diff --git a/inputs/lecture_1.tex b/inputs/lecture_1.tex index 10f049f..3de936a 100644 --- a/inputs/lecture_1.tex +++ b/inputs/lecture_1.tex @@ -46,7 +46,7 @@ First, let us recall some basic definitions: \end{fact} The converse to this fact is also true: \begin{theorem}[Kolmogorov's existence theorem / basic existence theorem of probability theory] - \label{kolmogorovxistence} + \label{kolmogorovexistence} Let $\cF(\R)$ be the set of all distribution functions on $\R$ and let $\cM(\R)$ be the set of all probability measures on $\R$. Then there is a one-to-one correspondence between $\cF(\R)$ and $\cM(\R)$ diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index d65077f..a485088 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -1,255 +1,162 @@ -\lecture{14}{2023-06-06}{} +\lecture{14}{2023-05-25}{Conditional expectation} -We want to derive some properties of conditional expectation. +\section{Conditional expectation} -\begin{theorem}[Law of total expectation] % Thm 1 - \label{ceprop1} - \label{totalexpectation} - \[ - \bE[\bE[X | \cG ]] = \bE[X]. - \] -\end{theorem} -\begin{proof} - Apply (b) from the definition for $G = \Omega \in \cG$. -\end{proof} -\begin{theorem} % Thm 2 - \label{ceprop2} - If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s.. -\end{theorem} -\begin{proof} - Suppose $\bP[X \neq Y] > 0$. - Without loss of generality $\bP[X > Y] > 0$. - Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$. - Let $A \coloneqq \{X > Y + \frac{1}{n}\}$. - % TODO -\end{proof} +\subsection{Introduction} -\begin{example} - Suppose $X \in L^1(\bP)$, $\cG \coloneqq \sigma(X)$. - Then $X$ is measurable with respect to $\cG$. - Hence $\bE[X | \cG] = X$. -\end{example} - -\begin{theorem}[Linearity] - \label{ceprop3} - \label{celinearity} - For all $a,b \in \R$ - we have - \[ - \bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG]. - \] -\end{theorem} -\begin{proof} - Trivial % TODO -\end{proof} - -\begin{theorem}[Positivity] - \label{ceprop4} - % 4 - \label{cpositivity} - If $X \ge 0$, then $\bE[X | \cG] \ge 0$ a.s. -\end{theorem} -\begin{proof} - Let $W $ be a version of $\E[X | \cG]$. - Suppose $\bP[ W < 0] > 0$. - Then $G \coloneqq \{W < -\frac{1}{n}\} \in \cG$ - For some $n \in \N$, we have $\bP[G] > 0$. - However it follows that - \[ - \int_G \bP[X | \cG] \dif \bP \le -\frac{1}{n} \bP[G] < 0 \le \int_G X \dif \bP. - \] -\end{proof} -\begin{theorem}[Conditional monotone convergence theorem] - \label{ceprop5} - % 5 - \label{mcmt} - Let $X_n,X \in L^1(\Omega, \cF, \bP)$. - Suppose $X_n \ge 0$ with $X_n \uparrow X$. - Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$. - -\end{theorem} -\begin{proof} - Let $Z_n$ be a version of $\bE[X_n | Y]$. - Since $X_n \ge 0$ and $X_n \uparrow$, - by \autoref{cpositivity}, - we have - \[ - \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0 - \] - and - \[ - \bE[X_n | \cG] \uparrow \text{a.s.} - \] - (consider $X_{n+1} - X_n$ ). - - Define $Z \coloneqq \limsup_{n \to \infty} Z_n$. - Then $Z$ is $\cG$-measurable - and $Z_n \uparrow Z$ a.s. - - Take some $G \in \cG$. - We know by (b) % TODO REF - that $\be[Z_n \One_G] = \bE[X_n \One_G]$. - The LHS increases to $\bE[Z \One_G]$ by the monotone - convergence theorem. - Again by MCT, $\bE[X_n \One_G]$ increases to - $\bE[X \One_G]$. - Hence $Z$ is a version of $\bE[X | \cG]$. -\end{proof} - -\begin{theorem}[Conditional Fatou] - \label{ceprop6} - \label{cfatou} - Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$. - Then - \[ - \bE[ \liminf_{n \to \infty} X_n | \cG] \le \liminf_{n \to \infty} \bE[X_n | \cG]. - \] -\end{theorem} -\begin{proof} - \todo{in the notes} -\end{proof} -\begin{theorem}[Conditional dominated convergence theorem] - \label{ceprop7} - \label{cdct} - Let $X_n,X \in L^1(\Omega, \cF, \bP)$. - Suppose $|X_n(\omega)| < X(\omega)$ a.e.