Timeline for How can I efficiently model the sum of Bernoulli random variables?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2013 at 11:13 | history | edited | mpiktas | CC BY-SA 3.0 |
fix a bug in math
|
| Oct 2, 2013 at 14:27 | history | edited | Scortchi♦ | CC BY-SA 3.0 |
fixed typos
|
| Dec 14, 2010 at 19:14 | comment | added | mpiktas | You're welcome :) I lifted the approximation from Petrov's book: amazon.co.uk/Limit-Theorems-Probability-Theory-Independent/dp/… | |
| Dec 14, 2010 at 16:54 | comment | added | whuber♦ | @mpiktas +1 Thank you for quantifying the approximation error. It's good to see this kind of analysis. | |
| Dec 13, 2010 at 8:02 | history | edited | mpiktas | CC BY-SA 2.5 |
added the error of approximation to account for the comments
|
| Dec 12, 2010 at 22:04 | comment | added | whuber♦ | @mpiktas The convergence to normality can be slow. For example, with p_i = 1/(i+1) (a sequence for which B_n diverges) and n = 2^16 = about 33000, the skewness is still 0.26 and a normal approximation assigns 0.1% probability to negative totals, an impossibility. Clearly we're starting to approach normality but we're not there yet. | |
| Dec 10, 2010 at 21:05 | comment | added | user1108 | @whuber, well said. | |
| Dec 10, 2010 at 18:53 | comment | added | whuber♦ | @G. Jay Kerns I agree that the analogy to the Poisson is imperfect, but I think it gives good guidance. Imagine a sequence of p's, p_i = 10^{-j}, where j is the order of magnitude of i (equal to 1 for i <= 10, to 2 for i <= 100, etc.). When n = 10^k, 90% of the p's equal 10^{-k} and their sum looks Poisson with expectation 0.9. Another 9% equal 10^{1-k} and their sum looks Poisson (with the same expectation). Thus the distribution looks approximately like a sum of k Poisson variates. It's obviously nowhere near Normal. Whence the need for the "monstrous condition." | |
| Dec 10, 2010 at 18:22 | comment | added | user1108 | By the way, I didn't actually check that monstrous condition in the second paragraph. | |
| Dec 10, 2010 at 18:12 | comment | added | user1108 | @mpiktas is right. The analogy to the Poisson distribution doesn't quite fit, here. | |
| Dec 10, 2010 at 17:58 | comment | added | mpiktas | That is why it must be $B_n\to\infty$. If $p_i$ approach zero at rate faster than $1/i$, $\lim B_n<\infty$. | |
| Dec 10, 2010 at 16:14 | comment | added | whuber♦ | This is not true when the p_i approach zero as i increases. Otherwise, you have just proven that the Poisson distribution is Normal! | |
| Dec 10, 2010 at 12:06 | history | answered | mpiktas | CC BY-SA 2.5 |