some cleanup
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@ -46,7 +46,7 @@ We will not prove the \autoref{lindebergclt} 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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details can be found in the notes.\notes
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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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@ -120,7 +120,7 @@ A generalized version of \autoref{levycontinuity} is the following:
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We will prove \autoref{levycontinuity} assuming
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\autoref{lec10_thm1}.
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\autoref{lec10_thm1} will be shown in the notes.\todo{TODO}
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\autoref{lec10_thm1} will be shown in the notes.\notes
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We will need the following:
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\begin{lemma}
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\label{lec13_lem1}
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@ -239,7 +239,7 @@ We still need to show that $\mu_n \implies \mu$.
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Then $a_n \to a$.
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\end{fact}
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\begin{subproof}
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\todo{in the notes}
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\notes
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\end{subproof}
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Assume that $\mu_n$ does not converge to $\mu$.
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By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
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@ -87,7 +87,8 @@ We now want to generalize this to arbitrary random variables.
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\]
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\end{definition}
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\paragraph{Plan}
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\subsection{Existence of conditional probability}
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We will give two different proves of \autoref{conditionalexpectation}.
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The first one will use orthogonal projections.
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The second will use the Radon-Nikodym theorem.
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@ -108,7 +109,7 @@ and then do the harder proof.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\todo{Notes}
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\notes
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\end{proof}
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\begin{refproof}{conditionalexpectation}
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@ -162,7 +163,7 @@ and then do the harder proof.
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$0 \overset{\text{a.s.}}{\le} Z_n \uparrow$.
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\end{claim}
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\begin{subproof}
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\todo{Notes}
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\notes
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\end{subproof}
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Define $Z(\omega) \coloneqq \limsup_{n \to \infty} Z_n(\omega)$.
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@ -1,4 +1,5 @@
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\lecture{15}{2023-06-06}{}
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\subsection{Properties of conditional expectation}
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We want to derive some properties of conditional expectation.
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@ -12,16 +13,20 @@ We want to derive some properties of conditional expectation.
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\begin{proof}
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Apply (b) from the definition for $G = \Omega \in \cG$.
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\end{proof}
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\begin{theorem} % Thm 2
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\begin{theorem}
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\label{ceprop2}
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If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s..
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If $X$ is $\cG$-measurable, then $X \overset{\text{a.s.}}{=}\bE[X | \cG]$.
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\end{theorem}
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\begin{proof}
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Suppose $\bP[X \neq Y] > 0$.
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Without loss of generality $\bP[X > Y] > 0$.
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Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
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Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
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% TODO
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Then
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\[
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\int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
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\]
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contradicting property (b) from \autoref{conditionalexpectation}.
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\end{proof}
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\begin{example}
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@ -40,7 +45,7 @@ We want to derive some properties of conditional expectation.
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\]
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\end{theorem}
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\begin{proof}
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Trivial % TODO
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trivial\todo{add details}
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\end{proof}
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\begin{theorem}[Positivity]
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@ -52,7 +57,7 @@ We want to derive some properties of conditional expectation.
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\begin{proof}
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Let $W $ be a version of $\bE[X | \cG]$.
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Suppose $\bP[ W < 0] > 0$.
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Then $G \coloneqq \{W < -\frac{1}{n}\} \in \cG$
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Then \[G \coloneqq \{W < -\frac{1}{n}\} \in \cG.\]
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For some $n \in \N$, we have $\bP[G] > 0$.
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However it follows that
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\[
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@ -106,7 +111,7 @@ We want to derive some properties of conditional expectation.
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\]
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\end{theorem}
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\begin{proof}
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\todo{in the notes}
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\notes
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\end{proof}
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\begin{theorem}[Conditional dominated convergence theorem]
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\label{ceprop7}
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@ -118,14 +123,14 @@ We want to derive some properties of conditional expectation.
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\end{theorem}
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\begin{proof}
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\todo{in the notes}
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\notes
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\end{proof}
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Recall
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\begin{theorem}[Jensen's inequality]
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\begin{fact}[Jensen's inequality]
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If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$,
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then $\bE[c \circ X] \ge c(\bE[X])$.