~ - and $\int |X| \dif \bP < \infty$. - Then $X_n(\omega) \to X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$. - -\end{theorem} -\begin{proof} - \todo{in the notes} -\end{proof} - -Recall -\begin{theorem}[Jensen's inequality] - If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$, - then $\bE[c \circ X] \ge c(\bE[X])$. -\end{theorem} - -For conditional expectation, we have -\begin{theorem}[Conditional Jensen's inequality] - \label{ceprop8} - \label{cjensen} - Let $X \in L^1(\Omega, \cF, \bP)$. - If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$, - then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s. -\end{theorem} -\begin{fact} - \label{convapprox} - If $c$ is convex, then there are two sequences of real numbers - $a_n, b_n \in \R$ - such that - \[ - c(x) = \sup_n(a_n x + b_n). - \] -\end{fact} -\begin{refproof}{cjensen} - By \autoref{convapprox}, $c(x) \ge a_n X + b_n$ - for all $n$. - Hence - \[ - \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG] - = a_n \bE[X | \cG] + b_n \text{a.s.} - \] - for all $n$. - Using that a countable union of sets o f measure zero has measure zero, - we conclude that a.s~this happens simultaneously for all $n$. - Hence - \[ - \bE[c(X) | \cG] \ge \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)). - \] -\end{refproof} - -Recall -\begin{theorem}[Hölder's inequality] - Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$. - Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$. - Then - \[ - \bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}. - \] -\end{theorem} - -\begin{theorem}[Conditional Hölder's inequality] - \label{ceprop9} - \label{choelder} - Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$. - Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$. - Then - \[ - \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}. - \] -\end{theorem} -\begin{proof} - Similar to the proof of Hölder's inequality. - \todo{Exercise} -\end{proof} - -\begin{theorem}[Tower property] - % 10 - \label{ceprop10} - \label{ctower} - Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras. - Then - \[ - \bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH]. - \] -\end{theorem} -\begin{proof} - \todo{Exercise} -\end{proof} - -\begin{theorem}[Taking out what is known] - % 11 - \label{ceprop11} - \label{takingoutwhatisknown} - - If $Y$ is $\cG$-measurable and bounded, then - \[ - \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG]. - \] -\end{theorem} -\begin{proof} - Assume w.l.o.g.~$X \ge 0$. -Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded). - \todo{Exercise} -\end{proof} +Consider a probability space $(\Omega, \cF, \bP)$ +and two events $A, B \in \cF$ with $\bP(B) > 0$. \begin{definition} - Let $\cG$ and $\cH$ be $\sigma$-algebras. - We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent}, - if % TODO + The \vocab{conditional probability} of $A$ given $B$ is defined as + \[ + \bP(A | B) \coloneqq \frac{\bP(A \cap B)}{\bP(B)}. + \] \end{definition} -\begin{theorem}[Role of independence] - \label{ceprop12} - \label{roleofindependence} - If $\cH$ is a sub-$\sigma$-algebra