The Berry-Esseen theorem

1 Cumulative distribution functions

For a random variable $X:(\mathrm{\Omega },ℱ,P)\to ℝ$, we define its cumulative distribution function $F_X\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$ by

$$F_X(x)=P(X\le x)={\int }_{\{X\le x\}}dP={\int }_{t\le x}d(X_{*}P)(t)=(X_{*}P)((-\mathrm{\infty },x]).$$

A distribution function is a function $F:ℝ\to [0,1]$ such that (i) $F(-\mathrm{\infty })={lim}_{x\to -\mathrm{\infty }}F(x)=0$, (ii) $F(\mathrm{\infty })={lim}_{x\to \mathrm{\infty }}F(x)=1$, (iii) $F$ is nondecreasing, (iv) $F$ is right-continuous: for each $x\in ℝ$,

$$F(x+)=\underset{t\downarrow x}{lim}F(t)=F(x).$$

It is a fact that the cumulative distribution function of a random variable is a distribution function and that for any distribution function $F$ there is a random variable $X$ for which $F=F_X$.

Let ${\gamma }_1$ be the standard Gaussian measure on $\mathrm{R}$: ${\gamma }_1$ has density

$$p(t,0,1)=\frac{1}{\sqrt{2\pi }}{e}^{-t^2/2}$$

with respect to Lebesgue measure on $ℝ$. Let $\mathrm{\Phi }$ be the cumulative distribution function of ${\gamma }_1$:

$$\mathrm{\Phi }(x)={\gamma }_1((-\mathrm{\infty },x])={\int }_{-\mathrm{\infty }}^x𝑑{\gamma }_1(t)={\int }_{-\mathrm{\infty }}^x\frac{1}{\sqrt{2\pi }}{e}^{-t^2/2}𝑑t.$$

We first prove the following lemma about distribution functions.

Lemma 1.

Suppose that $F$ is a distribution function, that $G:ℝ\to ℝ$ satisfies

$$G(-\mathrm{\infty })=\underset{x\to -\mathrm{\infty }}{lim}G(x)=0,G(\mathrm{\infty })=\underset{x\to \mathrm{\infty }}{lim}G(x)=1,$$

and that $G$ is differentiable and its derivative satisfies

(1)

Writing

$$\mathrm{\Delta }=\frac{1}{2M}\underset{x\in ℝ}{sup}|F(x)-G(x)|,$$

there is some $a\in ℝ$ such that for all $T>0$,

Proof.

Because $G(-\mathrm{\infty })=0$ and $G(\mathrm{\infty })=1$, there is some compact interval $K$ such that for $x\in ℝ\setminus K$. Then, because $G$ is continuous it is bounded on $K$, showing that $G$ is bounded on $ℝ$, and because $M>0$ we get .

Write $H=F-G$. Because $H(\mathrm{\infty })=0$ and $H(-\mathrm{\infty })=0$, there is a compact interval $K$ for which

$$2M\mathrm{\Delta }=\underset{x\in ℝ}{sup}|H(x)|=\underset{x\in K}{sup}|H(x)|.$$

By the Bolzano-Weierstrass theorem, either there is a sequence $u_n\in K$ increasing to some $u\in K$ such that $|H(u_n)|\uparrow 2M\mathrm{\Delta }$ or there is a sequence $u_n\in K$ decreasing to some $u\in K$ such that $|H(u_n)|\uparrow 2M\mathrm{\Delta }$. In the first case, either there is a subsequence $v_n$ of $u_n$ such that $|H(v_n)|=H(v_n)$ or there is a subsequence $v_n$ of $u_n$ such that $|H(v_n)|=-H(v_n)$. In the first subcase we get $H(u-)=2M\mathrm{\Delta }$, thus

$$F(u-)-G(u)=2M\mathrm{\Delta }.$$ (2)

In the second subcase we get $H(u-)=-2M\mathrm{\Delta }$, thus

$$F(u-)-G(u)=-2M\mathrm{\Delta }.$$ (3)

In the second case, either there is a subsequence $v_n$ of $u_n$ such that $|H(v_n)|=H(v_n)$ or there is a subsequence $v_n$ of $u_n$ such that $|H(v_n)=-H(v_n)$. In the first subcase we get $H(u+)=2M\mathrm{\Delta }$, thus

$$F(u)-G(u)=2M\mathrm{\Delta }.$$ (4)