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\end{theorem}
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then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
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\end{fact}
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For conditional expectation, we have
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\begin{theorem}[Conditional Jensen's inequality]
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@ -150,7 +155,7 @@ For conditional expectation, we have
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Hence
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\[
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\bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
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= a_n \bE[X | \cG] + b_n \text{a.s.}
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= a_n \bE[X | \cG] + b_n \text{ a.s.}
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\]
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for all $n$.
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Using that a countable union of sets o f measure zero has measure zero,
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@ -162,14 +167,14 @@ For conditional expectation, we have
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\end{refproof}
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Recall
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\begin{theorem}[Hölder's inequality]
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\begin{fact}[Hölder's inequality]
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Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
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Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
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Then
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\[
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\bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}.
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\]
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\end{theorem}
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\end{fact}
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\begin{theorem}[Conditional Hölder's inequality]
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\label{ceprop9}
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@ -193,7 +198,7 @@ Recall
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Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
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Then
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\[
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\bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH].
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\bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
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\]
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\end{theorem}
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\begin{proof}
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@ -219,7 +224,8 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
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\begin{definition}
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Let $\cG$ and $\cH$ be $\sigma$-algebras.
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We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
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\todo{TODO}
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if $\bP(G \cap H) = \bP(G) \bP(H)$
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for all events $G \in \cG$, $H \in \cH$.
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\end{definition}
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\begin{theorem}[Role of independence]
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@ -251,7 +257,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
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For $\bE[S_{n+1} | \cF_n]$ we obtain
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\begin{IEEEeqnarray*}{rCl}
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\bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}&
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\bE[S_{n+1} | \cF_n] &\overset{\text{\autoref{celinearity}}}{=}&
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\bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
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&\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
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&\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\
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@ -68,7 +68,8 @@ The Radon Nikodym theorem is the converse of that:
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Such a $Z$ is called the \vocab{Radon-Nikodym derivative}.
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\end{theorem}
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\begin{definition}
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Whenever the property $\forall A \in \cF, \mu(A)= 0 \implies \nu(A) = 0$,
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Whenever the property $\forall A \in \cF, \mu(A) = 0 \implies \nu(A) = 0$
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holds for two measures $\mu$ and $\nu$,
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we say that $\nu$ is \vocab{absolutely continuous}
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w.r.t.~$\mu$.
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This is written as $\nu \ll \mu$.
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@ -86,16 +87,19 @@ of conditional expectation:
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\begin{refproof}{radonnikodym}
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We will only sketch the proof. A full proof can be found in the notes.
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We will only sketch the proof.
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A full proof can be found in the official notes.
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\paragraph{Step 1: Uniqueness}
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See notes.
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\paragraph{Step 1: Uniqueness} \notes
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\paragraph{Step 2: Reduction to the finite measure case}
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See notes.
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\notes
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\paragraph{Step 3: Getting hold of $Z$}
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Assume now that $\mu$ and $\nu$ are two finite measures.
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Let
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\[\cC \coloneqq \{f: \Omega \to [0,\infty] | \forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\}.\]
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\[
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\cC \coloneqq \left\{f: \Omega \to [0,\infty] \middle| %
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\forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\right\}.
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\]
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We have $\cC \neq \emptyset$ since $0 \in \cC$.
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The goal is to find a maximal function $Z$ in $\cC$.
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@ -145,7 +149,7 @@ of conditional expectation:
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but it need not satisfy (ii).
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The tricky part is to make $A$ smaller such that it also
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satisfies (ii).\todo{Copy from notes}
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satisfies (ii).\notes
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\end{subproof}
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\end{refproof}
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@ -153,6 +157,8 @@ of conditional expectation:
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\section{Martingales}
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\subsection{Definition}
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We have already worked with martingales, but we will define them
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rigorously now.
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@ -168,15 +174,16 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
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Let $(\cF_n)$ be a filtration and
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$X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$.