of $\cF$ and $\cH$ is independent - of $\sigma(\sigma(X), \cG)$, then - \[ - \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG]. - \] -\end{theorem} -\begin{example} - If $X$ is independent of $\cG$, - then $\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X]$. -\end{example} -\begin{example}[Martingale property of the simple random walk] - Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$. - Let $S_n \coloneqq \sum_{i=1}^n X_i$ be the \vocab{simple random walk}. - Let $\cF$ denote the $\sigma$-algebra on the product space. - Define $\cF_n \coloneqq \sigma(X_1,\ldots)$. - Intuitively, $\cF_n$ contains all the information gathered until time $n$. - We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$ +Suppose we have two random variables $X$ and $Y$ on $\Omega$, +such that $X$ takes distinct values $x_1, x_2,\ldots, x_{m}$ +and $Y$ takes distinct values $y_1,\ldots, y_n$. +Then for this case, define the \vocab{conditional expectation} +of $X$ given $Y = y_j$ as +\[ +\bE[X | Y = y_j] \coloneqq \sum_{i=1}^m x_i \bP[X=x_i | Y = y_j]. +\] - For $\bE[S_{n+1} | \cF_n]$ we obtain - \begin{IEEEeqnarray*}{rCl} - \bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}& - \bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\ - &\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\ - &\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\ - &=& S_n - \end{IEEEeqnarray*} - -\end{example} +The random variable $Z = \bE[X | Y]$ +is defined as follows: +If $Y(\omega) = y_j$ then +\[ +Z(\omega) \coloneqq \underbrace{\bE[X | Y = y_j]}_{\text{\reflectbox{$\coloneqq$}} z_j}. +\] +Note that $\Omega_j \coloneqq \{\omega : Y(\omega) = y_j\}$ +defines a partition of $\Omega$ and on each $\Omega_j$ +(``the $j^{\text{th}}$ $Y$-atom'') +$ Z$ is constant. + + +Let $\cG \coloneqq \sigma(Y)$. +Then $Z$ is measurable with respect to $\cG$. +Furthermore +\begin{IEEEeqnarray*}{rCl} + \int_{\{Y = y_j\} } Z \dif \bP &=& z_j \int_{\{Y = y_j\}} \dif \bP\\ + &=& z_j \bP[Y=y_j]\\ + &=&\sum_{i=1}^m x_i \bP[X = x_i | Y = y_j] \bP[Y = y_j]\\ + &=&\sum_{i=1}^m x_i \bP[X = x_i, Y = y_j]\\ + &=& \int_{\{Y = y_j\}} X \dif \bP. +\end{IEEEeqnarray*} +Hence +\[ +\int_{G} Z \dif \bP = \int_{G} X \dif \bP +\] +for all $G \in \cG$. + +We now want to generalize this to arbitrary random variables. +\begin{theorem} + \label{conditionalexpectation} + Let $(\Omega, \cF, \bP)$ be a probability space, $X \in L^1(\bP)$ + and $\cG \subseteq \cF$ a sub-$\sigma$-algebra. + Then there exists a random variable $Z$ + such that + \begin{enumerate}[(a)] + \item $Z$ is $\cG$-measurable and $Z \in L^1(\bP)$, + \item $\int_G Z \dif \bP = \int_G X \dif \bP$ + for all $G \in \cG$. + \end{enumerate} + + Such a $Z$ is unique up to sets of measure $0$ and is + called the \vocab{conditional expectation} of $X$ given + the $\sigma$-algebra $\cG$ and written + $Z = \bE[X | \cG]$. +\end{theorem} +\begin{remark} + Suppose $\cG = \{\emptyset, \Omega\}$, + then + \[ + \bE[X | \cG] = (\omega \mapsto \bE[X]) + \] + is a constant random variable. +\end{remark} + +\paragraph{Plan} +We will give two different proves of \autoref{conditionalexpectation}. +The first one will use orthogonal projections. +The second will use the Radon-Nikodym theorem. +We'll first do the easy proof, derive some properties +and then do the harder proof. + +\begin{lemma} + \label{orthproj} + Suppose $H$ is a \vocab{Hilbert space}, + i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$ + making $H$ a complete metric space. + + For any $x \in H$ and $K \subseteq H$ closed, + there exists a unique $z \in K$ such that the following equivalent conditions hold: + \begin{enumerate}[(a)] + \item $\forall y \in K : \langle x-z, y\rangle_H = 0$, + \item $\forall y \in K: \|z-x\|_H \le \|z-x\|_H$. + \end{enumerate} +\end{lemma} +\begin{proof} + \todo{Notes} +\end{proof} + +\begin{refproof}{conditionalexpectation} + + Almost sure uniqueness of $Z$: + + Suppose $X \in L^1$ and $Z$ and $Z'$ satisfy (a) and (b). + We need to show that $\bP[Z \neq Z'] = 0$. + By (a), we have $Z, Z' \in L^1(\Omega, \cG, \bP)$. + By (b), $\bE[(Z - Z') \One_G] = 0$ for all $G \in \cG$. + + Assume that $\bP[Z > Z'] > 0$. + Since $\{Z > Z' + \frac{1}{n}\} \uparrow \{Z > Z'\}$, + we see that $\bP[Z > Z' + \frac{1}{n}] > 0$ for some $n$. + However $\{Z > Z' + \frac{1}{n}\} \in \cG$, + which is a contradiction, since + \[ + \bE[(Z - Z') \One_{Z - Z' > \frac{1}{n}}] \ge \frac{1}{n} \bP[ Z - Z' > \frac{1}{n}] > 0. + \] + + + \bigskip + Existence of $\bE(X | \cG)$ for $X \in L^2$: + + Let $H = L^2(\Omega, \cF, \bP)$ + and $K = L^2(\Omega, \cG, \bP)$. + + $K$ is closed, since a pointwise limit of $\cG$-measurable + functions is $\cG$ measurable (if it exists). + By \autoref{orthproj}, + there exists $z \in K$ such that + \[\bE[(X - Z)^2] = \inf \{ \bE[(X- W)^2] ~|~ W \in L^2(\cG)\}\] + and + \begin{equation} + \forall Y \in L^2(\cG) : \langle X - Z, Y\rangle = 0. + \label{lec13_boxcond} + \end{equation} + Now, if $G \in \cG$, then $Y \coloneqq \One_G \in L^2(\cG)$ + and by \eqref{lec13_boxcond} $\bE[Z \One_G] = \bE[X \One_G]$. + + + \bigskip + Existence of $\bE(X | \cG)$ for $X \in L^1$ : + + Let $X = X^+ - X^-$. + It suffices to show (a) and (b) for $X^+$. + Choose bounded random variables $X_n \ge 0$ such that $X_n \uparrow X$. + Since each $X_n \in L^2$, we can choose a version $Z_n$ of $\bE(X_n | \cG)$. + + \begin{claim} + $0 \overset{\text{a.s.}}{\le} Z_n \uparrow$. + \end{claim} + \begin{subproof} + \todo{Notes} + \end{subproof} + + Define $Z(\omega) \coloneqq \limsup_{n \to \infty} Z_n(\omega)$. + Then $Z$ is $\cG$-measurable and since $Z_n \uparrow Z$, + by MCT, $\bE(Z \One_G) = \bE(X \One_G)$ for all $G \in \cG$. +\end{refproof} diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex new file mode 100644 index 0000000..0eaf742 --- /dev/null +++ b/inputs/lecture_15.tex @@ -0,0 +1,255 @@ +\lecture{15}{2023-06-06}{} + +We want to derive some properties of conditional expectation. + +\begin{theorem}[Law of total expectation] % Thm 1 + \label{ceprop1} + \label{totalexpectation} + \[ + \bE[\bE[X | \cG ]] = \bE[X]. + \] +\end{theorem} +\begin{proof} + Apply (b) from the definition for $G = \Omega \in \cG$. +\end{proof} +\begin{theorem} % Thm 2 + \label{ceprop2} + If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s.. +\end{theorem} +\begin{proof} + Suppose $\bP[X \neq Y] > 0$. + Without loss of generality $\bP[X > Y] > 0$. + Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$. + Let $A \coloneqq \{X > Y + \frac{1}{n}\}$. + % TODO +\end{proof} + +\begin{example} + Suppose $X \in L^1(\bP)$, $\cG \coloneqq \sigma(X)$. + Then $X$ is measurable with respect to $\cG$. + Hence $\bE[X | \cG] = X$. +\end{example} + +\begin{theorem}[Linearity] + \label{ceprop3} + \label{celinearity} + For all $a,b \in \R$ + we have + \[ + \bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG]. + \] +\end{theorem} +\begin{proof} + Trivial % TODO +\end{proof} + +\begin{theorem}[Positivity] + \label{ceprop4} + % 4 + \label{cpositivity} + If $X \ge 0$, then $\bE[X | \cG] \ge 0$ a.s. +\end{theorem} +\begin{proof} + Let $W $ be a version of $\E[X | \cG]$. + Suppose $\bP[ W < 0] > 0$. + Then $G \coloneqq \{W < -\frac{1}{n}\} \in \cG$ + For some $n \in \N$, we have $\bP[G] > 0$. + However it follows that + \[ + \int_G \bP[X | \cG] \dif \bP \le -\frac{1}{n} \bP[G] < 0 \le \int_G X \dif \bP. + \] +\end{proof} +\begin{theorem}[Conditional monotone convergence theorem] + \label{ceprop5} + % 5 + \label{mcmt} + Let $X_n,X \in L^1(\Omega, \cF, \bP)$. + Suppose $X_n \ge 0$ with $X_n \uparrow X$. + Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$. + +\end{theorem} +\begin{proof} + Let $Z_n$ be a version of $\bE[X_n | Y]$. + Since $X_n \ge 0$ and $X_n \uparrow$, + by \autoref{cpositivity}, + we have + \[ + \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0 + \] + and + \[ + \bE[X_n | \cG] \uparrow \text{a.s.} + \] + (consider $X_{n+1} - X_n$ ). + + Define $Z \coloneqq \limsup_{n \to \infty} Z_n$. + Then $Z$ is $\cG$-measurable + and $Z_n \uparrow Z$ a.s. + + Take some $G \in \cG$. + We know by (b) % TODO REF + that $\be[Z_n \One_G] = \bE[X_n \One_G]$. + The LHS increases to $\bE[Z \One_G]$ by the monotone + convergence theorem. + Again by MCT, $\bE[X_n \One_G]$ increases to + $\bE[X \One_G]$. + Hence $Z$ is a version of $\bE[X | \cG]$. +\end{proof} + +\begin{theorem}[Conditional Fatou] + \label{ceprop6} + \label{cfatou} + Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$. + Then + \[ + \bE[ \liminf_{n \to \infty} X_n | \cG] \le \liminf_{n \to \infty} \bE[X_n | \cG]. + \] +\end{theorem} +\begin{proof} + \todo{in the notes} +\end{proof} +\begin{theorem}[Conditional dominated convergence theorem] + \label{ceprop7} + \label{cdct} + Let $X_n,X \in L^1(\Omega, \cF, \bP)$. + Suppose $|X_n(\omega)| < X(\omega)$ a.e.~ + and $\int |X| \dif \bP < \infty$. + Then $X_n(\omega) \to X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$. + +\end{theorem} +\begin{proof} + \todo{in the notes} +\end{proof} + +Recall +\begin{theorem}[Jensen's inequality] + If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$, + then $\bE[c \circ X] \ge c(\bE[X])$. +\end{theorem} + +For conditional expectation, we have +\begin{theorem}[Conditional Jensen's inequality] + \label{ceprop8} + \label{cjensen} + Let $X \in L^1(\Omega, \cF, \bP)$. + If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$, + then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s. +\end{theorem} +\begin{fact} + \label{convapprox} + If $c$ is convex, then there are two sequences of real numbers + $a_n, b_n \in \R$ + such that + \[ + c(x) = \sup_n(a_n x + b_n). + \] +\end{fact} +\begin{refproof}{cjensen} + By \autoref{convapprox}, $c(x) \ge a_n X + b_n$ + for all $n$. + Hence + \[ + \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG] + = a_n \bE[X | \cG] + b_n \text{a.s.} + \] + for all $n$. + Using that a countable union of sets o f measure zero has measure zero, + we conclude that a.s~this happens simultaneously for all $n$. + Hence + \[ + \bE[c(X) | \cG] \ge \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)). + \] +\end{refproof} + +Recall +\begin{theorem}[Hölder's inequality] + Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$. + Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$. + Then + \[ + \bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}. + \] +\end{theorem} + +\begin{theorem}[Conditional