d for convergence in distribution
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@ -241,7 +241,7 @@ Unfortunately, we won't prove \autoref{thm:bochner} in this lecture.
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\begin{definition}
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We say that a series of random variables $X_n$
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\vocab[Convergence!in distribution]{converges in distribution}
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to $X$ (notation: $X_n \xrightarrow{\text{dist}} X$), iff
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to $X$ (notation: $X_n \xrightarrow{\text{d}} X$), iff
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$\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$
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and $\bP$ is the distribution of $X$.
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\end{definition}
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@ -256,7 +256,7 @@ for all $f \in C_b(\R)$.
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\end{example}
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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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$X_n \xrightarrow{\text{d}} X$ 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{proof}\footnote{This proof was not done in the lecture,
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@ -328,7 +328,7 @@ for all $f \in C_b(\R)$.
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% \end{proof}
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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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$X_n \xrightarrow{\text{d}} X$ 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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We will assume these two theorems for now and derive the central limit theorem.
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@ -59,7 +59,7 @@ and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
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Let $S_n \coloneqq \sum_{i=1}^n X_i$.
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Then
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\[
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\frac{S_n - n \nu}{\sigma \sqrt{n} } \xrightarrow{dist.} \cN(0,1),
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\frac{S_n - n \nu}{\sigma \sqrt{n} } \xrightarrow{\text{d}} \cN(0,1),
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\]
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i.e.~$\forall x \in \R:$
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\[
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@ -105,14 +105,14 @@ we need the following theorem, which we won't prove here:
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(Note that this argument does not work for $p = 1$,
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because $(L^\infty)^\ast \not\cong L^1$).
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Let $A \in \cF_m$ for some fixed $m$ and write
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$Y = \One_A$.
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Let $A \in \cF_m$ for some fixed $m$
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and choose $Y = \One_A$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\int_A X \dif \bP
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&=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
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&=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
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&\overset{\text{for }n_k \ge m}{=}& \lim_{k \to \infty} \bE[X_m \One_A].
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&\overset{\text{for }n_k \ge m}{=}& \bE[X_m \One_A].
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\end{IEEEeqnarray*}
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Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
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and by \autoref{ceismartingale},
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@ -37,7 +37,7 @@ from the lecture on stochastic.
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but has been added here for completeness;
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see \autoref{def:weakconvergence}.
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}
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($X_n \xrightarrow{\text{dist}} X$)
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($X_n \xrightarrow{\text{d}} X$)
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iff for every continuous, bounded $f: \R \to \R$
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\[
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\bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
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@ -58,7 +58,7 @@ from the lecture on stochastic.
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\begin{tikzpicture}
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\node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$};
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\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
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\node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{dist}} X$};
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\node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{d}} X$};
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%\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
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\node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$};
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\node at (3,3) (Lq) { $X_n \xrightarrow{L^q} X$};
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@ -121,7 +121,7 @@ from the lecture on stochastic.
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hence $X_n \xrightarrow{\bP} X$.
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\end{subproof}
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\begin{claim} %+
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$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
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$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{d}} X$.
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\end{claim}
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\begin{subproof}
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Let $F$ be the distribution function of $X$
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@ -187,14 +187,14 @@ from the lecture on stochastic.
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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{\text{dist}} X \notimplies X_n \xrightarrow{\bP} X$.
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$X_n \xrightarrow{\text{d}} X \notimplies X_n \xrightarrow{\bP} X$.
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\end{claim}
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\begin{subproof}
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Note that $X_n \xrightarrow{\text{dist}} X$ only makes a statement
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Note that $X_n \xrightarrow{\text{d}} X$ only makes a statement
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about the distributions of $X$ and $X_1,X_2,\ldots$
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For example, take some $p \in (0,1)$ and let
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$X$, $X_1, X_2,\ldots$ be i.i.d.~with $X \sim \Bin(1,p)$.
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Trivially $X_n \xrightarrow{\text{dist}} X$.
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Trivially $X_n \xrightarrow{\text{d}} X$.
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However
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\[
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\bP[|X_n - X| = 1]
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