Steven R. Dunbar
Department of Mathematics
203 Avery Hall
Lincoln, NE 68588-0130
http://www.math.unl.edu
Voice: 402-472-3731
Fax: 402-472-8466

Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

__________________________________________________________________________

Analytic Proof of the DeMoivre-Laplace Central Limit Theorem

_______________________________________________________________________

Note: These pages are prepared with MathJax. MathJax is an open source JavaScript display engine for mathematics that works in all browsers. See http://mathjax.org for details on supported browsers, accessibility, copy-and-paste, and other features.

_______________________________________________________________________________________________

### Rating

Mathematicians Only: prolonged scenes of intense rigor.

_______________________________________________________________________________________________

### Section Starter Question

Derive the Fourier transform of the rectangle function $\Pi \left(t\right)$, where

$\Pi \left(t\right)=\left\{\begin{array}{cc}0\phantom{\rule{1em}{0ex}}\hfill & \left|t\right|>1∕2\hfill \\ 1∕2\phantom{\rule{1em}{0ex}}\hfill & \left|t\right|=1∕2\hfill \\ 1\phantom{\rule{1em}{0ex}}\hfill & \left|t\right|<1∕2\hfill \end{array}\right\.$

_________________________________________________________________________________

### Key Concepts

1. Let $X$ be a random variable with probability density function $f$. The Fourier transform, also known as the characteristic function of $f$ (and also of $X$) is
$\varphi \left(\xi \right)=\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi x}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

2. Let $\varphi \left(\xi \right)$ be the Fourier transform (also known as the characteristic function) of a probability distribution such that $\underset{-\infty }{\overset{\infty }{\int }}\left|\varphi \left(\xi \right)\right|\phantom{\rule{0.3em}{0ex}}d\xi <\infty$. Then there is a bounded continuous density $f\left(x\right)$ given by
$f\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi x}\varphi \left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .$

3. For every real $\xi$,
$\underset{n\to \infty }{lim}{\left(cos\left(\frac{\xi }{\sqrt{n}}\right)\right)}^{n}={\mathrm{e}}^{-{\xi }^{2}∕2}.$

4. $\begin{array}{llll}\hfill \mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)& =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }\left({\mathrm{e}}^{-{\xi }^{2}∕2}\right)\phantom{\rule{0.3em}{0ex}}d\xi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{\sqrt{2\pi }}\underset{{\omega }_{1}}{\overset{{\omega }_{2}}{\int }}{\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

__________________________________________________________________________

### Vocabulary

1. The indicator function for the interval $\left[{\omega }_{1},{\omega }_{2}\right]$ is
$g\left(x\right)=\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & {\omega }_{1}

2. Let $X$ be a random variable with probability density function $f$. The Fourier transform, also known as the characteristic function of $f$ (and also of $X$) is
$\varphi \left(\xi \right)=\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi x}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

3. Let $f\left(x\right)$ and $g\left(x\right)$ be two probability density functions with Fourier transforms $\varphi \left(\xi \right)$ and $\psi \left(\xi \right)$ respectively. Parseval’s relation is
$\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)g\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\psi \left(x-t\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

4. Fourier’s integral formula is
$g\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}g\left(y\right){\mathrm{e}}^{-\mathrm{i}\xi \left(x-y\right)}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}d\xi .$

__________________________________________________________________________

### Mathematical Ideas

#### Fourier Transforms and Characteristic Functions

Deﬁnition. Let $X$ be a random variable with probability density function $f$. The Fourier transform, also known as the characteristic function of $f$ (and also of $X$) is

$\varphi \left(\xi \right)=\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi x}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

Example. The uniform random variable with density function $f\left(x\right)=\frac{1}{2a}$ on $\left[-a,a\right]$ has Fourier transform

