added definition of conditional probability

inputs/lecture_14.tex

@@ -78,6 +78,15 @@ We now want to generalize this to arbitrary random variables.
          is a constant random variable.
 \end{remark}
 
+\begin{definition}[Conditional probability]
+    Let $A \subseteq \Omega$ be an event and $\cG \subseteq  \cF$
+    a sub-$\sigma$-algebra.
+    We define the \vocab{conditional probability} of $A$ given $\cG$ by
+    \[
+    \bP[A | \cG] \coloneqq \bE[\One_A | \cG].
+    \]
+\end{definition}
+
 \paragraph{Plan}
 We will give two different proves of \autoref{conditionalexpectation}.
 The first one will use orthogonal projections.

inputs/lecture_21.tex

@@ -1,5 +1,4 @@
 \lecture{21}{2023-06-29}{}
-% TODO: replace bf
 
 This is the last lecture relevant for the exam.
 (Apart from lecture 22 which will be a repetion).
@@ -12,18 +11,18 @@ This is the last lecture relevant for the exam.
 \begin{notation}
     Let $E$ be a complete, separable metric space (e.g.~$E = \R$).
     Suppose that for all $x \in  E$ we have a probability measure
-    $\bfP(x, \dif y)$ on $E$.
+    $\mathbf{P}(x, \dif y)$ on $E$.
     % i.e. $\mu(A) \coloneqq \int_A \bP(x, \dif y)$ is a probability measure.
     Such a probability measure is a called
     a \vocab{transition probability measure}.
 \end{notation}
-\begin{examle}
+\begin{example}
     $E =\R$,
-    \[\bfP(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
+    \[\mathbf{P}(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
     is a transition probability measure.
-\end{examle}
+\end{example}
 \begin{example}[Simple random walk as a transition probability measure]
-    $E = \Z$, $\bfP(x, \dif y)$
+    $E = \Z$, $\mathbf{P}(x, \dif y)$
     assigns mass $\frac{1}{2}$ to $y = x+1$ and $y = x -1$.
 \end{example}
 
@@ -32,26 +31,26 @@ This is the last lecture relevant for the exam.
     $x \in E$
     define
     \[
-        (\bfP f)(x) \coloneqq  \int_E f(y) \bfP(x, \dif y).
+        (\mathbf{P} f)(x) \coloneqq  \int_E f(y) \mathbf{P}(x, \dif y).
     \]
-    This $\bfP$ is called a \vocab{transition operator}.
+    This $\mathbf{P}$ is called a \vocab{transition operator}.
 \end{definition}
 \begin{fact}
-    If $f \ge 0$, then $(\bfP f)(\cdot ) \ge 0$.
+    If $f \ge 0$, then $(\mathbf{P} f)(\cdot ) \ge 0$.
 
-    If $f \equiv 1$, we have $(\bfP f) \equiv 1$.
+    If $f \equiv 1$, we have $(\mathbf{P} f) \equiv 1$.
 \end{fact}
 
 \begin{notation}
-    Let $\bfI$ denote the \vocab{identity operator},
+    Let $\mathbf{I}$ denote the \vocab{identity operator},
     i.e.
     \[
-        (\bfI f)(x) = f(x)
+        (\mathbf{I} f)(x) = f(x)
     \]
     for all $f$.
-    Then for a transition operator $\bfP$ we write
+    Then for a transition operator $\mathbf{P}$ we write
     \[
-    \bfL \coloneqq  \bfI - \bfP.
+    \mathbf{L} \coloneqq  \mathbf{I} - \mathbf{P}.
     \]
 \end{notation}
 
@@ -71,16 +70,16 @@ is the unique solution to this problem.
     Let $(\Omega, \cF, \{\cF_n\}_n, \bP_x)$
     be a filtered probability space, where for every $x \in \R$,
     $\bP_x$ is a probability measure.
-    Let $\bE_x$ denote expectation with respect to $\bfP(x, \cdot )$.
+    Let $\bE_x$ denote expectation with respect to $\mathbf{P}(x, \cdot )$.
     Then $(X_n)_{n \ge 0}$ is a \vocab{Markov chain} starting at $x \in \R$
     with \vocab[Markov chain!Transition probability]{transition probability}
-    $\bfP(x, \cdot )$ if
+    $\mathbf{P}(x, \cdot )$ if
     \begin{enumerate}[(i)]
         \item $\bP_x[X_0 = x] = 1$,
         \item for all bounded, measurable $f: \R \to  \R$,
             \[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
                 \bE_{x}[f(X_{n+1}) | X_n] = %
-            \int f(y) \bfP(X_n, \dif y).\]
+            \int f(y) \mathbf{P}(X_n, \dif y).\]
     \end{enumerate}
     (Recall $\cF_n = \sigma(X_1,\ldots, X_n)$.)
 \end{definition}
@@ -92,12 +91,6 @@ is the unique solution to this problem.
     \]
 \end{example}
 
-\begin{definition}[Conditional probability]
-    \[
-    \bP[A | \cG] \coloneqq \bE[\One_A | \cG].
-    \]
-\end{definition}
-
 \begin{example}
     Let $\xi_i$ be i.i.d.~with$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$
     and define $X_n \coloneqq \sum_{i=1}^{n} \xi_i$.
@@ -105,14 +98,8 @@ is the unique solution to this problem.
     Intuitively, conditioned on $X_n$, $X_{n+1}$ should
     be independent of $\sigma(X_1,\ldots, X_{n-1})$.
 
-    For a set $B$, we have
-    \[
-    \bP_0[X_{n+1} \in B| \sigma(X_1,\ldots, X_n)]
-    = \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_1,\ldots, X_n)]
-    = \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_n)].
-    \]
-
     \begin{claim}
+        For a set $B$, we have
         $\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
     \end{claim}
     \begin{subproof}
@@ -120,16 +107,7 @@ is the unique solution to this problem.
     \end{subproof}
 \end{example}
 
-
+{   \large\color{red}
-%TODO
-
-
-
-
-
-
-
-{   \huge\color{red}
     New information after this point is not relevant for the exam.
 }
 Stopping times and optional stopping are very relevant for the exam,