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Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale}
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if
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if the following hold:
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\begin{itemize}
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\item $X_n$ is $\cF_n$-measurable for all $n$
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($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
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\item $X_n$ is $\cF_n$-measurable for all $n$.
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($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
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\item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$
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for all $n$.
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\end{itemize}
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$(X_n)$ is called a \vocab{sub-martingale},
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$(X_n)_n$ is called a \vocab{sub-martingale},
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if it is adapted to $\cF_n$ but
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\[
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\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
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@ -185,12 +192,12 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
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if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
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\end{definition}
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\begin{corollary}
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Suppose that $f: \R \to \R$ is a convex function such that $f(xn) \in L^1(\bP)$.
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Suppose that $(X_n)_n$ is a martingal\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
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Suppose that $f: \R \to \R$ is a convex function such that $f(X_n) \in L^1(\bP)$.
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Suppose that $(X_n)_n$ is a martingale\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
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Then $(f(X_n))_n$ is a sub-martingale.
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\end{corollary}
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\begin{proof}
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Apply \autoref{cejensensinequality}.
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Apply \autoref{cjensen}.
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\end{proof}
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\begin{corollary}
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@ -1,5 +1,8 @@
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\lecture{17}{2023-06-15}{}
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\subsection{Doob's martingale convergence theorem}
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\begin{definition}[Stochastic process]
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A \vocab{stochastic process} is a collection of random
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variables $(X_t)_{t \in T}$ for some index set $T$.
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@ -64,6 +67,7 @@ such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le j \le
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Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases.
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It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \infty} U_N([a,b])$ exists pointwise.
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\begin{lemma} % Lemma 1
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\label{lec17l1}
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\[
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\{\omega | \liminf_{N \to \infty} Z_N(\omega) < a < b <
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\limsup_{N \to \infty} Z_N(\omega)\} \subseteq
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\end{lemma}
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\begin{lemma} % 2
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\label{lec17l2}
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Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
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Then $Y_N \ge (b -a) U_N([a,b]) - (X_N - a)^{-}$.
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Then \[Y_N \ge (b -a) U_N([a,b]) - (X_N - a)^{-}.\]
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\end{lemma}
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\begin{proof}
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Every upcrossing of $[a,b]$ increases the value of $Y$ by $(b-a)$,
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\end{proof}
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\begin{lemma} %3
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\label{lec17l3}
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Suppose $(X_n)_n$ is a supermartingale.
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Then in the above setup, $(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-]$.
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Then in the above setup
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\[(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].\]
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\end{lemma}
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\begin{proof}
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Obvious from lemma 2 % TODO REF
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This is obvious from \autoref{lec17l2}
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and the supermartingale property.
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\end{proof}
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\begin{corollary}
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Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$ },
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i.e.~$\sup_n \bE[|X_n| ] < \infty$.
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Let $(X_n)_n$ be a
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\vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$},
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i.e.~$\sup_n \bE[|X_n|] < \infty$.
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Then $(b-a) \bE(U_\infty) \le |a| + \sup_n \bE(|X_n|)$.
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In particular, $\bP[U_\infty = \infty] = 0$.
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\end{corollary}
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\begin{proof}
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By lemma 3 % TODO REF
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By \autoref{lec17l3}
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we have that
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\[(b-a) \bE[U_N([a,b])] \le \bE[ | X_N| ] + |a| \le \sup_n \bE[|X_n|] + |a|.\]
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Since $U_N(\cdot) \ge 0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$,
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@ -1,10 +1,9 @@
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\lecture{18}{2023-06-20}{}
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Recall our key lemma for supermartingales from last time:
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Recall our key lemma \ref{lec17l3} for supermartingales from last time:
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\[
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(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].
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\]
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% TODO Ref
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What happens for submartingales?
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If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
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@ -14,11 +13,14 @@ Hence the same holds for submartingales, i.e.
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a.s.~to a finite limit, which is a.s.~finite.
