Timeline for estimate the error term in CLT
Current License: CC BY-SA 2.5
5 events
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Sep 20, 2010 at 17:23 | comment | added | Mark Meckes | Okay, so you only expect $1/m$ for a specific $f$ you're interested in. You should edit the original question to make that clear, since it sounds like you're saying you expect $1/m$ for an arbitrary smooth bounded $f$. (I don't have an answer offhand but I'll think about it when I have some time.) | |
Sep 20, 2010 at 16:36 | comment | added | gondolier | I agree that there exists $f$ such that $1/sqrt{m}$ is tight. In fact I am not looking for uniform estimates but rather for a specific function $f(x) = x^2 e^(-x^2/4)$. In my OP, I said the scaling constant will depend on the function $f$. Based also on numerical result, I believe its rate is $1/m$. Do you think there is any method to produce a sharp expansion of the form $c/m + o(1/m)$? Any lower bound idea? | |
Sep 20, 2010 at 16:09 | comment | added | Mark Meckes | Just to clarify: You are correct about the $1/m$ rate for certain very nice functions, but I think that for the class of all bounded smooth functions the correct rate is $1/\sqrt{m}$, and that the loss is somehow due to the lack of uniformity in that $1/m$ rate for very nice functions. | |
Sep 20, 2010 at 16:05 | comment | added | Mark Meckes | I'm not convinced by your heuristic for the characteristic function because you only get the $1/m$ rate for convergence of the log once $m \gg t^2$. So your argument shows that $\mathbb{E} e^{itX_m} \to \mathbb{E} e^{itX}$ at rate $1/m$, but with an implicit constant that may depend on $t$. Likewise, if $f$ is a polynomial, then $\mathbb{E} f(X_m) \to \mathbb{E} f(X)$ at rate $1/m$ with an implicit constant that may depend on the degree and coefficients of $f$. If, as you suggest above, you do a truncation and Taylor expansion you may lose some of that rate. | |
Sep 20, 2010 at 15:42 | history | answered | gondolier | CC BY-SA 2.5 |