$\begin{array}{llll}\hfill \varphi \left(\xi \right)& =\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi x}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx=\underset{-a}{\overset{a}{\int }}\frac{{\mathrm{e}}^{\mathrm{i}\xi x}}{2a}\phantom{\rule{0.3em}{0ex}}dx={\frac{{\mathrm{e}}^{\mathrm{i}\xi x}}{2a\cdot \mathrm{i}\xi }|}_{-a}^{a}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{\mathrm{e}}^{\mathrm{i}\xi a}}{2a\cdot \mathrm{i}\xi }-\frac{{\mathrm{e}}^{-\mathrm{i}\xi a}}{2a\cdot \mathrm{i}\xi }=\frac{sin\left(a\xi \right)}{a\xi }.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The Fourier transform has many properties, but the subsequent proofs only need one, the Fourier inversion formula:

$f\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi x}\varphi \left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .$

(valid if $\underset{-\infty }{\overset{\infty }{\int }}\left|\varphi \left(\xi \right)\right|\phantom{\rule{0.3em}{0ex}}d\xi <\infty$).

Lemma 1 (Parseval relation). Let $f\left(x\right)$ and $g\left(x\right)$ be two probability density functions with Fourier transforms $\varphi \left(\xi \right)$ and $\psi \left(\xi \right)$ respectively. Then

$\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)g\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\psi \left(x-t\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

Proof.

1. Start with the deﬁnition of $\varphi \left(\xi \right)$ and multiply by ${\mathrm{e}}^{-\mathrm{i}\xi t}$:
${\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)=\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi \left(x-t\right)}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

2. Integrate both sides with respect to $g\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi$ to get
$\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)g\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\xi \left(x-t\right)}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dxg\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .$

3. Interchange the order of integration and apply the deﬁnition of $\psi \left(\xi \right)$
$\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)g\left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\psi \left(x-t\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

Remark. The proof is easily generalized to cumulative distribution functions $F\left(x\right)$ and $G\left(x\right)$ with associated probability measures $\phantom{\rule{0.3em}{0ex}}dF$ and $\phantom{\rule{0.3em}{0ex}}dG$. See [3].

Theorem 2. Let $\varphi \left(\xi \right)$ be the Fourier transform (also known as the characteristic function) of a probability distribution such that $\underset{-\infty }{\overset{\infty }{\int }}\left|\varphi \left(\xi \right)\right|\phantom{\rule{0.3em}{0ex}}d\xi <\infty$. Then there is a bounded continuous density $f\left(x\right)$ given by

$f\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi x}\varphi \left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .$

Proof.

1. Consider ${n}_{a}\left(x\right)=an\left(ax\right)$, the scaled normal density function, where $n\left(x\right)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-{x}^{2}∕2}$ is the standard normal probability density.
2. The Fourier transform (characteristic function) of the scaled normal density function is ${\psi }_{a}\left(\xi \right)=\sqrt{2\pi }n\left(\xi ∕a\right)$, see the Problems for a proof.
3. Then consider the Parseval relation applied to ${n}_{a}\left(x\right)$:
$\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)\frac{a}{\sqrt{2\pi }}{\mathrm{e}}^{-{a}^{2}{\xi }^{2}∕2}\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\sqrt{2\pi }n\left(\frac{x-t}{a}\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$

Rearranging the constants and using the evenness of $n\left(\cdot \right)$

 $\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right){\mathrm{e}}^{-{a}^{2}{\xi }^{2}∕2}\phantom{\rule{0.3em}{0ex}}d\xi =\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{a}n\left(\frac{t-x}{a}\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx.$ (1)
4. Let
 ${f}_{a}\left(x\right)=\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{a}n\left(\frac{t-x}{a}\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx$ (2)

so that ${f}_{a}\left(x\right)$ is the probability density that is the convolution of $f$ with the scaled normal density, ${f}_{a}={n}_{1∕a}\ast f$.

