diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 5bd845d..a3b53b8 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -229,7 +229,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou 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 $\cG$, @@ -249,7 +249,6 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou \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 + &=& S_n. \end{IEEEeqnarray*} - \end{example} diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex new file mode 100644 index 0000000..a76abc1 --- /dev/null +++ b/inputs/lecture_16.tex @@ -0,0 +1,247 @@ +\lecture{16}{2023-06-13}{} + +\subsection{Conditional expectation} + +\begin{theorem} + \label{ceprop11} + \label{ceroleofindependence} + Let $X$ be a random variable, + and let $\cG, \cH$ be $\sigma$-algebras. + + If $\cH$ is independent of $\sigma\left( \sigma(X), \cG \right)$, + then + \[ + \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG]. + \] + + In particular, if $X$ is independent of $\cG$, + then + \[ + \bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X]. + \] +\end{theorem} + +\todo{Definition of independence wrt a $\sigma$-algebra} + +\begin{proof} + Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$. + Then for all $H \in \cH$, we have that $\One_H$ + and any random variable measurable with respect to either $\sigma(X)$ + or $\cG$ must be independent. + + It suffices to consider the case of $X \ge 0$. + Let $G \in \cG$ and $H \in \cH$. + By assumption, $X \One_G$ and $\One_H$ are independent. + Let $Z \coloneqq \bE[X | \cG]$. + Then + \begin{IEEEeqnarray*}{rCl} + \underbrace{\bE[X;G \cap H]}_{\coloneqq \int_{G \cap H} X \dif \bP} &=& \bE[(X \One_G) \One_H]\\ + &=& \bE[X \One_G] \bE[\One_H]\\ + &=& \bE[Z \One_G] \bP(H)\\ + &=& \bE[Z; G \cap H] + \end{IEEEeqnarray*} + + The identity above means, that the measures $A \mapsto \bE[X; A]$ + and $A \mapsto \bE[Z; A]$ + agree on the $\sigma$-algebra $\sigma(\cG, \cH)$ for events + of the form $G \cap H$. + Since sets of this form generate $\sigma(\cG, \cH)$, + these two measures must agree on $\sigma(\cG, \cH)$. + The claim of the theorem follows by the uniqueness of conditional expectation. + + To deduce the second statement, choose $\cG = \{\emptyset, \Omega\}$. +\end{proof} + + +\subsection{The Radon Nikodym theorem} + +First, let us recall some basic facts: +\begin{fact} + Let $(\Omega, \cF, \mu)$ be a \vocab[Measure space!$\sigma$-finite]{$\sigma$-finite + measure space}, + i.e.~$\Omega$ can be decomposed into countably many subsets of finite measure. + Let $f: \Omega \to [0, \infty)$ be measurable. + Define $\nu(A) \coloneqq \int_A f \dif \mu$. + Then $\nu$ is also a $\sigma$-finite measure on $(\Omega, \cF)$.\todo{Application of mct} + Moreover, $\nu$ is finite iff $f$ is integrable. +\end{fact} +Note that in this setting, if $\mu(A) = 0$ it follows that $\nu(A) = 0$. + +The Radon Nikodym theorem is the converse of that: +\begin{theorem}[Radon-Nikodym] + \label{radonnikodym} + + Let $\mu$ and $\nu$ be two $\sigma$-finite measures + on $(\Omega, \cF)$. + Suppose + \[ + \forall A \in \cF . ~ \mu(A) = 0 \implies \nu(A) = 0. + \] + Then + \begin{enumerate}[(1)] + \item there exists $Z: \Omega \to [0, \infty)$ measurable, + such that + \[\forall A \in \cF . ~ \nu(A) = \int_A Z \dif \mu.\] + \item Such a $Z$ is unique up to equality a.e.~(w.r.t. $\mu$). + \item $Z$ is integrable w.r.t.~$ \mu$ iff $\nu$ is a finite measure. + \end{enumerate} + + Such a $Z$ is called the \vocab{Radon-Nikodym derivative}. +\end{theorem} +\begin{definition} + Whenever the property $\forall A \in \cF, \mu(A)= 0 \implies \nu(A) = 0$, + we say that $\nu$ is \vocab{absolutely continuous} + w.r.t.~$\mu$. + This is written as $\nu \ll \mu$. +\end{definition} + +With \autoref{radonnikodym} we get a very short proof of the existence +of conditional expectation: +\begin{proof}[Second proof of \autoref{conditionalexpectation}] + Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq \cF$. + It suffices to consider the case of $X \ge 0$. + For all $G \in \cG$, define $\nu(G) \coloneqq \int_G X \dif \bP$. + Obviously, $\nu \ll \bP$ on $\cG$. + Then apply \autoref{radonnikodym}. +\end{proof} + + +\begin{refproof}{radonnikodym} + We will only sketch the proof. A full proof can be found in the notes. + + \paragraph{Step 1: Uniqueness} + See notes. + \paragraph{Step 2: Reduction to the finite measure case} + See notes. + \paragraph{Step 3: Getting hold of $Z$} + Assume now that $\mu$ and $\nu$ are two finite measures. + Let + \[\cC \coloneqq \{f: \Omega \to [0,\infty] | \forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\}.