experiment or how the data was obtained (which is of the utmost importance when manipulating statistical data).

• The 95% confidence interval is, for the particular case,

$$ I = \lambda \pm 1.96 \space stderr = \lambda \pm 1.96 \space \sqrt{\lambda} = 47.18182 \pm 1.96 \space \sqrt{47.18182} \approx [33.72, 60.64] $$

Hence, as the measurement (n=88 events) is outside the 95% confidence interval, we conclude that,

1. The process does not follow a Poisson process, or,

2. The $\lambda$ we have been given is not correct.

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Important note: the first accepted answer above is wrong, as it incorrectly states that the standard error for a Poisson observation is $\sqrt{\lambda/n}$. That is the standard error for a Sample Mean (Survey Sample) process.

Answer 5

The Poisson distribution Wikipedia page gives an answer. Suppose $$ K \sim \text{Pois}(\lambda) $$ with $\lambda$ unknown and we sample $K$ $n$ times getting $k_1, k_2, \ldots, k_n$. We let $$ \begin{align*} k =& \sum_{i=1}^n k_i\\ \end{align*} $$

Suppose we want a confidence interval with confidence level $0\le c \le 1$. Let $\alpha = 1-c$. Then $$ CI_{\alpha} = \left\{\lambda: \frac{1}{n}\text{PPF}_{\chi^2_{2k}}(\alpha/2) \le \lambda \le \frac{1}{n}\text{PPF}_{\chi^2_{2k+2}}(1-\alpha/2)\right\} $$ Where $\chi^2_D$ is a $\chi^2$-distribution with $D$ degrees of freedom and $\text{PPF}_{\chi^2_D}$ is the probability point function (PPF) or quantile function for that distribution.

I've done some simulations and the coverage seems to be good as long as $k=n\mu > 1$. I haven't compared to any other methods.