In the second subcase we get $H(u+)=2M\mathrm{\Delta }$, thus

$$F(u)-G(u)=-2M\mathrm{\Delta }.$$ (5)

We now deal with the subcase (3). Let $a=u-\mathrm{\Delta }$. For , by (1) we have

$$|G(x+a)-G(u)|=\left|{\int }_u^{u+x-\mathrm{\Delta }}{G}^{\prime }(y)𝑑y\right|\le |x-\mathrm{\Delta }|M=(\mathrm{\Delta }-x)M,$$

whence

$$G(x+a)\ge G(u)+(x-\mathrm{\Delta })M.$$

Because and as $F$ is nondecreasing and using (3),

$F(x+a)-G(x+a)$ $\le F(u-)-G(x+a)$

$\le F(u-)-(G(u)+(x-\mathrm{\Delta })M)$

$=-M(x+\mathrm{\Delta }).$

Then, because $x\mapsto \frac{1-\cos Tx}{x^2}x$ is an odd function,

${\displaystyle {\int }_{-\mathrm{\Delta }}^{\mathrm{\Delta }}}{\displaystyle \frac{1-\cos Tx}{x^2}}(F(x+a)-G(x+a))𝑑x$ $\le -M{\displaystyle {\int }_{-\mathrm{\Delta }}^{\mathrm{\Delta }}}{\displaystyle \frac{1-\cos Tx}{x^2}}(x+\mathrm{\Delta })𝑑x$

$=-2M\mathrm{\Delta }{\displaystyle {\int }_0^{\mathrm{\Delta }}}{\displaystyle \frac{1-\cos Tx}{x^2}}𝑑x.$

On the other hand,

Thus

which yields the claim of the lemma, for the subcase (3). ∎

We now prove a lemma that gives an inequality for characteristic functions. We remark that because $F$ is a distribution function, it makes sense to speak about the measure induced by $F$, and because $G$ is of bounded variation and is continuous, its variation function $V_G$ is continuous and the functions $V_G-G$ and $V_G$ are nondecreasing, and it thus makes sense to speak about the signed measure induced by $G=V_G-(V_G-G)$, which is equal to the difference of the measures induced by $V_G$ and $V_G-G$.

Lemma 2.

Suppose that $F$ is a distribution function, that $G:ℝ\to ℝ$ satisfies

$$G(-\mathrm{\infty })=\underset{x\to -\mathrm{\infty }}{lim}G(x)=0,G(\mathrm{\infty })=\underset{x\to \mathrm{\infty }}{lim}G(x)=1,$$

that $G$ is differentiable and of bounded variation and that its derivative satisfies

and that

Write

$$\mathrm{\Delta }=\frac{1}{2M}\underset{x\in ℝ}{sup}|F(x)-G(x)|$$

and

$$f(t)={\int }_{ℝ}{e}^{itx}𝑑F(x),g(t)={\int }_{ℝ}{e}^{itx}𝑑G(x).$$

Then for all $T>0$,

$$\mathrm{\Delta }\le \frac{1}{\pi M}{\int }_0^T\frac{|f(t)-g(t)|}{t}𝑑t+\frac{12}{\pi T}.$$

Proof.

For any $t\in ℝ$, because $(F-G)(-\mathrm{\infty })=0$ and $(F-G)(\mathrm{\infty })=0$ and because , integrating by parts gives

$f(t)-g(t)$ $={\displaystyle {\int }_{ℝ}}{e}^{itx}𝑑F(x)-{\displaystyle {\int }_{ℝ}}{e}^{itx}𝑑G(x)$

$={\displaystyle {\int }_{ℝ}}{e}^{itx}d(F-G)(x)$

$=-it{\displaystyle {\int }_{ℝ}}(F-G)(x)e^{itx}𝑑x.$

Take $a$ to be the real number that Lemma 1 yields. As

$$\frac{f(t)-g(t)}{-it}{e}^{-ita}(T-|t|)=(T-|t|){\int }_{ℝ}(F(x+a)-G(x+a))e^{itx}𝑑x,$$

we obtain, using Fubini’s theorem,

Therefore, because $F$ and $G$ are real valued and thus $|f(-t)-g(-t)|=|\overline{f(t)-g(t)}|=|f(t)-g(t)|$,

