estimate the error term in CLT - MathOverflow

estimate the error term in CLT

2010-09-19T10:02:28Z

Let $X_m = \frac{1}{\sqrt{m}}\sum_{k=1}^m Z_k$ where $Z_k$ are iid equally likely on $\{\pm 1\}$. Then $X_m$ convergens to $X \sim \mathcal{N}(0,1)$ in distribution by CLT.

Let $f$ be a smooth bounded function on $\mathbb{R}$. Then $\mathbb{E}[f(X_m)] \to \mathbb{E}[f(X)]$. I wonder if there is any general method to give sharp asymptotic estimate of the error term $\mathbb{E}[f(X_m)] - \mathbb{E}[f(X)]$, which I expect to be $\Theta(1/m)$. The scaling constant should depend on $f$ (as well as the distribution of $Z_k$ if they are not binary).

For law of large number, this type of estimate can be done via the Delta method (e.g., to estimate $\mathbb{E}[f(\bar{Z})] - f(0)$). There must be a counterpart for CLT... I haven't found the Edgeworth expansion useful because it seems to work with distribution with densities.

Edited: To be clear, I am only interested in some specific nice function (e.g., $f(x) = x^2 e^{-x^2/4}$) and finding a sharp expansion for the error term of the form, say, $c/m + o(1/m)$, where $c$ will depend n $f$. As pointed by Mark, the worst-case rate of all bounded smooth function $f$ is $1/\sqrt{m}$, which agrees with the upper bound given by Stein's method.

Answer by Nate Eldredge for estimate the error term in CLT

2010-09-19T16:36:33Z

The Berry-Esseen theorem is a classical result of this sort. It predicts errors on the order of $m^{-1/2}$, however.

Answer by Byron Schmuland for estimate the error term in CLT

2010-09-20T01:07:20Z

If $X_m$ has cumulative distribution function $F_m$, and $X$ has cumulative distribution function $F$, then (at least formally) integration by parts gives you $$E(f(X_m))-E(f(X))=\int (F_m(x)-F(x)) df(x).$$ Now you can apply the Berry-Esseen bound.

Answer by Mark Meckes for estimate the error term in CLT

2010-09-20T14:54:19Z

Stein's method typically gives good Berry-Esseen type bounds for smooth test functions. See Chapter III of Stein's book (entirely viewable in Google Books). For example, specializing to your case of symmetric Bernoulli summands, equation (37) on p. 38 gives $$ \vert \mathbb{E}f(X_m)-\mathbb{E}f(X)\vert \le \frac{2\Vert f' \Vert_\infty}{\sqrt{m}}. $$ For more general summands, there is some simple dependence on the third and fourth moments as well as $\Vert f \Vert_\infty$.

Also, I'm pretty sure that $m^{-1/2}$ is the correct rate here even for Bernoullis, although I can't find a reference for a lower bound at the moment. Why do you expect better?

Answer by mr.gondolier for estimate the error term in CLT

2010-09-20T15:42:53Z

Sorry this is NOT an answer to my question... just some clarafications.

The reason I think $m^{-1/2}$ is not tight is as follows. For example, take $f$ to be the characteristic function, we have

$\mathbb{E}[e^{itX_m}] = (\mathbb{E}[e^{it Z/\sqrt{m}}])^m = (1 - t^2/(2m) + o(1/m))^m \to e^{-t^2/2} = \mathbb{E}[e^{itX}]$

at rate $1/m$, because $m\log(1-1/m) \to -1$ at rate $1/m$.

Also, it seems all moments of $X_m$ converge to the moments of $X$ at rate $1/m$. Doing a Taylor expansion for those nice $f$ should also yield a rate of $1/m$?