Timeline for What conditions on a probability distribution defined by long-time averaging do I need to satisfy a central limit theorem?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2011 at 14:14 | comment | added | Jess Riedel | Ha, very embarrassing on my part, David. Good thing this is saved on the internet for perpetuity. Thanks so much for the help. | |
| Sep 13, 2011 at 1:58 | comment | added | David Moews | Setting $Y_M:=Y$, there is no way to pick scaling constants $a_M$ such that $a_M Y_M$ converges to something nontrivial. $|Y_M|^{1/\sqrt{M}}$ will converge if rescaled appropriately. | |
| Sep 13, 2011 at 1:07 | comment | added | David Moews | To compute the variance of $Y$, observe that $Y^2$ is the product of $M$ independent random variables, each distributed as $\cos^2 W$, where $W$ is uniform. Since $\cos^2$ has average value $\frac{1}{2}$, and the r.v.s are independent, ${\bf E}[Y^2]$ is the product of $M$ copies of $\frac{1}{2}$, which is $2^{-M}$. | |
| Sep 12, 2011 at 17:19 | vote | accept | Jess Riedel | ||
| Sep 12, 2011 at 17:19 | comment | added | Jess Riedel | Thanks very much for this compact and clear explanation. You are right about $Y$ not approaching a normal distribution. I discovered an an error in my code this morning, and the true distribution it approaches has much heavier tails than a Gaussian. Could you please elaborate on how you know that Var[$Y$] = $2^{-M}$? Is it correct to express Var[$Y$] in terms of the moments of the r.v. ln($Y$) (which is normally distributed), yielding a taylor series? Are there convergence issues I should worry about as a physicist? | |
| Sep 9, 2011 at 22:50 | history | edited | David Moews | CC BY-SA 3.0 |
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| Sep 9, 2011 at 22:36 | history | answered | David Moews | CC BY-SA 3.0 |