5. From (2), $\underset{a\to 0}{lim}{f}_{a}\left(x\right)=f\left(x\right)$.
6. From (1)

$\begin{array}{llll}\hfill \underset{a\to 0}{lim}\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{a}n\left(\frac{t-x}{a}\right)f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx& =\underset{a\to 0}{lim}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right){\mathrm{e}}^{-{a}^{2}{\xi }^{2}∕2}\phantom{\rule{0.3em}{0ex}}d\xi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi t}\varphi \left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
7. Using these limits in (1)
$f\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-\mathrm{i}\xi x}\varphi \left(\xi \right)\phantom{\rule{0.3em}{0ex}}d\xi .$

Remark. This proof is adapted from Feller [3, pages 507-511]. This is one of several proofs possible for the inversion theorem. For alternative proofs see [1, pages 177-178], [2, page 155], or [5, pages 20-21].

#### The Main Idea of the Proof

Let

 $g\left(x\right)=\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & {\omega }_{1} (3)

Lemma 3 (Fourier’s Formula). As a Fourier integral

 $g\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }{\mathrm{e}}^{-\mathrm{i}x\xi }\phantom{\rule{0.3em}{0ex}}d\xi .$ (4)

with the slight exception that $g\left({\omega }_{1}\right)=\frac{1}{2}$ and $g\left({\omega }_{2}\right)=\frac{1}{2}$.

Proof. Left as an exercise. □

Using the analytic expression for the probability of the sum of Bernoulli trials

$\begin{array}{llll}\hfill & \mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}=\underset{0}{\overset{1}{\int }}g\left(\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}=\underset{0}{\overset{1}{\int }}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}×exp\left(-\mathrm{i}\xi \frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}d\xi \phantom{\rule{0.3em}{0ex}}dt.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Changing the exponential into a product of cosines comes from equations (2) and (3) in Analytic Model of Coin Flipping.

Interchange the order of integration, justiﬁed since ${r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)$ assumes only a ﬁnite set of values.

$\begin{array}{c}\mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\\ =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }\left[\underset{0}{\overset{1}{\int }}exp\left(-\mathrm{i}\xi \frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt\right]\phantom{\rule{0.3em}{0ex}}dx\\ =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }{\left(cos\left(\frac{\xi }{\sqrt{n}}\right)\right)}^{n}\phantom{\rule{0.3em}{0ex}}d\xi .\end{array}$

Lemma 4. For every real $\xi$,

$\underset{n\to \infty }{lim}{\left(cos\left(\frac{\xi }{\sqrt{n}}\right)\right)}^{n}={\mathrm{e}}^{-{\xi }^{2}∕2}$

Proof. This follows from

$cos\left(\frac{\xi }{\sqrt{n}}\right)=1-\frac{{\xi }^{2}}{2n}+o\left(\frac{1}{n}\right)$

so that

$\underset{n\to \infty }{lim}{\left(cos\left(\frac{\xi }{\sqrt{n}}\right)\right)}^{n}=\underset{n\to \infty }{lim}{\left(1-\frac{{\xi }^{2}}{2n}+o\left(\frac{1}{n}\right)\right)}^{n}={\mathrm{e}}^{-{\xi }^{2}∕2}.$

Alternatively, taking logarithms and using L’Hospital’s Rule

$\begin{array}{llll}\hfill \underset{n\to \infty }{lim}nlog\left(cos\left(\frac{\xi }{\sqrt{n}}\right)\right)& =\underset{n\to \infty }{lim}\frac{1}{cos\left(\frac{\xi }{\sqrt{n}}\right)}\cdot -sin\left(\frac{\xi }{\sqrt{n}}\right)\cdot \frac{-\xi }{2{n}^{3∕2}}/\frac{-1}{{n}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{cos\left(\frac{\xi }{\sqrt{n}}\right)}sin\left(\frac{\xi }{\sqrt{n}}\right)\cdot \frac{\sqrt{n}}{\xi }\cdot \frac{-{\xi }^{2}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-{\xi }^{2}∕2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Using this limit