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\end{lemma}
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\subsection{Doob's $L^p$ inequality}
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\begin{question}
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What about $L^p$ convergence of martingales?
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\end{question}
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\begin{example}[Branching process]
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\begin{example}[\vocab{Branching process}]
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Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
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$\bP[Z_n = 1] = p \in (0,1)$.
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@ -26,11 +28,11 @@ Hence the same holds for submartingales, i.e.
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Define $X_{n+1} \coloneqq u^{Z_{n+1}} X_{n}$.
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\paragraph{Exercise}
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Given $u \ge 0$, find $p = p(u)$
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such that $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
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% TODO
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\begin{exercise}
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Given $u \ge 0$, find $p = p(u)$
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such that $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
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\end{exercise}
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\todo{TODO}
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By \autoref{doobmartingaleconvergence}, there is an
|
||||
|
@ -51,10 +53,11 @@ Hence the same holds for submartingales, i.e.
|
|||
$N_0(\epsilon)$ (possibly random)
|
||||
such that for all $n > N_0(\epsilon)$
|
||||
\[
|
||||
\left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
|
||||
\left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) %
|
||||
\implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
|
||||
\]
|
||||
Hence it can not converge in $L^1$.
|
||||
% TODO Confusion
|
||||
Thus it can not converge in $L^1$.
|
||||
% TODO Make this less confusing
|
||||
|
||||
\end{example}
|
||||
|
||||
|
@ -75,16 +78,21 @@ consider $L^2$.
|
|||
Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s.,
|
||||
by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le n$.
|
||||
Play with conditional expectation.
|
||||
% TODO Exercise
|
||||
\todo{Exercise}
|
||||
\end{proof}
|
||||
|
||||
\begin{fact}[Parallelogram identity]
|
||||
% TODO
|
||||
\begin{fact}[\vocab{Parallelogram identity}]
|
||||
Let $X, Y \in L^2$.
|
||||
Then
|
||||
\[
|
||||
2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
|
||||
\]
|
||||
\end{fact}
|
||||
|
||||
\begin{theorem}
|
||||
\begin{theorem}\label{martingaleconvergencel2}
|
||||
Suppose that $(X_n)_n$ is a martingale bounded in
|
||||
$L^2$, i.e.~$\sup_n \bE[X_n^2] < \infty$.
|
||||
$L^2$,\\
|
||||
i.e.~$\sup_n \bE[X_n^2] < \infty$.
|
||||
Then there is a random variable $X_\infty$ such that
|
||||
\[
|
||||
X_n \xrightarrow{L^2} X_\infty.
|
||||
|
@ -100,7 +108,7 @@ consider $L^2$.
|
|||
\bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
|
||||
\]
|
||||
by \autoref{martingaleincrementsorthogonal}
|
||||
(this is known as the \vocab{parallelogram identity}). % TODO Move
|
||||
% (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
|
||||
In particular,
|
||||
\[
|
||||
\sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
|
||||
|
@ -141,7 +149,7 @@ First, we need a very important inequality:
|
|||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
We first need
|
||||
In order to prove \autoref{dooblp}, we first need
|
||||
\begin{lemma}
|
||||
\label{dooplplemma}
|
||||
Let $p > 1$ and $X,Y$ non-negative random variable
|
||||
|
@ -176,7 +184,7 @@ We first need
|
|||
|
||||
Suppose now $Y \not\in L^p$.
|
||||
Then look at $Y_M = Y \wedge M$.
|
||||
Apply the case of $Y \in L^p$ and use the monotone convergence theorem.
|
||||
Apply the above to $Y_M \in L^p$ and use the monotone convergence theorem.
|
||||
\end{proof}
|
||||
|
||||
\begin{refproof}{dooblp}
|
||||
|
@ -185,9 +193,13 @@ We first need
|
|||
\[
|
||||
E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
|
||||
\]
|
||||
Then $\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP (\ast\ast)$.
|
||||
Then
|
||||
\begin{equation}
|
||||
\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
|
||||
\label{lec18eq2star}
|
||||
\end{equation}
|
||||
Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
|
||||
(by \autoref{jensen}).
