According to Wikipedia,

the Poisson binomial distribution is the discrete probability distribution of a sum of independent Bernoulli trials that are not necessarily identically distributed

In other words, the Bernoulli trials have different probabilities $p_{1},\dots,p_{n}$.

Suppose I have a task where the participants are presented stimuli that fall into either category A or category B. The experiment goes on for $N$ trials, and for each trial the participants report which category is the correct one.

The trials are independent but there is probably a different probability for each trial that they give a correct response. If I have a large number of trials (e.g. 500), is it possible to find meaningful maximum likelihood estimates of the distribution parameters (the $p_{i}$'s)?

How many participants would I need?

I would be interested to test whether the ordinary Binomial or the Poisson binomial is a better fit to the data I have at hand. However, I was wondering whether it is theoretically possible to find a plausible answer to this question.

  • 4
    $\begingroup$ It does not seem very meaningful from a statistical perspective to use a different $p_i$ for each trial if in the end you only observe the sum. Those parameters are not identifiable and I doubt you can test for this "Poisson binomial" distribution, versus a Binomial one. $\endgroup$
    – Xi'an
    Sep 30, 2016 at 13:00
  • 1
    $\begingroup$ As @Xi'an points out, the only time you can tell them apart is if each observation is the result of multiple Bernoulli trials. Then you may be able to see that the variance is too large for the mean. $\endgroup$ Sep 30, 2016 at 13:22
  • $\begingroup$ I am not sure if I understand what you mean by observing only the sum. For each trial, I know the presented stimulus (A vs. B) and the participant's responses (A vs. B). I have $20$ participants. $\endgroup$
    – anonymous
    Sep 30, 2016 at 13:24
  • 3
    $\begingroup$ In this case, you can estimate the $p_i$'s separately and your statistical problem is to test whether or not the $p_i$'s are equal, isn't it? $\endgroup$
    – Xi'an
    Sep 30, 2016 at 13:43
  • 2
    $\begingroup$ Bounty for interesting question. Small difference in assumptions leads to very different models (1 vs 500) parameters $\endgroup$
    – mmh
    Oct 2, 2016 at 13:13

1 Answer 1


As I see your problem, you have $K$ individuals completing $N$ trials, that result in binary outcomes (success or failure). So you are dealing with $N\times K$ random variables $X_{ij}$. You are interested in computing probabilities of success for each trial $p_i$.

So the first thing to notice is that you assume in here that participants are exchangeable, so there is no more or less skilled participants - is this assumption correct for your data? The first thing that comes to my mind is that for the kind of data as yours Item Response Theory models would be better suited. Using such models you could estimate model assuming that the tasks vary in their difficulty and that the participants vary in their skills (e.g. using simple Rasch model).

But let's stick to what you said and assume that participants are exchangeable and you are interested only in probabilities per task. As others already noticed, you are not dealing here with Poisson binomial distribution, since we use such distribution for sums of successes from $N$ independent Bernoulli trials with different probabilities $p_1,\dots,p_N$. For this you would have to introduce new random variable defined as $Y_j = \sum_{i=1}^N X_{ij}$, i.e. total number of successes per participant. As noted by Xi'an, parameters $p_i$ are not identifiable in here and if you have data on result of each trial by each participant, it is better to think about it as of Bernoulli variables parametrized by $p_i$.

From what you are saying, you would like to test if Poisson binomial distribution fits the data better then ordinary binomial. As I read it, you want to test if the trials differ in probabilities of success, versus if probability of success is the same for each trial (since $p_1 = p_2 = \dots = p_N$ is an ordinary binomial distribution). Saying it other way around, your null hypothesis would be that not only participants, but also trials are exchangeable, so identifying particular trials tells us nothing about the data since they all have the same probability of success. If we have null hypothesis stated like this, it instantly leads to permutation test, where you would randomly "shuffle" your $N\times K$ matrix and compare the statistic computed on such permuted data to statistic computed on unshuffled data. For the statistic to compare I would use combined variance's

$$ \sum_{i=1}^N \hat p_i(1-\hat p_i) $$

where $\hat p_i$ is probability of success estimated from the data for the $i$-th participant (columnwise means). In case of equal $\hat p_i$'s would reduce to $N \hat p(1-\hat p)$.

To illustrate it I conducted a simulation with three different scenarios: (a) all $p_i = 0.5$, (b) they come from Beta(1000,1000) distribution, (c) they come from uniform distribution. In the first case $p_i$'s are all equal; in the second case they are "random", but grouped around common mean; and in the third case they are totally "random". The plot below shows distributions of test statistic under null hypothesis (i.e. computed on shuffled data), red lines show the variance computed on unshuffled data. As you can see, the combined variances of unshuffled data crosses with the null distribution in the first case (test is not significant) and slightly approach the distribution in the second case (significant difference from the null). In the third case the red line is even not visible on the plot since it is that far from the null distribution (significant difference).

Results of simulation

So while the test correctly identified "all the same $p_i$'s" scenario (a), but didn't find the "similar but not the same" scenario (b) to fulfill the criteria of equality. The question is if you want to be that strict about it? Nonetheless, this is a direct implementation of test for your hypothesis. It compares the basic criteria that would enable you to distinguish ordinary binomial from Poisson binomial (their variances).

There is of course lot's of other possibilities, more or less appropriate depending on your problem, e.g.: comparing the individual confidence intervals, pairwise $z$-tests, ANOVA, using some kind of logistic regression model etc. However, as I said before, this sounds rather like a problem for Item Response Theory models and assuming equal skills of participants sounds risky.


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