$\left|{\displaystyle {\int }_{ℝ}}(F(x+a)-G(x+a)){\displaystyle \frac{1-\cos Tx}{x^2}}𝑑x\right|$ $\le {\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-T}^T}{\displaystyle \frac{|f(t)-g(t)|}{|t|}}(T-|t|)𝑑t$

$={\displaystyle {\int }_0^T}{\displaystyle \frac{|f(t)-g(t)|}{t}}(T-t)𝑑t$

$\le T{\displaystyle {\int }_0^T}{\displaystyle \frac{|f(t)-g(t)|}{t}}𝑑t.$

Using this with Lemma 1,

$2MT\mathrm{\Delta }\left(3{\displaystyle {\int }_0^{T\mathrm{\Delta }}}{\displaystyle \frac{1-\cos x}{x^2}}𝑑x-\pi \right)$ $\le T{\displaystyle {\int }_0^T}{\displaystyle \frac{|f(t)-g(t)|}{t}}𝑑t.$

But

$3{\displaystyle {\int }_0^{T\mathrm{\Delta }}}{\displaystyle \frac{1-\cos x}{x^2}}𝑑x-\pi$ $=3{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{1-\cos x}{x^2}}𝑑x-3{\displaystyle {\int }_{T\mathrm{\Delta }}^{\mathrm{\infty }}}{\displaystyle \frac{1-\cos x}{x^2}}𝑑x-\pi$

$\ge 3{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{1-\cos x}{x^2}}𝑑x-6{\displaystyle {\int }_{T\mathrm{\Delta }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{x^2}}𝑑x-\pi$

$=3\cdot {\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{6}{T\mathrm{\Delta }}}-\pi$

$={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{6}{T\mathrm{\Delta }}},$

with which we have

$$2MT\mathrm{\Delta }\left(3{\int }_0^{T\mathrm{\Delta }}\frac{1-\cos x}{x^2}𝑑x-\pi \right)\ge 2MT\mathrm{\Delta }\cdot \left(\frac{\pi }{2}-\frac{6}{T\mathrm{\Delta }}\right)=MT\mathrm{\Delta }\pi -12M,$$

and hence

$$MT\mathrm{\Delta }\pi -12M\le T{\int }_0^T\frac{|f(t)-g(t)|}{t}𝑑t,$$

i.e.

$$\mathrm{\Delta }\le \frac{12}{\pi T}+\frac{1}{\pi M}{\int }_0^T\frac{|f(t)-g(t)|}{t}𝑑t,$$

proving the claim. ∎

2 Berry-Esseen theorem

Let $X_{n,j}$, $n\ge 1$, $1\le j\le k_n$, be $L^3$ random variables, with $k_n\to \mathrm{\infty }$, such that for each $n$, the random variables $X_{n,j}$, $1\le j\le k_n$, are independent, and such that for all $n$ and $j$,

$$E(X_{n,j})=0.$$

Let $F_{n,j}$ be the cumulative distribution function of $X_{n,j}$:

$$F_{n,j}(x)=P(X_{n,j}\le x).$$

Let $f_{n,j}$ be the characteristic function of $X_{n,j}$ (equivalently, the characteristic function of $F_{n,j}$):

$$f_{n,j}(t)={\int }_{ℝ}{e}^{itx}d(X_{n,j}{}_{*}P)(x)={\int }_{ℝ}{e}^{itx}𝑑F_{n,j}(x).$$

Write, for $n\ge 1$,

$$S_n=\sum _{j=1}^{k_n}{X}_{n,j},$$

and let $F_n$ be the cumulative distribution function of $S_n$:

$$F_n(x)=P(S_n\le x)$$

Also, let $f_n$ be the characteristic function of $S_n$ (equivalently, the characteristic function of $F_n$). Because $X_{n,j}$, $1\le j\le k_n$, are independent, we have $S_n{}_{*}P=(X_{n,1}{}_{*}P)*\mathrm{\cdots }*(X_{n,k_n}{}_{*}P)$ and hence

$$f_n(t)={\int }_{ℝ}{e}^{itx}d(S_n{}_{*}P)(x)=\prod _{j=1}^{k_n}{f}_{n,j}(t).$$

For $n\ge 1$ and $1\le j\le k_n$, write

$${\sigma }_{n,j}^2=E(X_{n,j}^2),s_n^2=\sum _{j=1}^{k_n}{\sigma }_{n,j}^2$$

and

$${\gamma }_{n,j}=E({|X_{n,j}|}^3),{\mathrm{\Gamma }}_n=\sum _{j=1}^{k_n}{\gamma }_{n,j}.$$

We further assume that for each $n$,

$$s_n^2=\sum _{j=1}^{k_n}{\sigma }_{n,j}^2=1.$$ (6)

We will use the following inequality which we state separately because it is of general use.