|
||||
(by \autoref{cjensen}).
|
||||
Hence
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]
|
||||
|
@ -196,14 +208,14 @@ We first need
|
|||
\end{IEEEeqnarray*}
|
||||
By the law of total expectation, \autoref{totalexpectation},
|
||||
it follows that
|
||||
\[
|
||||
\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0 (\ast\ast\ast).
|
||||
\]
|
||||
\begin{equation}
|
||||
\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
|
||||
\end{equation}
|
||||
|
||||
Now
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
|
||||
&\overset{(\ast\ast) (\ast\ast\ast)}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
|
||||
&\overset{\eqref{lec18eq2star}, \eqref{lec18eq3star}}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
|
||||
&=& \frac{1}{\ell} \int_E |X_n| \dif \bP
|
||||
\end{IEEEeqnarray*}
|
||||
|
||||
|
|
|
@ -54,7 +54,7 @@ However, some subsets can be easily described, e.g.
|
|||
where we have applied Markov's inequality. % TODO REF
|
||||
|
||||
Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
|
||||
we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen. % TODO REF
|
||||
we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{cjensen}).
|
||||
Hence, choose $k$ large enough to make the relevant
|
||||
term less than $\epsilon$.
|
||||
\end{proof}
|
||||
|
@ -94,7 +94,7 @@ However, some subsets can be easily described, e.g.
|
|||
&\ge & \epsilon
|
||||
\end{IEEEeqnarray*}
|
||||
where the assumption that $X$ is in $L^1$ was used to apply
|
||||
the reverse of Fatou's lemma.
|
||||
the reverse of Fatou's lemma. % TODO reverse fatou
|
||||
|
||||
This yields a contradiction since $\bP(F) = 0$.
|
||||
\item We want to apply part (a) to $F = \{ |X| > k\}$.
|
||||
|
@ -120,7 +120,10 @@ However, some subsets can be easily described, e.g.
|
|||
\begin{proof}
|
||||
Fix $\epsilon > 0$.
|
||||
Choose $\delta > 0$ such that
|
||||
\[\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon. (\ast)\]
|
||||
\begin{equation}
|
||||
\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon.
|
||||
\label{lec19eqstar}
|
||||
\end{equation}
|
||||
Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
|
||||
Then, by \autoref{condjensen}, $|Y| \le \bE[ |X| | \cG]$.
|
||||
Hence $\bE[|Y|] \le \bE[|X|]$.
|
||||
|
@ -132,7 +135,7 @@ However, some subsets can be easily described, e.g.
|
|||
\begin{IEEEeqnarray*}{rCl}
|
||||
\bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
|
||||
\end{IEEEeqnarray*}
|
||||
by $(\ast)$, since $\bP[|Y| > k] < \delta$.
|
||||
by \eqref{lec19eqstar}, since $\bP[|Y| > k] < \delta$.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}
|
||||
|
@ -155,7 +158,7 @@ However, some subsets can be easily described, e.g.
|
|||
\end{cases}
|
||||
\end{IEEEeqnarray*}
|
||||
|
||||
$\phi$ is $1$-Lipshitz. % TODO
|
||||
$\phi$ is $1$-Lipschitz. % TODO
|
||||
|
||||
We have
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
|
@ -168,7 +171,7 @@ However, some subsets can be easily described, e.g.
|
|||
\autoref{lec19f4} part (b).
|
||||
Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
|
||||
|
||||
Since $\phi$ is Lipshitz,
|
||||
Since $\phi$ is Lipschitz,
|
||||
$ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
|
||||
By the bounded convergence theorem % TODO
|
||||
$|\phi(X_n)| \le k \implies \int | \phi(X_n) - \phi(X)| \dif \bP \to 0$.
|
||||
|
@ -232,6 +235,3 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
|
|||
Then there exists a random variable $X \in L^p$, such that
|
||||
$X_n = \bE[X | \cF_n]$ for all $n$.
