Probability distribution of Bernoulli trial of independent events with arithmetic progression probability [closed]

Question

If $n$ tests are independent of each other, and the probability of an event in each test form an arithmetic progression, that is, the probability of the event in the first test is $p$, and in the following tests it is $p + d, p + 2d, p + 3d,\ldots ,p + (n - 1)d$. The question is how to calculate the probability $P(X=k)$ where $X$ is the total number of events which occurs in all of the tests combined.

We use the approximation below to deal with this question,

$$p(X = k) = \left( \begin{array}{l} n \\ k \\ \end{array} \right)q^k (1 - q)^{n - k} ,q = p + \frac{{d(n - 1)}}{2} \, . $$

This approximation in fact uses the mean probability. The result as below is somewhat good, However, we don't know how to estimate the deviation/error from the exact probability. I think that the result will be much better if $d/p$ is smaller. The parameters used for the result figure are $n=24,p=0.2,d=0.1/23$. In this situation the result is very good.

In total, the question is whether there is a simple formula to compute $p(X=k)$ and how to estimate the deviation/error from the exact probability if only approximation solution available.

Answer 1

The sum of independent bernoulli variables with different probabilities is a Poisson-Binomial distribution. Searching on that name gives many articles for approximating it in terms of other distributions. I read (but didn't check) that your particular case is discussed by Thomas, M. A., and Taub, A. E. (1982). Calculating binomial probabilities when the trial probabilities are unequal, Journal of Statistical Computation and Simulation, 14, 125-131.

Answer 2

What is your question? In the situation you describe, both distributions are approximately Poisson with the same parameter, so it is no wonder they also approximate each other.