Lemma 3.

For $n\ge 1$ and ,

$$\left|\log (1+z)-\sum _{m=1}^{n-1}\frac{{(-1)}^{m-1}{z}^m}{m}\right|\le \frac{{|z|}^n}{n(1-|z|)}.$$

We now prove an inequality for $f_n$, the characteristic function of $S_n$.

Lemma 4.

For $n\ge 1$, if then

$$|f_n(t)-e^{-t^2/2}|\le {\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/2}.$$

Proof.

For $1\le j\le k_n$ and $l\ge 0$ and $v\in ℝ$,

$$f_{n,j}^{(l)}(v)={(i)}^{l}E(X_{n,j}^l{e}^{iv{X}_{n,j}}).$$

Thus

$$f_{n,j}(0)=1,f_{n,j}^{\prime }(0)=iE(X_{n,j})=0,f_{n,j}^{\prime \prime }(0)=-E(X_{n,j}^2)=-{\sigma }_{n,j}^2,$$

and

$$f_{n,j}^{\prime \prime \prime }(v)=-iE(X_{n,j}^3{e}^{iv{X}_{n,j}}).$$

Then by Taylor’s theorem, there is some $s$ between $0$ and $t$ such that

$$f_{n,j}(t)=1-\frac{{\sigma }_{n,j}^2}{2}{t}^2-\frac{iE(X_{n,j}^3{e}^{is{X}_{n,j}})}{6}{t}^3.$$

Put

$$-iE(X_{n,j}^3{e}^{is{X}_{n,j}})=\theta {\gamma }_{n,j},$$

for which

$$|\theta |=\frac{|E(X_{n,j}^3{e}^{is{X}_{n,j}})|}{E({|X_{n,j}|}^3)}\le 1.$$

Because the $L^2$ norm is upper bounded by the $L^3$ norm and because ,

and hence

$|f_{n,j}(t)-1|$ $=\left|-{\displaystyle \frac{{\sigma }_{n,j}^2}{2}}{t}^2+{\displaystyle \frac{\theta {\gamma }_{n,j}{t}^3}{6}}\right|$

$\le {\displaystyle \frac{1}{2}}{|{\sigma }_{n,j}t|}^2+{\displaystyle \frac{{\gamma }_{n,j}}{48{\mathrm{\Gamma }}_n}}$

Lemma 3 and the inequality ${|a+b|}^2\le 2({|a|}^2+{|b|}^2)$ then tell us that

$\left|\log f_{n,j}(t)-(f_{n,j}(t)-1)\right|$ $\le {\displaystyle \frac{{|f_{n,j}(t)-1|}^2}{2(1-|f_{n,j}(t)-1|)}}$

$={\displaystyle \frac{2}{3}}{\left|-{\displaystyle \frac{{\sigma }_{n,j}^2}{2}}{t}^2+{\displaystyle \frac{\theta {\gamma }_{n,j}}{6}}{t}^3\right|}^2$

$\le {\displaystyle \frac{4}{3}}\left({\displaystyle \frac{{\sigma }_{n,j}^4}{4}}{t}^4+{\displaystyle \frac{{|\theta |}^2{\gamma }_{n,j}^2}{36}}{t}^6\right).$

Because ${\sigma }_{n,j}\le {\gamma }_{n,j}^{1/3}$ and $|\theta |\le 1$,

$\left|\log f_{n,j}(t)-(f_{n,j}(t)-1)\right|$ $\le {\displaystyle \frac{4}{3}}\left({\displaystyle \frac{{\sigma }_{n,j}{\gamma }_{n,j}}{4}}{t}^4+{\displaystyle \frac{{\gamma }_{n,j}^2}{36}}{t}^6\right)$