$\begin{array}{llll}\hfill & \underset{n\to \infty }{lim}\mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }\left({\mathrm{e}}^{-{\xi }^{2}∕2}\right)\phantom{\rule{0.3em}{0ex}}d\xi =\frac{1}{\sqrt{2\pi }}\underset{{\omega }_{1}}{\overset{{\omega }_{2}}{\int }}{\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The interchange of the operations of integration and taking the limit as $n\to \infty$ needs justiﬁcation. However, the limits of integration are $-\infty$ and $+\infty$ and the function

$\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }$

is not absolutely integrable. Recall that a function is integrable on $E$ if $\left|{\int }_{E}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx\right|<\infty$. A function is absolutely integrable on $E$ if ${\int }_{E}\left|f\left(x\right)\right|\phantom{\rule{0.3em}{0ex}}dx<\infty$. A theorem from Lebesgue integration theory says that a function $f\left(x\right)$ is integrable if and only if $f\left(x\right)$ is absolutely integrable. The interchange of the operations of integration and taking the limit as $n\to \infty$ is justiﬁed if the integrand is integrable, hence absolutely integrable. However, the function

$\frac{{\mathrm{e}}^{\mathrm{i}\omega \xi }}{\mathrm{i}\xi }=\frac{cos\left(\omega \xi \right)+\mathrm{i}sin\left(\omega \xi \right)}{\mathrm{i}\xi }$

is not absolutely integrable on $\left(-\infty ,\infty \right)$, see the exercises. Therefore, although this proof of the Central Limit Theorem is fairly direct with Fourier transforms, it is not rigorously justiﬁed.

#### Rigorous justiﬁcation

The equation (4) is a special case of Fourier’s general formula

$g\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}g\left(y\right){\mathrm{e}}^{-\mathrm{i}\xi \left(x-y\right)}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}dx$

applied to the indicator function (3). To make the proof above rigorous, introduce two auxiliary functions ${g}_{𝜖}^{-}\left(x\right)$ and ${g}_{𝜖}^{+}\left(x\right)$, both graphed in Figure 1.

By construction, ${g}_{𝜖}^{-}\left(x\right) and then

$\begin{array}{c}\underset{0}{\overset{1}{\int }}{g}_{𝜖}^{-}\left(\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt\\ \le \mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\\ \le \underset{0}{\overset{1}{\int }}{g}_{𝜖}^{+}\left(\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt.\\ \end{array}$

Now

are absolutely integrable as functions of $\xi$ on $\left(-\infty ,\infty \right)$. (See the exercises for a proof.) Now the same argument as above yields rigorously

$\begin{array}{c}\underset{n\to \infty }{lim}\underset{0}{\overset{1}{\int }}{g}_{𝜖}^{-}\left(\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt\\ =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-{y}^{2}∕2}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{-}\left(y\right){\mathrm{e}}^{\mathrm{i}y\xi }\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}d\xi \\ =\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{-}\left(y\right){\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy\end{array}$

and

$\begin{array}{c}\underset{n\to \infty }{lim}\underset{0}{\overset{1}{\int }}{g}_{𝜖}^{+}\left(\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}\right)\phantom{\rule{0.3em}{0ex}}dt\\ =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{-{y}^{2}∕2}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{+}\left(y\right){\mathrm{e}}^{\mathrm{i}y\xi }\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}d\xi \\ =\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{+}\left(y\right){\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy.\end{array}$

Combining the two previous expressions with the inequality for ${g}_{𝜖}^{-}$ and ${g}_{𝜖}^{+}$

$\begin{array}{c}\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{-}\left(y\right){\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy\\ \le \underset{n\to \infty }{liminf}\mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\\ \le \underset{n\to \infty }{limsup}\mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\\ \le \frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{g}_{𝜖}^{-}\left(y\right){\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy.\end{array}$