Hölder's inequality] + \label{ceprop9} + \label{choelder} + Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$. + Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$. + Then + \[ + \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}. + \] +\end{theorem} +\begin{proof} + Similar to the proof of Hölder's inequality. + \todo{Exercise} +\end{proof} + +\begin{theorem}[Tower property] + % 10 + \label{ceprop10} + \label{ctower} + Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras. + Then + \[ + \bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH]. + \] +\end{theorem} +\begin{proof} + \todo{Exercise} +\end{proof} + +\begin{theorem}[Taking out what is known] + % 11 + \label{ceprop11} + \label{takingoutwhatisknown} + + If $Y$ is $\cG$-measurable and bounded, then + \[ + \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG]. + \] +\end{theorem} +\begin{proof} + Assume w.l.o.g.~$X \ge 0$. +Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded). + \todo{Exercise} +\end{proof} + +\begin{definition} + Let $\cG$ and $\cH$ be $\sigma$-algebras. + We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent}, + if % TODO +\end{definition} + +\begin{theorem}[Role of independence] + \label{ceprop12} + \label{roleofindependence} + If $\cH$ is a sub-$\sigma$-algebra of $\cF$ and $\cH$ is independent + of $\sigma(\sigma(X), \cG)$, then + \[ + \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG]. + \] +\end{theorem} +\begin{example} + If $X$ is independent of $\cG$, + then $\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X]$. +\end{example} +\begin{example}[Martingale property of the simple random walk] + Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$. + Let $S_n \coloneqq \sum_{i=1}^n X_i$ be the \vocab{simple random walk}. + Let $\cF$ denote the $\sigma$-algebra on the product space. + Define $\cF_n \coloneqq \sigma(X_1,\ldots)$. + Intuitively, $\cF_n$ contains all the information gathered until time $n$. + We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$ + + For $\bE[S_{n+1} | \cF_n]$ we obtain + \begin{IEEEeqnarray*}{rCl} + \bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}& + \bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\ + &\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\ + &\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\ + &=& S_n + \end{IEEEeqnarray*} + +\end{example} diff --git a/inputs/lecture_2.tex b/inputs/lecture_2.tex index 5360dc4..1c38d36 100644 --- a/inputs/lecture_2.tex +++ b/inputs/lecture_2.tex @@ -1,3 +1,4 @@ +\lecture{2}{}{} \section{Independence and product measures} In order to define the notion of independence, we first need to construct diff --git a/inputs/lecture_3.tex b/inputs/lecture_3.tex index 2e8c762..8708d78 100644 --- a/inputs/lecture_3.tex +++ b/inputs/lecture_3.tex @@ -1,3 +1,4 @@ +\lecture{3}{}{} \todo{Lecture 3 needs to be finished} \begin{notation} Let $\cB_n$ denote $\cB(\R^n)$. diff --git a/inputs/lecture_5.tex b/inputs/lecture_5.tex index 5364619..b2f53e1 100644 --- a/inputs/lecture_5.tex +++ b/inputs/lecture_5.tex @@ -1,4 +1,4 @@ -% Lecture 5 2023-04-21 +\lecture{5}{2023-04-21}{} \subsection{The laws of large numbers} diff --git a/wtheo.sty b/wtheo.sty index e42ca30..1aab83a 100644 --- a/wtheo.sty +++ b/wtheo.sty @@ -103,3 +103,4 @@ \DeclareSimpleMathOperator{Exp} \newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand\lecture[3]{{\color{gray}\hfill Lecture #1 (#2)}}