$={\displaystyle \frac{4}{3}}\left({\displaystyle \frac{|{\sigma }_{n,j}t|}{4}}+{\displaystyle \frac{{|{\gamma }_{n,j}^{1/3}t|}^3}{36}}\right){\gamma }_{n,j}{|t|}^3$

$\le {\displaystyle \frac{4}{3}}\left({\displaystyle \frac{1}{2\cdot 4}}+{\displaystyle \frac{1}{8\cdot 36}}\right){\gamma }_{n,j}{|t|}^3$

$={\displaystyle \frac{37}{216}}{\gamma }_{n,j}{|t|}^3$

Combining this with $f_{n,j}(t)=1-\frac{{\sigma }_{n,j}^2}{2}{t}^2+\frac{\theta {\gamma }_{n,j}}{6}{t}^3$,

$$\left|\log f_{n,j}(t)+\frac{{\sigma }_{n,j}^2}{2}{t}^2\right|\le \left|\frac{\theta {\gamma }_{n,j}}{6}{t}^3\right|+\frac{1}{5}{\gamma }_{n,j}{|t|}^3\le \frac{1}{6}{\gamma }_{n,j}{|t|}^3+\frac{1}{5}{\gamma }_{n,j}{|t|}^3\le \frac{1}{2}{\gamma }_{n,j}{|t|}^3.$$

Because this is true for each $1\le j\le k_n$ and because, according to (6), ${\sum }_{j=1}^{k_n}{\sigma }_{n,j}^2=1$,

$$\left|\log f_n(t)+\frac{t^2}{2}\right|\le \frac{|t|3^2}{2}\sum _{j=1}^{k_n}{\gamma }_{n,j}=\frac{{|t|}^3}{2}{\mathrm{\Gamma }}_n.$$

For any $z\in ℂ$ it is true that $|e^z-1|\le |z|e^{|z|}$, so the above yields

$|f_n(t)e^{t^2/2}-1|$ $=|\exp (\log (f_n(t)e^{t^2/2}))-1|$

$\le |\log (f_n(t)e^{t^2/2})|\exp \left(\left|\log (f_n(t)e^{t^2/2})\right|\right)$

$=\left|\log f_n(t)+{\displaystyle \frac{t^2}{2}}\right|\exp \left(\left|\log (f_n(t)e^{t^2/2})\right|\right)$

$\le {\displaystyle \frac{{|t|}^3}{2}}{\mathrm{\Gamma }}_n\exp \left({\displaystyle \frac{{|t|}^3}{2}}{\mathrm{\Gamma }}_n\right).$

But , so

$$|f_n(t)e^{t^2/2}-1|\le \frac{{|t|}^3}{2}{\mathrm{\Gamma }}_n{e}^{1/16}\le {|t|}^3{\mathrm{\Gamma }}_n,$$

which completes the proof. ∎

The next lemma gives a different bound on the characteristic function of $S_n$.

Lemma 5.

For $n\ge 1$, if then

$$|f_n(t)|\le e^{-t^2/3}.$$

Proof.

First, for a distribution function $F$ with characteristic function $f$,

${|f(t)|}^2$ $=f(t)\overline{f(t)}$

$={\displaystyle {\int }_{ℝ}}{e}^{itx}𝑑F(x)\cdot {\displaystyle {\int }_{ℝ}}{e}^{-itx}𝑑F(y)$

$={\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{ℝ}}{e}^{it(x-y)}𝑑F(x)\right)𝑑F(y)$

$={\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{ℝ}}\cos t(x-y)+i\sin t(x-y)dF(x)\right)𝑑F(y).$

Because ${|f(t)|}^2$ is real it follows that

$${|f(t)|}^2={\int }_{ℝ}\left({\int }_{ℝ}\cos t(x-y)𝑑F(x)\right)𝑑F(y).$$

Using

$$\left|\cos u-\left(1-\frac{u^2}{2}\right)\right|\le \frac{{|u|}^3}{6},{|a+b|}^p\le 2^{p-1}({|a|}^p+{|b|}^p),$$

we have

$$\left|\cos t(x-y)-\left(1-\frac{{(t(x-y))}^2}{2}\right)\right|\le \frac{2}{3}\left({|tx|}^3+{|ty|}^3\right)=\frac{2{|t|}^3}{3}({|x|}^3+{|y|}^3)$$

and then

${|f(t)|}^2$ $\le {\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{ℝ}}1-{\displaystyle \frac{{(t(x-y))}^2}{2}}+{\displaystyle \frac{2{|t|}^3}{3}}({|x|}^3+{|y|}^3)dF(x)\right)𝑑F(y).$

