229 lines
9.5 KiB
TeX
229 lines
9.5 KiB
TeX
\lecture{12}{2023-05-16}{Proof of the CLT}
|
|
|
|
We now want to prove the \yaref{clt}.
|
|
The plan is to do the following:
|
|
\begin{enumerate}[1.]
|
|
\item Identify the characteristic function of a standard normal
|
|
\item Show that the characteristic functions of the $V_n$ converge pointwise
|
|
to that of $\cN$.
|
|
\item Apply \yaref{levycontinuity}
|
|
\end{enumerate}
|
|
|
|
First, we need to prove some properties of characteristic functions.
|
|
|
|
\begin{lemma}
|
|
\label{charfprops}
|
|
For every real random variable $X$, we have
|
|
\begin{enumerate}[(i)]
|
|
\item $\phi_X(0) = 1$ and $|\phi_X(t)| \le 1$ for all $t \in \R$.
|
|
\item $\phi_X$ is uniformly continuous.
|
|
\item If $\bE[|X|^n] < \infty$ for any $n \in \N$, then $\phi_X$ i
|
|
$n$-times continuously differentiable
|
|
and $\bE[X^n] = (-\i)^n \phi_X^{(n)}(0)$.
|
|
\item For independent random variables $X$ and $Y$, we have
|
|
\[
|
|
\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t).
|
|
\]
|
|
\end{enumerate}
|
|
\end{lemma}
|
|
\begin{refproof}{charfprops}
|
|
\begin{enumerate}[(i)]
|
|
\item $\phi_X(0) = \bE[e^{\i 0 X}] = \bE[1] = 1$.
|
|
For $t \in \R$, we have $|\phi_X(t)| = |\bE[e^{\i t X}]| \overset{\yaref{jensen}}{\le} \bE[|e^{\i t X}|] = 1$.
|
|
|
|
\item Let $t, h \in \R$.
|
|
Then
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
|\phi_X(t+h) - \phi_X(t)| &=& |\bE[e^{\i (t+h) X} - e^{\i t X}]|\\
|
|
&=& |\bE[e^{\i t X} (e^{\i h X} - 1)]|\\
|
|
&\overset{\yaref{jensen}}{\le}&
|
|
\bE[|e^{\i t X}| \cdot |e^{\i h X} -1|]\\
|
|
&=& \bE[|e^{\i h X} - 1|]
|
|
\text{\reflectbox{$\coloneqq$}} g(h)\\
|
|
\end{IEEEeqnarray*}
|
|
|
|
Hence $\sup_{t \in \R} |\phi_X(t + h) - \phi_X(t) | \le g(h)$.
|
|
We show that $\lim_{h \to 0} g(h) = 0$.
|
|
|
|
For all $\omega \in \Omega$, we realize
|
|
\[
|
|
\lim_{h \to 0} |e^{\i h X(\omega)}- 1| = 0.
|
|
\]
|
|
Thus $|e^{\i h X} - 1| \xrightarrow{h \to 0} 0$ almost surely.
|
|
Since also for all $h \in \R$ we have $|e^{\i h X} - 1| \le 2$,
|
|
it follow that $|e^{\i h X} - 1|$ is dominated for all $h \in \R$.
|
|
Thus, we can apply the dominated convergence theorem % TODO REF
|
|
and obtain
|
|
\[
|
|
\lim_{h \to 0} g(h) = \lim_{h \to 0} \bE[|e^{\i h X} - 1|]
|
|
= \bE[\lim_{h \to 0} |e^{\i h X} - 1|] = 0.
|
|
\]
|
|
It follows that
|
|
\[
|
|
\lim_{n \to 0} \sup_{t \in \R} | \phi_X(t + h) - \phi_X(t)| = 0,
|
|
\]
|
|
which means that $\phi_X$ is uniformly continuous.
|
|
\item
|
|
\begin{claim}
|
|
\label{charfprop:c1}
|
|
For $y \in \R$, we have $|e^{\i y} - 1| \le |y|$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
For $y \ge 0$, we have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
|e^{\i y} - 1| &=& |\int_0^y \cos(s) \dif s + \i \int_0^y \sin(s) \dif s|\\
|
|
&=& |\int_0^y e^{\i s} \dif s|\\
|
|
&\overset{\yaref{jensen}}{\le}& \int_0^y |e^{\i s}| ds = y.
|
|
\end{IEEEeqnarray*}
|
|
For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$
|
|
and we can apply the above to $-y$.
|
|
\end{subproof}
|
|
|
|
First, we look at $n = 1$. Then $\bE[|X|] < \infty$.
|
|
Consider
|
|
\[
|
|
\frac{\phi_X(t + h) - \phi_X(t)}{h} = \bE\left[e^{\i t X} \frac{e^{\i h X} - 1}{h}\right].
|
|
\]
|
|
We have $e^z = \sum_{n = 0}^\infty \frac{z^k}{n!}$.
|
|
Hence
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\lim_{n \to \infty} e^{\i t X} \left( \frac{1 + \i h X + \frac{(\i h X)^2}{2} + o(h^2) - 1}{h} \right)
|
|
&=& e^{\i t X} \i X \text{ almost surely.}
|
|
\end{IEEEeqnarray*}
|
|
|
|
For arbitrary $h \in \R$, we have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
|e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\
|
|
&\overset{\yaref{charfprop:c1}}{\le}& \left|\frac{1}{h} \i h X\right| = |X|.