|
||||
\end{theorem}
|
||||
|
||||
|
||||
|
||||
|
|
|
@ -1,3 +1,4 @@
|
|||
\lecture{20}{2023-06-27}{}
|
||||
\begin{refproof}{ceismartingale}
|
||||
By the tower property (\autoref{cetower})
|
||||
it is clear that $(\bE[X | \cF_n])_n$
|
||||
|
@ -5,10 +6,11 @@
|
|||
|
||||
First step:
|
||||
Assume that $X$ is bounded.
|
||||
Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
|
||||
Then, by \autoref{cjensen}, $|X_n| \le \bE[|X| | \cF_n]$,
|
||||
hence $\sup_{\substack{n \in \N \\ \omega \in \Omega}} | X_n(\omega)| < \infty$.
|
||||
Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
|
||||
By the convergence theorem for martingales in $L^2$ % TODO REF
|
||||
By the convergence theorem for martingales in $L^2$
|
||||
(\autoref{martingaleconvergencel2})
|
||||
there exists a random variable $Y$,
|
||||
such that $X_n \xrightarrow{L^2} Y$.
|
||||
|
||||
|
@ -42,28 +44,31 @@
|
|||
\begin{IEEEeqnarray*}{rCl}
|
||||
\int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
|
||||
\end{IEEEeqnarray*}
|
||||
as $\bP$ is \vocab{regular}, \todo{Definition?}
|
||||
i.e.~$\forall \epsilon > 0 \exists k . \bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
|
||||
as $\bP$ is \vocab[Measure!regular]{regular}, \todo{Make this a definition?}
|
||||
i.e.~$\forall \epsilon > 0 . ~\exists k . ~\bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
|
||||
|
||||
% Take some $\epsilon > 0$ and $M$ large enough such that
|
||||
% \[
|
||||
% \int |X - X'| \dif \bP < \epsilon.
|
||||
% \]
|
||||
Take some $\epsilon > 0$ and $M$ large enough such that
|
||||
\[
|
||||
\int |X - X'| \dif \bP < \epsilon.
|
||||
\]
|
||||
|
||||
% Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
|
||||
% Then $X_n' \xrightarrow{L^p} X'$ by the first step.
|
||||
Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
|
||||
Then $X_n' \xrightarrow{L^p} X'$ by the first step.
|
||||
|
||||
% It is
|
||||
% \begin{IEEEeqnarray*}{rCl}
|
||||
% \|X_n - X_n'\|_{L^p}^p &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
|
||||
% &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
|
||||
% &=& \|X - X'\|_{L^p}^p\\
|
||||
% &<& \epsilon.
|
||||
% \end{IEEEeqnarray*}
|
||||
It is
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\|X_n - X_n'\|_{L^p}^p
|
||||
&=& \bE[\bE[X - X' | \cF_n]^{p}]\\
|
||||
&\overset{\text{Jensen}}{\le}& \bE[\bE[(X - X')^p | \cF_n]]\\
|
||||
&=& \|X - X'\|_{L^p}^p\\
|
||||
&<& \epsilon.
|
||||
\end{IEEEeqnarray*}
|
||||
|
||||
Hence
|
||||
\[
|
||||
\|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} + | X - X'|_{L^p} \le 3 \epsilon.
|
||||
\|X_n - X\|_{L^p} %
|
||||
\le \|X_n - X_n'\|_{L^p} + \|X_n' - X'\|_{L^p} + \|X - X'\|_{L^p} %
|
||||
\le 3 \epsilon.
|
||||
\]
|
||||
Thus $X_n \xrightarrow{L^p} X$.
