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 - The Fusion of Science and Community**

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

Loading...

Similar Threads - Limiting dist dependent | Date |
---|---|

A Averaging over the upper sum limit of a discrete function | Jan 7, 2018 |

I Central limit theorem, panel study | Mar 25, 2017 |

A Scientific method to calculate the time limits of a task | Feb 26, 2017 |

I Validity of replacing X by E[X] in a formula | Dec 16, 2016 |

**Physics Forums - The Fusion of Science and Community**