UPDATE I edited my original question to make it as clear as possible.

My goal is to find a reliable goodness-of-fit test for Poisson-distributed samples. There are a few discussions here related to goodness-of-fit tests for discrete distributions, e.g., the Poisson distribution (for example, here and here). I have created a simulation to understand what happens to the type I error in the case of the chi-squared test. I am working with a sum of Poisson-distributed variables (which is in turn a Poisson-distributed variable itself):


n <- 100000
alpha <- 0.05 # significance level
n_sim <- 10
res_chi2 <- vector(mode = "list", length = n_sim)
res_ks <- vector(mode = "list", length = n_sim)

lambda_i <- 10^sample(-10:-2, 100, replace = TRUE) # 100 Poisson-distributed variables
total_lambda <- sum(lambda_i) # the random variable of interest is a sum of Poisson-distributed variables

for (i in 1:n_sim){
  # observed frequencies
  my_sample <- rowSums(sapply(lambda_i, function(x) rpois(n, x))) # generate a sample by aggregating event counts of subsamples
  sample_freq <- table(my_sample)
  # expected frequencies
  # calculated using the density function for the aggregate Poisson distribution
  theor_freq <- dpois(as.numeric(names(sample_freq)), total_lambda)*n
  # add missing count for (n,+ inf) to the last bin
  # now frequencies are normalized to n (sample size)
  theor_freq[length(theor_freq)] <- theor_freq[length(theor_freq)] + n - sum(theor_freq)
  # test statistic, the first formula below
  #  https://www.itl.nist.gov/div898/handbook/eda/section3/eda35f.htm
  test_statistic <- sum((theor_freq - sample_freq)^2/theor_freq)
  # no estimated parameters, df = number of categories - 1
  p_value <- 1 - pchisq(test_statistic, df = length(theor_freq)-1)
  # if TRUE, the null is accepted
  res_chi2[[i]] <- p_value >= alpha

sum_passed_chi2 <- Reduce(`+`,res_chi2)

# 1000 simulations
> 1000 - sum_passed_chi2
> 92
# the null was rejected 92 times

The type I error is equal to 9% for the chi-squared test. Why is it overestimated? Can I assume that a well-suited goodness-of-fit test will give an error of approximately 5% (my significance level)? How do I implement/design a proper goodness-of-fit to test whether a sample is distributed according to a Poisson distribution with known parameters?

UPDATE 2 I also ran a simulation with a single sample drawn from a Poisson distribution, i.e.:

my_sample <- rpois(n, total_lambda)

In this case, the type I error rate is 8%.

  • 1
    $\begingroup$ I am having a hard time seeing a valid, well-tested implementation of a chi-squared test here. Consider the possibility of bugs in your program. $\endgroup$
    – whuber
    Aug 3, 2020 at 19:55
  • $\begingroup$ @whuber thanks for looking at my question. I am polishing it and will sumbit an edited version of the question with well-documented code shortly $\endgroup$ Aug 3, 2020 at 20:20
  • $\begingroup$ Another possibility is that you appear to be applying the chi-squared test to data where it's known to be invalid. In particular, if I'm reading your code correctly, you are generating datasets with incredibly low Poisson parameters, which in many cases (despite the large n) will generate a lot of zeros, a few ones, and (very rarely) anything larger. $\endgroup$
    – whuber
    Aug 3, 2020 at 20:30
  • $\begingroup$ @whuber What are circumstances in which the chi-squared test is known to be invalid? My simulation is based on a practical problem where one deals with a bunch of random variables, some of which may have low Poisson parameters (the rate lambda is such that an event may or may not happen during the simulation, e.g., 0.001 and simulation size is for example 100). The sum of such variables is Poisson-distributed, and one can calculate its total rate. I am trying to find a robust test, and most well-known tests seem to be not suited here (e.g. the KS test has a continuity requirement) $\endgroup$ Aug 3, 2020 at 23:26
  • 1
    $\begingroup$ Yes. Indeed, the standard function chisq.test in R offers that as an option. I tend to circumvent the problem by using the specified (null) distribution to determine the bins. For instance, I will group all extreme counts into a bin with an expected count of around 5. This is perfectly legitimate because it isn't based on the data, it's based only on the hypothesis. $\endgroup$
    – whuber
    Aug 5, 2020 at 18:24


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