d for convergence in distribution

inputs/lecture_10.tex

@@ -241,7 +241,7 @@ Unfortunately, we won't prove \autoref{thm:bochner} in this lecture.
 \begin{definition}
     We say that a series of random variables $X_n$
      \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 $X_n$
      and  $\bP$ is the distribution of $X$.
 \end{definition}
@@ -256,7 +256,7 @@ for all $f \in  C_b(\R)$.
 \end{example}
 \begin{theorem} % Theorem 1
     \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 $F$.
 \end{theorem}
 % \begin{proof}\footnote{This proof was not done in the lecture,
@@ -328,7 +328,7 @@ for all $f \in  C_b(\R)$.
 % \end{proof}
 \begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
     % 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 $t \in \R$.
 \end{theorem}
 We will assume these two theorems for now and derive the central limit theorem.

inputs/lecture_11.tex

@@ -59,7 +59,7 @@ and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
     Let $S_n \coloneqq  \sum_{i=1}^n X_i$.
     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.~$\forall  x \in \R:$
    \[

inputs/lecture_20.tex

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

inputs/prerequisites.tex

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