Using this for $f_{n,j}$, and using that $E(X_{n,j})=0$,

${|f_{n,j}(t)|}^2$ $\le {\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{ℝ}}1-{\displaystyle \frac{{(t(x-y))}^2}{2}}+{\displaystyle \frac{2{|t|}^3}{3}}({|x|}^3+{|y|}^3)d{F}_{n,j}(x)\right)𝑑F_{n,j}(y)$

$={\displaystyle {\int }_{ℝ}}1-{\displaystyle \frac{t^2{\sigma }_{n,j}^2}{2}}-{\displaystyle \frac{t^2{y}^2}{2}}+{\displaystyle \frac{2{|t|}^3{\gamma }_{n,j}}{3}}+{\displaystyle \frac{2{|t|}^3{|y|}^3}{3}}dF(y)$

$=1-{\displaystyle \frac{t^2{\sigma }_{n,j}^2}{2}}-{\displaystyle \frac{t^2{\sigma }_{n,j}^2}{2}}+{\displaystyle \frac{2{|t|}^3{\gamma }_{n,j}}{3}}+{\displaystyle \frac{2{|t|}^3{\gamma }_{n,j}}{3}}$

$=1-t^2{\sigma }_{n,j}^2+{\displaystyle \frac{4{|t|}^3{\gamma }_{n,j}}{3}}.$

Because $1+u\le e^u$ for all $u\in ℝ$,

$${|f_{n,j}(t)|}^2\le \exp \left(-t^2{\sigma }_{n,j}^2+\frac{4{|t|}^3{\gamma }_{n,j}}{3}\right).$$

Then, by (6),

${|f_n(t)|}^2$ $={\displaystyle \prod _{j=1}^{k_n}}{|f_{n,j}(t)|}^2$

$\le {\displaystyle \prod _{j=1}^{k_n}}\exp \left(-t^2{\sigma }_{n,j}^2+{\displaystyle \frac{4{|t|}^3{\gamma }_{n,j}}{3}}\right)$

$=\exp \left(-t^2{\displaystyle \sum _{j=1}^{k_n}}{\sigma }_{n,j}^2+{\displaystyle \frac{4{|t|}^3}{3}}{\displaystyle \sum _{j=1}^{k_n}}{\gamma }_{n,j}\right)$

$=\exp \left(-t^2+{\displaystyle \frac{4{|t|}^3}{3}}{\mathrm{\Gamma }}_n\right).$

As ,

$$|f_n(t)|\le \exp \left(-\frac{t^2}{2}+\frac{2{|t|}^3}{3}{\mathrm{\Gamma }}_n\right)\le \exp \left(-\frac{t^2}{2}+\frac{2{|t|}^2}{12}\right)=e^{-t^2/3},$$

proving the claim. ∎

We now combine Lemma 4 and Lemma 5.

Lemma 6.

For $n\ge 1$, if then

$$|f_n(t)-e^{-t^2/2}|\le 16{\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/3}.$$

Proof.

Either or . In the first case, Lemma 4 tells us

$$|f_n(t)-e^{-t^2/2}|\le {\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/2}\le {\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/3}\le 16{\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/3}.$$

In the second case, Lemma 5 tells us

$$|f_n(t)|\le e^{-t^2/3},$$

and so, as in this case we have $1\le 8{\mathrm{\Gamma }}_n{|t|}^3$,

$$|f_n(t)-e^{-t^2/2}|\le |f_n(t)|+e^{-t^2/2}\le e^{-t^2/3}+e^{-t^2/2}\le 2e^{-t^2/3}\le 16{\mathrm{\Gamma }}_n{|t|}^3{e}^{-t^2/3},$$

showing that the claim is true in both cases. ∎

We finally prove the Berry-Esseen theorem.

Theorem 7 (Berry-Esseen theorem).

There is some such that for each $n\ge 1$,

$$\underset{x\in ℝ}{sup}|F_n(x)-\mathrm{\Phi }(x)|\le A_0{\mathrm{\Gamma }}_n.$$

Proof.

