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d for convergence in distribution

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Josia Pietsch 2023-07-15 02:00:04 +02:00
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6
inputs/lecture_10.tex
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@ -241,7 +241,7 @@ Unfortunately, we won't prove \autoref{thm:bochner} in this lecture.
\begin{definition} \begin{definition}
We say that a series of random variables $X_n$ We say that a series of random variables $X_n$
\vocab[Convergence!in distribution]{converges in distribution} \vocab[Convergence!in distribution]{converges in distribution}
to $X$ (notation: $X_n \xrightarrow{\text{dist}} X$), iff to $X$ (notation: $X_n \xrightarrow{\text{d}} X$), iff
$\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$ $\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$
and $\bP$ is the distribution of $X$. and $\bP$ is the distribution of $X$.
\end{definition} \end{definition}
@ -256,7 +256,7 @@ for all $f \in C_b(\R)$.
\end{example} \end{example}
\begin{theorem} % Theorem 1 \begin{theorem} % Theorem 1
\label{lec10_thm1} \label{lec10_thm1}
$X_n \xrightarrow{\text{dist}} X$ iff $X_n \xrightarrow{\text{d}} X$ iff
$F_n(t) \to F(t)$ for all continuity points $t$ of $F$. $F_n(t) \to F(t)$ for all continuity points $t$ of $F$.
\end{theorem} \end{theorem}
% \begin{proof}\footnote{This proof was not done in the lecture, % \begin{proof}\footnote{This proof was not done in the lecture,
@ -328,7 +328,7 @@ for all $f \in C_b(\R)$.
% \end{proof} % \end{proof}
\begin{theorem}[Levy's continuity theorem]\label{levycontinuity} \begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
% Theorem 2 % Theorem 2
$X_n \xrightarrow{\text{dist}} X$ iff $X_n \xrightarrow{\text{d}} X$ iff
$\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$. $\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
\end{theorem} \end{theorem}
We will assume these two theorems for now and derive the central limit theorem. We will assume these two theorems for now and derive the central limit theorem.

2
inputs/lecture_11.tex
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@ -59,7 +59,7 @@ and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
Let $S_n \coloneqq \sum_{i=1}^n X_i$. Let $S_n \coloneqq \sum_{i=1}^n X_i$.
Then Then
\[ \[
\frac{S_n - n \nu}{\sigma \sqrt{n} } \xrightarrow{dist.} \cN(0,1), \frac{S_n - n \nu}{\sigma \sqrt{n} } \xrightarrow{\text{d}} \cN(0,1),
\] \]
i.e.~$\forall x \in \R:$ i.e.~$\forall x \in \R:$
\[ \[

6
inputs/lecture_20.tex
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@ -105,14 +105,14 @@ we need the following theorem, which we won't prove here:
(Note that this argument does not work for $p = 1$, (Note that this argument does not work for $p = 1$,
because $(L^\infty)^\ast \not\cong L^1$). because $(L^\infty)^\ast \not\cong L^1$).
Let $A \in \cF_m$ for some fixed $m$ and write Let $A \in \cF_m$ for some fixed $m$
$Y = \One_A$. and choose $Y = \One_A$.
Then Then
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\int_A X \dif \bP \int_A X \dif \bP
&=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\ &=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
&=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\ &=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
&\overset{\text{for }n_k \ge m}{=}& \lim_{k \to \infty} \bE[X_m \One_A]. &\overset{\text{for }n_k \ge m}{=}& \bE[X_m \One_A].
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
and by \autoref{ceismartingale}, and by \autoref{ceismartingale},

12
inputs/prerequisites.tex
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@ -37,7 +37,7 @@ from the lecture on stochastic.
but has been added here for completeness; but has been added here for completeness;
see \autoref{def:weakconvergence}. see \autoref{def:weakconvergence}.
} }
($X_n \xrightarrow{\text{dist}} X$) ($X_n \xrightarrow{\text{d}} X$)
iff for every continuous, bounded $f: \R \to \R$ iff for every continuous, bounded $f: \R \to \R$
\[ \[
\bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)]. \bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
@ -58,7 +58,7 @@ from the lecture on stochastic.
\begin{tikzpicture} \begin{tikzpicture}
\node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$}; \node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$};
\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$}; \node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
\node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{dist}} X$}; \node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{d}} X$};
%\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$}; %\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
\node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$}; \node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$};
\node at (3,3) (Lq) { $X_n \xrightarrow{L^q} X$}; \node at (3,3) (Lq) { $X_n \xrightarrow{L^q} X$};
@ -121,7 +121,7 @@ from the lecture on stochastic.
hence $X_n \xrightarrow{\bP} X$. hence $X_n \xrightarrow{\bP} X$.
\end{subproof} \end{subproof}
\begin{claim} %+ \begin{claim} %+
$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$. $X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{d}} X$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Let $F$ be the distribution function of $X$ Let $F$ be the distribution function of $X$
@ -187,14 +187,14 @@ from the lecture on stochastic.
\end{subproof} \end{subproof}
\begin{claim} \begin{claim}
$X_n \xrightarrow{\text{dist}} X \notimplies X_n \xrightarrow{\bP} X$. $X_n \xrightarrow{\text{d}} X \notimplies X_n \xrightarrow{\bP} X$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Note that $X_n \xrightarrow{\text{dist}} X$ only makes a statement Note that $X_n \xrightarrow{\text{d}} X$ only makes a statement
about the distributions of $X$ and $X_1,X_2,\ldots$ about the distributions of $X$ and $X_1,X_2,\ldots$
For example, take some $p \in (0,1)$ and let For example, take some $p \in (0,1)$ and let
$X$, $X_1, X_2,\ldots$ be i.i.d.~with $X \sim \Bin(1,p)$. $X$, $X_1, X_2,\ldots$ be i.i.d.~with $X \sim \Bin(1,p)$.
Trivially $X_n \xrightarrow{\text{dist}} X$. Trivially $X_n \xrightarrow{\text{d}} X$.
However However
\[ \[
\bP[|X_n - X| = 1] \bP[|X_n - X| = 1]