lecture 12
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@ -65,6 +65,7 @@ and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
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\lim_{n \to \infty} \bP\left[\frac{S_n - n \mu}{\sigma \sqrt{n} } \le x\right] = \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2 \pi} } e^{\frac{-t^2}{2}}dt.
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\lim_{n \to \infty} \bP\left[\frac{S_n - n \mu}{\sigma \sqrt{n} } \le x\right] = \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2 \pi} } e^{\frac{-t^2}{2}}dt.
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\]
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\]
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\end{theorem}
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\end{theorem}
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We will abbreviate the central limit theorem by \vocab{CLT}.
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There exists a special case of this theorem, which was proved earlier:
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There exists a special case of this theorem, which was proved earlier:
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\begin{theorem}[de-Moivre (1730, $p = 0.5$), Laplace (1812, general $p$ )]
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\begin{theorem}[de-Moivre (1730, $p = 0.5$), Laplace (1812, general $p$ )]
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226
inputs/lecture_12.tex
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226
inputs/lecture_12.tex
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% Lecture 12 2023-05-16
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We now want to prove \autoref{clt}.
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The plan is to do the following:
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\begin{enumerate}[1.]
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\item Identify the characteristic function of a standard normal
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\item Show that the characteristic functions of the $V_n$ converge pointwise
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to that of $\cN$.
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\item Apply \autoref{levycontinuity}
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\end{enumerate}
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First, we need to prove some properties of characteristic functions.
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\begin{lemma}
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\label{charfprops}
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For every real random variable $X$, we have
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\begin{enumerate}[(i)]
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\item $\phi_X(0) = 1$ and $|\phi_X(t)| \le 1$ for all $t \in \R$.
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\item $\phi_X$ is uniformly continuous.
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\item If $\bE[|X|^n] < \infty$ for any $n \in \N$, then $\phi_X$ i
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$n$-times continuously differentiable
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and $\bE[X^n] = (-\i)^n \phi_X^{(n)}(0)$.
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\item For independent random variables $X$ and $Y$, we have
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\[
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\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t).
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\]
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\end{enumerate}
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\end{lemma}
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\begin{refproof}{charfprops}
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\begin{enumerate}[(i)]
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\item $\phi_X(0) = \bE[e^{\i 0 X}] = \bE[1] = 1$.
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For $t \in \R$, we have $|\phi_X(t)| = |\bE[e^{\i t X}]| \overset{\text{Jensen}}{\le} \bE|e^{\i t X}|] = 1$.
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\item Let $t, h \in \R$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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|\phi_X(t+h) - \phi_X(t)| &=& |\bE[e^{\i (t+h) X} - e^{\i t X}]|\\
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&=& |\bE[e^{\i t X} (e^{\i h X} - 1)]|\\
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&\overset{\text{Jensen}}{\le}&
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\bE[|e^{\i t X}| \cdot |e^{\i h X} -1|]\\
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&=& \bE[|e^{\i h X} - 1|]\\
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\end{IEEEeqnarray*}
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Hence $\sup_{t \in \R} |\phi_X(t + h) - \phi_X(t) | \le g(h)$.
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We show that $\lim_{h \to 0} g(h) = 0$.
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For all $\omega \in \Omega$, we realize
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\[
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\lim_{h \to 0} |e^{\i h X(\omega)}- 1| = 0.
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\]
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Thus $|e^{\i h X} - 1| \xrightarrow{h \to 0} 0$ almost surely.
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Since also for all $h \in \R$ we have $|e^{\i h X} - 1| \le 2$,
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it follow that $|e^{\i h X} - 1|$ is dominated for all $h \in \R$.
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Thus, we can apply the dominated convergence theorem % TODO REF
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and obtain
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\[
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\lim_{h \to 0} g(h) = \lim_{h \to 0} \bE[|e^{\i h X} - 1|]
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= \bE[\lim_{h \to 0} |e^{\i h X} - 1|] = 0.
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\]
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It follows that
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\[
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\lim_{n \to 0} \sup_{t \in \R} | \phi_X(t + h) - \phi_X(t)| = 0,
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\]
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which means that $\phi_X$ is uniformly continuous.
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\item
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\begin{claim}
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\label{charfprop:c1}
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For $y \in \R$, we have $|e^{\i y} - 1| \le |y|$.
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\end{claim}
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\begin{subproof}
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For $y \ge 0$, we have
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\begin{IEEEeqnarray*}{rCl}
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|e^{\i y} - 1| &=& |\int_0^y \cos(s) \d s + \i \int_0^y \sin(s) \d s|\\
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&=& |\int_0^y e^{\i s} \d s|\\
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&\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y.
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\end{IEEEeqnarray*}
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For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$
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and we can apply the above to $-y$.
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\end{subproof}
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First, we look at $n = 1$. Then $\bE[|X|] < \infty$.
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Consider
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\[
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\frac{\phi_X(t + h) - \phi_X(t)}{h} = \bE\left[e^{\i t X} \frac{e^{\i h X} - 1}{h}\right].
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\]
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We have $e^z = \sum_{n = 0}^\infty \frac{z^k}{n!}$.
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\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)
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&=& e^{\i t X} \i X \text{ almost surely.}
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\end{IEEEeqnarray*}
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For arbitrary $h \in \R$, we have
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\begin{IEEEeqnarray*}{rCl}
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|e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\
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&\overset{\text{\autoref{charfprop:c1}}}{\le }& \left|\frac{1}{h} \i h X\right| = |X|.
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\end{IEEEeqnarray*}
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Thus the dominated convergence theorem can be applied and we obtain
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\[
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\lim_{h \to 0} \frac{\phi_X(t + h) - \phi_X(t)}{h} = \lim_{h \to 0} \bE\left[ e^{\i t X} \left( \frac{e^{\i h X}-1}{h} \right) \right]
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= \bE[e^{\i t X} \i X].
