A Binomial distribution has a standard normal limiting distribution, i.e. (X-E[X])/se(X) -> N(0,1), where X is the sum of independent and identically distributed Bernoulli variables.(adsbygoogle = window.adsbygoogle || []).push({});

Does this hold even when

i) the Bernoulli variables are independent but non-identically distributed? That is, say that each Bernoulli variable have different survivor intensity and define X as the sum of these non-identical variables. I believe this distribution is called the Poisson-Binomial distribution. Do we have: (X-E[X])/se(X) -> N(0,1)?

ii) the Bernoulli variables are dependent and non-identically distributed? That is, say that each Bernoulli variable have different survivor intensity and that they are correlated. Define X as the sum of these non-identical variables. Do we have: (X-E[X])/se(X) -> N(0,1)?

The case ii) is what I'm mainly interested in. I'm pretty sure case i) holds, but isn't 100% case ii) holds. If it holds I would appreciate a reference to any paper or so since I need the conditions under which it holds.

Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Limiting dist for sum of dependent and non-identical Bernoulli vars

**Physics Forums | Science Articles, Homework Help, Discussion**