Timeline for CLT for the squares of unbounded non-identically independently distributed random variables
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2011 at 16:04 | vote | accept | Bullmoose | ||
| Nov 10, 2011 at 14:43 | answer | added | Mark Meckes | timeline score: 4 | |
| Nov 10, 2011 at 1:26 | comment | added | Bullmoose | Beautiful! Thanks! I somehow missed the Berry-Essen theorem for non-identically distributed summands. If you write your comments in an answer, I'll accept it. | |
| Nov 9, 2011 at 20:16 | comment | added | Mark Meckes | Yes, then you would have a bound on the 3rd absolute moment of $X_i^2$, and you could apply the Berry-Esseen theorem. | |
| Nov 9, 2011 at 19:18 | comment | added | Bullmoose | I apologize for the notation. $X_i$ is a random variable, and $A_i$ is its distribution. Hmm, so are you saying that if I bound the 6th central moment of $X_i$ (which would be equivalent to bounding the 6th absolute moment), CLT would apply? | |
| Nov 9, 2011 at 16:54 | comment | added | Mark Meckes | I'm somewhat confused by your notation. What is the difference between $A_i$ and $X_i$? Also, I'm not sure you actually do know the Lyapunov condition, which only requires a $2+\epsilon$ order absolute moment (and which follows from the Lindeberg-Feller condition using Markov/Chebychev). In particular, if for some $\epsilon > 0$, $\sup_i \mathbb{E} |X_i|^{4+\epsilon} < \infty$, then the CLT applies. | |
| Nov 9, 2011 at 4:56 | history | edited | Bullmoose | CC BY-SA 3.0 |
fixed a mistake in the normalized sum of squares formula
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| Nov 9, 2011 at 4:13 | history | asked | Bullmoose | CC BY-SA 3.0 |