lecture 13 part 1
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@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$.
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Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
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Then $\bP$ has a continuous probability density given by
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\[
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R(t) dt}.
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt.
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\]
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\end{theorem}
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@ -247,11 +247,13 @@ Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
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which is the distribution of $X \equiv 0$.
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But $F_n(0) \centernot\to F(0)$.
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\end{example}
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\begin{theorem}
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\begin{theorem} % Theorem 1
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\label{lec10_thm1}
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$X_n \xrightarrow{\text{dist}} X$ iff
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$F_n(t) \to F(t)$ for all continuity points $t$ of $F$.
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\end{theorem}
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\begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
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% Theorem 2
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$X_n \xrightarrow{\text{dist}} X$ iff
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$\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
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\end{theorem}
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@ -139,19 +139,19 @@ First, we need to prove some properties of characteristic functions.
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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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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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\]
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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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\]
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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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\]
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\end{refproof}
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@ -163,12 +163,12 @@ Now, we can finally prove the CLT:
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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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\]
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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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\]
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Let $t \in \R$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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@ -181,7 +181,7 @@ Now, we can finally prove the CLT:
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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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&=& 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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@ -189,13 +189,13 @@ Now, we can finally prove the CLT:
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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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\[
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\phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
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(1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
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\]
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\]
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where we have used the following:
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\begin{claim}
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@ -207,7 +207,7 @@ Now, we can finally prove the CLT:
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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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\]
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Using \autoref{levycontinuity}, we obtain \autoref{clt}.
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\end{refproof}
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@ -221,6 +221,3 @@ Now, we can finally prove the CLT:
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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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109
inputs/lecture_13.tex
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109
inputs/lecture_13.tex
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@ -0,0 +1,109 @@
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% Lecture 13 2023-05
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%The difficult part is to show \autoref{levycontinuity}.
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%This is the last lecture, where we will deal with independent random variables.
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We have seen, that
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if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
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$\sigma^2 = \Var(X_1)$,
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then $\frac{\sum_{i=1}^{n} (X_i - \mu)}{\sigma \sqrt{n} } \xrightarrow{(d)} \cN(0,1)$.
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\begin{question}
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What happens if $X_1, X_2,\ldots$ are independent, but not identically distributed? Do we still have a CLT?
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\end{question}
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\begin{theorem}[Lindeberg CLT]
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\label{lindebergclt}
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Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$.
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Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$
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and assume that $\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0$ for all $\epsilon > 0$
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(\vocab{Lindeberg condition}, ``The truncated variance is negligible compared to the variance.'').
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Then the CLT holds, i.e.~
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\[
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\frac{\sum_{i=1}^n (X_i - \mu_i)}{S_n} \xrightarrow{(d)} \cN(0,1).
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\]
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\end{theorem}
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\begin{theorem}[Lyapunov condition]
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\label{lyapunovclt}
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Let $X_1, X_2,\ldots$ be independent, $\mu_i = \bE[X_i] < \infty$,
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$\sigma_i^2 = \Var(X_i) < \infty$
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and $S_n \coloneqq \sqrt{\sum_{i=1}^n \sigma_i^2}$.
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Then, assume that, for some $\delta > 0$,
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\[
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\lim_{n \to \infty} \sum_{i=1}^{n} \bE[(X_i - \mu_i)^{2 + \delta}] = 0
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\]
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(\vocab{Lyapunov condition}).
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Then the CLT holds.
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\end{theorem}
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\begin{remark}
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The Lyapunov condition implies the Lindeberg condition.
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(Exercise).
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\end{remark}
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We will not prove the \autoref{linebergclt} or \autoref{lyapunovclt}
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in this lecture. However, they are quite important.
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We will now sketch the proof of \autoref{levycontinuity},
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details can be found in the notes.\todo{Complete this}
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A generalized version of \autoref{levycontinuity} is the following:
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\begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}]
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\label{genlevycontinuity}
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Suppose we have random variables $(X_n)_n$ such that
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$\bE[e^{\i t X_n}] \xrightarrow{n \to \infty} \phi(t)$ for all $t \in \R$
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for some function $\phi$ on $\R$.
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Then the following are equivalent:
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\begin{enumerate}[(a)]
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\item The distribution of $X_n$ is \vocab[Distribution!tight]{tight} (dt. ``straff''),
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i.e.~$\lim_{a \to \infty} \sup_{n \in \N} \bP[|X_n| > a] = 0$.
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\item $X_n \xrightarrow{(d)} X$ for some real-valued random variable $X$.
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\item $\phi$ is the characteristic function of $X$.
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\item $\phi$ is continuous on all of $\R$.
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\item $\phi$ is continuous at $0$.
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\end{enumerate}
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\end{theorem}
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\begin{example}
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Let $Z \sim \cN(0,1)$ and $X_n \coloneqq n Z$.
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We have $\phi_{X_n}(t) = \bE[[e^{\i t X_n}] = e^{-\frac{1}{2} t^2 n^2} \xrightarrow{n \to \infty} \One_{\{t = 0\} }$.
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$\One_{\{t = 0\}}$ is not continuous at $0$.
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By \autoref{genlevycontinuity}, $X_n$ can not converge to a real-valued
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random variable.
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Exercise: $X_n \xrightarrow{(d)} \overline{X}$,
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where $\bP[\overline{X} = \infty] = \frac{1}{2} = \bP[\overline{X} = -\infty]$.
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Similar examples are $\mu_n \coloneqq \delta_n$ and
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$\mu_n \coloneqq \frac{1}{2} \delta_n + \frac{1}{2} \delta_{-n}$.
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\end{example}
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\begin{example}
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Suppose that $X_1, X_2,\ldots$ are i.d.d.~with $\bE[X_1] = 0$.
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Let $\sigma^2 \coloneqq \Var(X_i)$.
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Then the distribution of $\frac{S_n}{\sigma \sqrt{n}}$ is tight:
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\begin{IEEEeqnarray*}{rCl}
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\bE\left[ \left( \frac{S_n}{\sqrt{n} }^2 \right)^2 \right] &=&
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\frac{1}{n} \bE[ (X_1+ \ldots + X_n)^2]\\
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&=& \sigma^2
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\end{IEEEeqnarray*}
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For $a > 0$, by Chebyshev's inequality, % TODO
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we have
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\[
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\bP\left[ \left| \frac{S_n}{\sqrt{n}} \right| > a \right] \leq \frac{\sigma^2}{a^2} \xrightarrow{a \to \infty} 0.
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\]
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verifying \autoref{genlevycontinuity}.
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\end{example}
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\begin{example}
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Suppose $C$ is a random variable which is Cauchy distributed, i.e.~$C$
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has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$.
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We know that $\bE[|C|] = \infty$.
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We have $\phi_C(t) = \bE[e^{\i t C}] = e^{-|t|}$.
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Suppose $C_1, C_2, \ldots, C_n$ are i.i.d.~Cauchy distributed
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and let $S_n \coloneqq C_1 + \ldots + C_n$.
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Exercise: $\phi_{S_n}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $S_n \sim C$.
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\end{example}
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