Normal approximation to multi Bernoulli distribution

Question

I know I can approximate a Binomial distribution of the size n with $binomial(n,p) \approx \mathcal N(np, np(1-p))$ and a binomial distribution can be seen as n separate Bernoulli distribution $\sum_{i=1}^n bernoulli(p) = binomial(n,p)$

My question is, suppose that we have n distinct Bernoulli distributions with different probabilities $\sum_{i=1}^n bernoulli(p_i)$. Is there a resealable normal approximation for this? I'm thinking about averaging $\bar{p} = mean(p_i)$ and use $\sum_{i=1}^n bernoulli(p_i) \approx \mathcal N(n\bar{p}, n\bar{p}(1-\bar{p}))$ but I don't know how accurate it that and I don't know how to approach it to show how close is this.

Answer 1

This is called Poisson binomial distribution. And the moments for normal approximation are as follows:

1. Mean: $\sum\limits_{i=1}^n p_i$
2. Variance: $\sigma^2 =\sum\limits_{i=1}^n (1 - p_i)p_i$
3. Skewness: $\frac{1}{\sigma^3}\sum\limits_{i=1}^n ( 1-2p_i ) ( 1-p_i ) p_i$

BUT:

If you want to approximate it, this paper suggests this: $$ G(x) = Φ(x) + s(1 – x^2) φ(x) / 6 \\ CDF(k; p) \approx G\bigg(\frac{k + 0.5 - μ}{σ}\bigg), k = 0, 1, ..., N \\ $$

Where $s =$ Skewness, $Φ$ and $φ$ are the CDF and PDF, respectively, of the standard normal distribution