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\]
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It follows that $\phi_X$ is differentiable and $\phi_X(t) = \bE[e^{\i t X} \i X]$.
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For $t = 0$ we get $\phi'_X(0) = \i \bE[X]$, i.e.~
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-$\i \phi'_X(0) = \bE[X]$.
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Adapting the proof of (ii) gives that
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$\phi'_X(t)$ is continuous.
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Adapting the proof of (iii) gives
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the statement for arbitrary $n \in \N$.
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\item Easy exercise.
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\end{enumerate}
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\end{refproof}
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\begin{lemma}\label{lec12_2} % Lemma 2
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For $X \sim \cN(0,1)$, we have $\phi_X(t) = e^{-\frac{t^2}{2}}$
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for all $t \in \R$.
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\end{lemma}
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\begin{refproof}{lec12_2}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \d x\\
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&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \d x\\
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&=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \d x,\\
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\end{IEEEeqnarray*}
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since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$
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is even, their product is odd, wich gives that the integral is $0$.
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\begin{IEEEeqnarray*}{rCl}
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\phi'_X(t) &=& \bE[\i X e^{\i t X}] \\
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&=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \d x\\
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&=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \d x\\
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&=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}} \d x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \d x\right)\\
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&=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \d x\\
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&=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi} e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0}
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- \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \d x\\
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&=& -t \phi_X(t)
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\end{IEEEeqnarray*}
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Thus, for all $t \in \R$
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\[
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(\log(\phi_X(t))' = \frac{\phi'_X(t)}{\phi_X(t)} = -t.
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\]
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Hence there exists $c \in \R$, such that
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\[
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\log(\phi_X(t)) = -\frac{t^2}{2} + c.
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\]
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Since $\phi_X(0) = 1$, we obtain $c = 0$.
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Thus
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\[
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\phi_X(t) = e^{-\frac{t^2}{2}}.
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\]
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\end{refproof}
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Now, we can finally prove the CLT:
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\begin{refproof}{clt}
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Let $X_1,X_2,\ldots$ be i.i.d.~random variables
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with $\bE[X_1] = \mu_1$, $\Var(X_1) = \sigma^2$.
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Let
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\[
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Y_i \coloneqq \frac{X_i - \mu}{\sigma}
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\]
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i.e.~we normalize to $\bE[Y_1] = 0$ and $\Var(Y_1) = 1$.
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We need to show that
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\[
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V_n \coloneqq \frac{S_n - n \mu}{ \sigma \sqrt{n}} = \frac{Y_1+ \ldots + Y_n}{\sqrt{n}} \xrightarrow{\omega, n\to \infty} \cN(0,1) % TODO
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\]
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Let $t \in \R$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\phi_{V_n}(t) = \bE[e^{\i t Y_n}] = \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
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&=& \bE[e^{\i t \frac{Y_1}{\sqrt{n}}}] \cdot \ldots \cdot \bE[e^{\i t \frac{Y_n}{\sqrt{n} }}]\\
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&=& \left( \phi(\frac{t}{\sqrt{n} } \right)^n.
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\end{IEEEeqnarray*}
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where $\phi(t) \coloneqq \phi_{Y_1}(t)$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{as $s \to 0$}\\
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&=& 1 - \underbrace{\i \bE[Y_1] s}_{=0}
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- \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\
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&=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $}
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\end{IEEEeqnarray*}
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Setting $s \coloneqq \frac{t}{\sqrt{n}}$ we obtain
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\[
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\phi\left(\frac{t}{ \sqrt{n} }\right) = 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \text{ as $n \to \infty$}
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\]
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\[
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\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
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(1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
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\]
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where we have used the following:
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\begin{claim}
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For a sequence $a_n, n\in \N$ with $\lim_{n \to \infty} n a_n = \lambda$,
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it holds that $\lim_{n \to \infty} (1 + a_n)^n = e^{\lambda}$.
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\end{claim}
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We have shown that
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\[
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\phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_N(t).
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\]
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Using \autoref{levycontinuity}, we obtain \autoref{clt}.
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\end{refproof}
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\begin{remark}
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If $X: \Omega \to \R^d$ with distribution $\nu$,
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we define
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\begin{IEEEeqnarray*}{rCl}
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\phi_X: \R^d &\longrightarrow & \C \\
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t &\longmapsto & \bE[e^{\i \langle t , X\rangle}]
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\end{IEEEeqnarray*}
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where $\langle t, X\rangle \coloneqq \sum_{i = 1}^d t_i X_i$.
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\end{remark}
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Exercise: Find out, which properties also hold for $d > 1$.
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@ -43,9 +43,8 @@ We'll get back to percolation later.
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\section{Characteristic functions, weak convergence and the central limit theorem}
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\section{Characteristic functions, weak convergence and the central limit theorem}
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Characteristic functions are also known as the \vocab{Fourier transform}.
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% Characteristic functions are also known as the \vocab{Fourier transform}.
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Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
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%Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
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We will abbreviate the central limit theorem by \vocab{CLT}.
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So far we have dealt with the average behaviour,
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So far we have dealt with the average behaviour,
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\[
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\[
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@ -35,6 +35,7 @@
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\input{inputs/lecture_9.tex}
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\input{inputs/lecture_9.tex}
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\input{inputs/lecture_10.tex}
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\input{inputs/lecture_10.tex}
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\input{inputs/lecture_11.tex}
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\input{inputs/lecture_11.tex}
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\input{inputs/lecture_12.tex}
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\cleardoublepage
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\cleardoublepage
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