|
|
\end{IEEEeqnarray*}
|
|
Thus the dominated convergence theorem can be applied and we obtain
|
|
\[
|
|
\lim_{h \to 0} \frac{\phi_X(t + h) - \phi_X(t)}{h} = \lim_{h \to 0} \bE\left[ e^{\i t X} \left( \frac{e^{\i h X}-1}{h} \right) \right]
|
|
= \bE[e^{\i t X} \i X].
|
|
\]
|
|
|
|
It follows that $\phi_X$ is differentiable and $\phi_X(t) = \bE[e^{\i t X} \i X]$.
|
|
For $t = 0$ we get $\phi'_X(0) = \i \bE[X]$, i.e.~
|
|
-$\i \phi'_X(0) = \bE[X]$.
|
|
|
|
Adapting the proof of (ii) gives that
|
|
$\phi'_X(t)$ is continuous.
|
|
|
|
Adapting the proof of (iii) gives
|
|
the statement for arbitrary $n \in \N$.
|
|
|
|
\item Easy exercise.
|
|
\end{enumerate}
|
|
\end{refproof}
|
|
|
|
\begin{lemma}\label{lec12_2} % Lemma 2
|
|
For $X \sim \cN(0,1)$, we have $\phi_X(t) = e^{-\frac{t^2}{2}}$
|
|
for all $t \in \R$.
|
|
\end{lemma}
|
|
\begin{refproof}{lec12_2}
|
|
We have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_X(t) &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\
|
|
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\
|
|
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\
|
|
\end{IEEEeqnarray*}
|
|
since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$
|
|
is even, their product is odd, wich gives that the integral is $0$.
|
|
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi'_X(t) &=& \bE[\i X e^{\i t X}] \\
|
|
&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
|
|
&=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
|
|
&=& \frac{1}{\sqrt{2 \pi}} \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\
|
|
&=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi}}(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\
|
|
&=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi} e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0}
|
|
- \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} \dif x\\
|
|
&=& -t \phi_X(t)
|
|
\end{IEEEeqnarray*}
|
|
Thus, for all $t \in \R$
|
|
\[
|
|
(\log(\phi_X(t)))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
|
|
\]
|
|
Hence there exists $c \in \R$, such that
|
|
\[
|
|
\log(\phi_X(t)) = -\frac{t^2}{2} + c.
|
|
\]
|
|
Since $\phi_X(0) = 1$, we obtain $c = 0$.
|
|
Thus
|
|
\[
|
|
\phi_X(t) = e^{-\frac{t^2}{2}}.
|
|
\]
|
|
\end{refproof}
|
|
|
|
|
|
Now, we can finally prove the \yaref{clt}:
|
|
\begin{refproof}{clt}
|
|
Let $X_1,X_2,\ldots$ be i.i.d.~random variables
|
|
with $\bE[X_1] = \mu_1$, $\Var(X_1) = \sigma^2$.
|
|
|
|
Let
|
|
\[
|
|
Y_i \coloneqq \frac{X_i - \mu}{\sigma}
|
|
\]
|
|
i.e.~we normalize to $\bE[Y_1] = 0$ and $\Var(Y_1) = 1$.
|
|
We need to show that
|
|
\[
|
|
V_n \coloneqq \frac{S_n - n \mu}{ \sigma \sqrt{n}} = \frac{Y_1+ \ldots + Y_n}{\sqrt{n}} \xrightarrow{\omega, n\to \infty} \cN(0,1) % TODO
|
|
\]
|
|
Let $t \in \R$.
|
|
Then
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_{V_n}(t) &=& \bE[e^{\i t Y_n}]\\
|
|
&=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n}} \right)}] \\
|
|
&=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot \ldots \cdot \bE\left[e^{\i t \frac{Y_n}{\sqrt{n}}}\right]\\
|
|
&=& \left( \phi\left(\frac{t}{\sqrt{n}}\right) \right)^n.
|
|
\end{IEEEeqnarray*}
|
|
where $\phi(t) \coloneqq \phi_{Y_1}(t)$.
|
|
|
|
We have
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{ as $s \to 0$}\\
|
|
&=& 1 - \underbrace{\i \bE[Y_1] s}_{=0}
|
|
- \bE[Y_1^2] \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
|
|
\\
|
|
&=& 1 - \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
|
|
\end{IEEEeqnarray*}
|
|
|
|
Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain
|
|
\[
|
|
\phi\left(\frac{t}{ \sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$}
|
|
\]
|
|
|
|
|
|
\[
|
|
\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n}} \right) \right)^n
|
|
= \left(1 - \frac{t^2}{2 n} + o\left( \frac{t^2}{n} \right)\right)^n
|
|
\xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
|
|
\]
|
|
where we have used the following:
|
|
|
|
\begin{claim}
|
|
For a sequence $a_n, n\in \N$ with $\lim_{n \to \infty} n a_n = \lambda$,
|
|
it holds that $\lim_{n \to \infty} (1 + a_n)^n = e^{\lambda}$.
|
|
\end{claim}
|
|
|
|
|
|
We have shown that
|
|
\[
|
|
\phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_{\cN(0,1)}(t).
|
|
\]
|
|
Using \yaref{levycontinuity}, we obtain the \yaref{clt}.
|
|
\end{refproof}
|
|
|
|
\begin{remark}
|
|
If $X: \Omega \to \R^d$ with distribution $\nu$,
|
|
we define
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\phi_X: \R^d &\longrightarrow & \C \\
|
|
t &\longmapsto & \bE[e^{\i \langle t , X\rangle}]
|
|
\end{IEEEeqnarray*}
|
|
where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
|
|
\end{remark}
|
|
Exercise: Find out, which properties also hold for $d > 1$.
|
|
\todo{TODO}
|