# Poisson Binomial Distribution - confidence intervals

I'm working on a project which involves multiple trials for which the probability of success is not the same across trials. Given the unequal probabilities per trial, I'm using the Poisson Binomial distribution.

I can easily model the probability of a given number of successes across all trials using the R package poisbinom. Looking at the problem the other way around, I would also like to provide confidence intervals on the probability of successes given an observed number of successes.

For the Binomial Distribution, I know this can be done using a normal approximation when $$p>0.1$$ and $$n>15$$, but I cannot find a reference for computing CIs for the Poisson Binomial Distribution. This is obviously complicated by the fact that each trial has a different probability of success

Any references to using a normal approximation or other methods is appreciated.

EDIT:

I've been doing some simulations to see if a normal approximation (Wald Method) would work, and generally I see the same results I would expect to see as with a Binomial Distribution. The empirical variability I see from the simulation is certainly non-normal, but normal approximation does a reasonably good job with 2 decimal points - 0.## +/- 0.## when the sample size is large.

From some background reading, the same deviation from normal is seen for the Binomial Distribution and various approximation methods are proposed. The Wald Method is simple but biased.

From what I've seen from simulations, I think similar CI methods as those used for a Binomial Distribution would work.

UPDATE:

I've tried multiple types of CI calculation that are applicable to the Binomial Distribution, or more specifically, the Binomial Proportion - Wald, Agresti-Coull, Wilson. see here for descriptions From my simulations, Wald performs best across the broadest range of values. The others are generally more often recommended, but I think that is primarily for proportions that are very small (< 0.01) or very large (> 0.99).

I feel I've answered my own question here, but I appreciate any feedback.