estimate the error term in CLT

Question

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 1

Your conjecture is correct. Suppose that (say) $f$ has a bounded $5$th derivative. Then
\begin{equation} Ef(X_m) - Ef(X)=-\frac{Ef''''(X)+o(1)}{12m}. \tag{1} \end{equation} Indeed, let \begin{equation} Z_{mj}:=\frac{Z_j}{\sqrt m},\quad Y_{mj}:=\frac{Y_j}{\sqrt m},\quad T_{mk}:=\sum_{j=1}^{k-1}Z_{mj}+\sum_{j=k+1}^m Y_{mj}, \end{equation} where $Z_1,\dots,Z_m,Y_1,\dots,Y_m$ are independent random variables and $Y_j\sim N(0,1)$. Note that $EZ_{mj}^p=EY_{mj}^p$ for $p=0,\dots,3$, $EZ_{mj}^4=\tfrac1{m^2}$, $EY_{mj}^4=\tfrac3{m^2}$. Also, $Z_{mk}$ is independent of $T_{mk}$ for each $k$, and so is $Y_{mk}$. Moreover, $Ef''''(T_{mk})\to Ef''''(X)$ uniformly in $k=1,\dots,m$ (as $m\to\infty$); this should be as easy to show as the convergence $Ef(X_m) \to Ef(X)$, mentioned in the question.

Now one has
\begin{equation} Ef(X_m) - Ef(X)=\sum_{k=1}^m D_k, \tag{2} \end{equation} where
\begin{equation} D_k:=E[f(T_{mk}+Z_{mk})-f(T_{mk}+Y_{mk})] \end{equation} \begin{equation} =\sum_{p=0}^4\frac1{p!}\,Ef^{(p)}(T_{mk})E(Z_{mk}^p-Y_{mk}^p)+O(E(|Z_{mk}|^5+|Y_{mk}|^5)) \end{equation} \begin{equation} =\tfrac1{4!}\,(Ef''''(X)+o(1))(\tfrac1{m^2}-\tfrac3{m^2})+O(m^{-5/2}) \end{equation} by Taylor's expansion, with $o(1)$ uniform in $k=1,\dots,m$. Now $(1)$ follows immediately from $(2)$.

Answer 2

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 3

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 4

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

Answer 5

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 $1/m$?