Since the inequality holds for every $𝜖>0$, it follows that

$\begin{array}{c}\underset{n\to \infty }{lim}\mu \left({\omega }_{1}<\frac{{r}_{1}\left(t\right)+{r}_{2}\left(t\right)+\cdots +{r}_{n}\left(t\right)}{\sqrt{n}}<{\omega }_{2}\right)\\ =\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}g\left(y\right){\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy=\frac{1}{\sqrt{2\pi }}\underset{{\omega }_{1}}{\overset{{\omega }_{2}}{\int }}{\mathrm{e}}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy.\end{array}$

### Section Ending Answer

${\mathsc{ℱ}}_{x}\left[\Pi \left(x\right)\right]\left(\omega \right)=\underset{-\infty }{\overset{\infty }{\int }}{\mathrm{e}}^{\mathrm{i}\omega x}\Pi \left(x\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{sinx}{x}=sinc\left(x\right).$

#### Sources

The deﬁnition of the Fourier transform and the inversion Theorem are adapted from William Feller, Introduction to Probability Theory and It Applications, Volume II, Second Edition, by W. Feller, J. Wiley and Sons, pages 507-511. The main results of the section are adapted from Statistical Independence in Probability, Analysis, and Number Theory by Mark Kac, Carus Mathematical Monographs, Number 12, Chapter 3, Sections 1 and 2, pages 36-41. [4]

_______________________________________________________________________________________________

### Algorithms, Scripts, Simulations

#### Scripts

__________________________________________________________________________

### Problems to Work for Understanding

1. Show that the Fourier transform (characteristic function) of the scaled normal density function $an\left(ax\right)$, where $n\left(x\right)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-{x}^{2}∕2}$, is $\psi \left(\xi \right)=\sqrt{2\pi }n\left(x∕a\right)$.
2. Show the function
$\frac{{\mathrm{e}}^{\mathrm{i}\omega \xi }}{\mathrm{i}\xi }=\frac{cos\left(\omega \xi \right)+\mathrm{i}sin\left(\omega \xi \right)}{\mathrm{i}\xi }$

is not absolutely integrable on $\left(-\infty ,\infty \right)$.

3. Prove Fourier’s Formula
$g\left(x\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mathrm{e}}^{\mathrm{i}{\omega }_{2}\xi }-{\mathrm{e}}^{\mathrm{i}{\omega }_{1}\xi }}{\mathrm{i}\xi }{\mathrm{e}}^{-\mathrm{i}x\xi }\phantom{\rule{0.3em}{0ex}}d\xi .$

4. Show that the function ${\mathsc{ℱ}}_{x}\left[{g}^{-}\right]\left(\xi \right)$ is absolutely integrable on $\left(-\infty ,\infty \right)$.

__________________________________________________________________________

### References

[1]   Leo Breiman. Probability. Addison Wesley, 1968.

[2]   Kai Lai Chung. A Course in Probability Theory. Academic Press, 1974.

[3]   William Feller. A Introduction to Probability Theory and It Applications, Volume II, Second Edition, volume II. John Wiley and Sons, second edition, 1971.

[4]   Mark Kac. Statistical Independence in Probability, Analysis and Number Theory, volume 12 of The Carus Mathematical Monographs. Mathematical Association of America, 1959.

[5]   S. R. S. Varadhan. Probability Theory. Number 7 in Courant Lecture Notes. American Mathematical Society, 2001.

__________________________________________________________________________

__________________________________________________________________________

I check all the information on each page for correctness and typographical errors. Nevertheless, some errors may occur and I would be grateful if you would alert me to such errors. I make every reasonable eﬀort to present current and accurate information for public use, however I do not guarantee the accuracy or timeliness of information on this website. Your use of the information from this website is strictly voluntary and at your risk.

I have checked the links to external sites for usefulness. Links to external websites are provided as a convenience. I do not endorse, control, monitor, or guarantee the information contained in any external website. I don’t guarantee that the links are active at all times. Use the links here with the same caution as you would all information on the Internet. This website reﬂects the thoughts, interests and opinions of its author. They do not explicitly represent oﬃcial positions or policies of my employer.

Information on this website is subject to change without notice.