|
||||
\end{refproof}
|
||||
|
@ -226,15 +231,16 @@ we need the following theorem, which we won't prove here:
|
|||
|
||||
We have
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]
|
||||
&=& \bE[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1} X_k \One_{\{ T = k\} } - X_{n-1}(\One_{T \ge n} + \One_{\{T = n-1\}})
|
||||
+ \sum_{k=1}^{n-2} X_k \One_{\{T = k\} } | \cF_{n-1}]\\
|
||||
&&\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]\\
|
||||
&=& \bE\left[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1} X_k \One_{\{ T = k\} }
|
||||
- X_{n-1}(\One_{T \ge n} + \One_{\{T = n-1\}})\right.\\
|
||||
&&\left.+ \sum_{k=1}^{n-2} X_k \One_{\{T = k\} } \middle| \cF_{n-1}\right]\\
|
||||
&=& \bE[(X_n - X_{n-1}) \One_{\{ T \ge n\} } | \cF_{n-1}]\\
|
||||
&=& \One_{\{ T \ge n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})\\
|
||||
&& \begin{cases}
|
||||
&=& \One_{\{ T \ge n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})
|
||||
\begin{cases}
|
||||
\le 0\\
|
||||
= 0 \text{ if $(X_n)_n$ is a martingale}.
|
||||
\end{cases}.
|
||||
\end{cases}
|
||||
\end{IEEEeqnarray*}
|
||||
|
||||
\end{proof}
|
||||
|
|
|
@ -1,4 +1,5 @@
|
|||
\lecture{21}{2023-06-29}{}
|
||||
\subsection{An Application of the Optional Stopping Theorem}
|
||||
|
||||
This is the last lecture relevant for the exam.
|
||||
(Apart from lecture 22 which will be a repetion).
|
||||
|
@ -60,7 +61,7 @@ Suppose that $A^c \subseteq \R$ is a bounded domain.
|
|||
Given a bounded function $f$ on $\R$,
|
||||
we want a function $u$ which is bounded,
|
||||
such that
|
||||
$Lu = 0$ on $A^c$ and $u = f$ on $A$.
|
||||
$\mathbf{L}u = 0$ on $A^c$ and $u = f$ on $A$.
|
||||
\end{goal}
|
||||
|
||||
We will show that $u(x) = \bE_x[f(X_{T_A})]$
|
||||
|
@ -98,12 +99,18 @@ is the unique solution to this problem.
|
|||
Intuitively, conditioned on $X_n$, $X_{n+1}$ should
|
||||
be independent of $\sigma(X_1,\ldots, X_{n-1})$.
|
||||
|
||||
\begin{claim}
|
||||
\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}
|
||||
\[
|
||||
\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
|
||||
= \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
|
||||
\end{claim*}
|
||||
\begin{subproof}
|
||||
The rest of the lecture was very chaotic...
|
||||
\todo{TODO}
|
||||
% We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
|
||||
% $\sigma(X_1,\ldots,X_{n-1})$
|
||||
% is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
|
||||
% Hence the claim follows from \autoref{ceroleofindependence}.
|
||||
\end{subproof}
|
||||
\end{example}
|
||||
|
||||
|
|
|
@ -1,5 +1,5 @@
|
|||
\lecture{3}{}{}
|
||||
\todo{Lecture 3 needs to be finished}
|
||||
\todo{My battery died during this lecture, so this still needs to be finished}
|
||||
\begin{notation}
|
||||
Let $\cB_n$ denote $\cB(\R^n)$.
|
||||
\end{notation}
|
||||
|
|
|
@ -121,6 +121,7 @@
|
|||
How do we prove that something happens almost surely?
|
||||
The first thing that should come to mind is:
|
||||
\begin{lemma}[Borel-Cantelli]
|
||||
\label{borelcantelli}
|
||||
If we have a sequence of events $(A_n)_{n \ge 1}$
|
||||
such that $\sum_{n \ge 1} \bP(A_n) < \infty$,
|
||||
then $\bP[ A_n \text{for infinitely many $n$}] = 0$
|
||||
|
|
|
@ -104,3 +104,4 @@
|
|||
|
||||
\newcommand*\dif{\mathop{}\!\mathrm{d}}
|
||||
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
|
||||
\newcommand\notes{\todo{TODO: copy from official notes}}
|
||||
|
|
Loading…
Reference in a new issue