Let $Z$ be a random variable with $Z_{*}P={\gamma }_1$, i.e. whose cumulative distribution function is $\mathrm{\Phi }$. By (6) and because $X_{n,j}$, $1\le j\le k_n$, are independent and satisfy $E(X_{n,j})=0$,

$$E(S_n^2)=\sum _{j=1}^{k_n}E(X_{n,j}^2)=\sum _{j=1}^{k_n}{\sigma }_{n,j}^2=1.$$

If then by Chebyshev’s inequality

$$F_n(x)=P(S_n\le x)=P(-S_n\ge -x)\le \frac{1}{x^2}E(|S_n{|}^2)=\frac{1}{x^2}$$

and

$$\mathrm{\Phi }(x)=P(Z\le x)=P(-Z\ge -x)\le \frac{1}{x^2}E(|Z{|}^2)=\frac{1}{x^2}.$$

If $x>0$ then also by Chebyshev’s inequality

$$1-F_n(x)=1-P(S_n\le x)=P(S_n>x)\le \frac{1}{x^2}$$

and

$$1-\mathrm{\Phi }(x)=1-P(Z\le x)=P(Z>x)\le \frac{1}{x^2}.$$

Therefore, because $F_n$ and $\mathrm{\Phi }$ are nonnegative and $1-F_n$ and $1-\mathrm{\Phi }$ are nonnegative, for all $x\in ℝ$ we have

$$|F_n(x)-\mathrm{\Phi }(x)|\le \frac{1}{x^2}.$$

Then, because $|F_n|\le 1$ and $|\mathrm{\Phi }|\le 1$,

${\mathrm{\Phi }}^{\prime }(x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}\le \frac{1}{\sqrt{2\pi }}$. We apply Lemma 2 with $F=F_n$, $G=\mathrm{\Phi }$, and $M=\frac{1}{\sqrt{2\pi }}$, and because the characteristic function of $\mathrm{\Phi }$ is $\varphi (t)=e^{-t^2/2}$, we obtain for $T=\frac{1}{4{\mathrm{\Gamma }}_n}$,

$\underset{x\in ℝ}{sup}|F_n(x)-\mathrm{\Phi }(x)|$ $\le {\displaystyle \frac{2}{\pi }}{\displaystyle {\int }_0^{\frac{1}{4{\mathrm{\Gamma }}_n}}}{\displaystyle \frac{|f_n(t)-\varphi (t)|}{t}}𝑑t+{\displaystyle \frac{96M{\mathrm{\Gamma }}_n}{\pi }}$

$={\displaystyle \frac{2}{\pi }}{\displaystyle {\int }_0^{\frac{1}{4{\mathrm{\Gamma }}_n}}}{\displaystyle \frac{|f_n(t)-e^{-t^2/2}|}{t}}+{\displaystyle \frac{96{\mathrm{\Gamma }}_n}{\pi \sqrt{2\pi }}}.$

Then applying Lemma 6,

$\underset{x\in ℝ}{sup}|F_n(x)-\mathrm{\Phi }(x)|$ $\le {\displaystyle \frac{2}{\pi }}{\displaystyle {\int }_0^{\frac{1}{4{\mathrm{\Gamma }}_n}}}{\displaystyle \frac{16{\mathrm{\Gamma }}_n{t}^3{e}^{-t^2/3}}{t}}𝑑t+{\displaystyle \frac{96{\mathrm{\Gamma }}_n}{\pi \sqrt{2\pi }}}$

$={\mathrm{\Gamma }}_n\left({\displaystyle \frac{32}{\pi }}{\displaystyle {\int }_0^{\frac{1}{4{\mathrm{\Gamma }}_n}}}{t}^2{e}^{-t^2/3}𝑑t+{\displaystyle \frac{96}{\pi \sqrt{2\pi }}}\right).$

This proves the claim with

$$A_0=\frac{32}{\pi }{\int }_0^{\mathrm{\infty }}{t}^2{e}^{-t^2/3}𝑑t+\frac{96}{\pi \sqrt{2\pi }}=\frac{32}{\pi }\cdot \frac{3\sqrt{3\pi }}{4}+\frac{96}{\pi \sqrt{2\pi }}=35.64\mathrm{\dots }.$$

∎