\] + + We have $\cC \neq \emptyset$ since $0 \in \cC$. + The goal is to find a maximal function $Z$ in $\cC$. + Obviously its integral will also be maximal. + + \begin{enumerate}[(a)] + \item If $f,g \in \cC$, than $f \lor g$ (the pointwise maximum) + s also in $\cC$. + \item Suppose $\{f_n\}_{n \ge 1}$ is an increasing sequence in $\cC$. + Let $f$ be the pointwise limit. + Then $f \in \cC$. + \item For all $f \in \cC$, we have + \[ + \int_\Omega f \dif \mu \le \nu(\Omega) < \infty. + \] + \end{enumerate} + + Define $\alpha \coloneqq \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$. + Let $f_n \in \cC, n\in \N$ be a sequence + with $\int f_n \dif \mu \to \alpha$. + Define $g_n \coloneqq \max \{f_1,\ldots,f_n\} \in \cC$. + Applying (b), we get that the pointwise limit, $Z$, + is an element of $\cC$. + + \paragraph{Step 4: Showing that our choice of $Z$ works} + Define $\lambda(A) \coloneqq \nu(A) - \int_A Z \dif \mu \ge 0$. + $\lambda$ is a measure. + \begin{claim} + $\lambda = 0$. + \end{claim} + \begin{subproof} + Call $G \in \cF$ \emph{good} if the following hold: + \begin{enumerate}[(i)] + \item $\lambda(G) - \frac{1}{k}\mu(G) > 0$. + \item $\forall B \subseteq G, B \in \cF. ~ \lambda(B) - \frac{1}{k}\mu(B) \ge 0$. + \end{enumerate} + Suppose we know that for all $A \in \cF, k \in \N$ + we have + $\lambda(A) \le \frac{1}{k} \mu(A)$. + Then $\lambda(A) = 0$ since $\mu$ is finite. + + Assume the claim does not hold. + Then there must be some $k \in \N$, $A \in \cF$ + such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$. + Fix this $A$ and $k$. + Then $A$ satisfies condition (i) of being good, + but it need not satisfy (ii). + + The tricky part is to make $A$ smaller such that it also + satisfies (ii).\todo{Copy from notes} + \end{subproof} +\end{refproof} + + + +\section{Martingales} + +We have already worked with martingales, but we will define them +rigorously now. + +\begin{definition}[Filtration] + A \vocab{filtration} is a sequence $(\cF_n)$ of $\sigma$-algebras + such that $\cF_n \subseteq \cF_{n+1}$ for all $n \ge 1$. +\end{definition} +Intuitively, we can think of a $\cF_n$ as the set of information +we have gathered up to time $n$. +Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables. + +\begin{definition} + Let $(\cF_n)$ be a filtration and + $X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$. + Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale} + if + \begin{itemize} + \item $X_n$ is $\cF_n$-measurable for all $n$ + ($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ). + \item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$ + for all $n$. + \end{itemize} + + $(X_n)$ is called a \vocab{sub-martingale}, + if it is adapted to $\cF_n$ but + \[ + \bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n. + \] + It is called a \vocab{super-martingale} + if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$. +\end{definition} +\begin{corollary} + Suppose that $f: \R \to \R$ is a convex function such that $f(xn) \in L^1(\bP)$. + 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}. + Then $(f(X_n))_n$ is a sub-martingale. +\end{corollary} +\begin{proof} + Apply \autoref{cejensensinequality}. +\end{proof} + +\begin{corollary} + If $(X_n)_n$ is a martingale, + then $\bE[X_n] = \bE[X_0]$. +\end{corollary} + + +\begin{example} + \begin{itemize} + \item The simple random walk: + + Let $\xi_1, \xi_2, ..$ iid, + $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$, + $X_n \coloneqq \xi_1 + \ldots + \xi_n$ + and $\cF_n \coloneqq \sigma(\xi_1, \ldots, \xi_n) = \sigma(X_1, \ldots, X_n)$. + Then $X_n$ is $\cF_n$-measurable. + Showing that $(X_n)_n$ is a martingale + is left as an exercise. + \item See exercise sheet 9. + \item The branching process (next lecture). + \end{itemize} +\end{example} + + + + + + + + + + diff --git a/probability_theory.tex b/probability_theory.tex index 1cdff28..87de2d8 100644 --- a/probability_theory.tex +++ b/probability_theory.tex @@ -39,6 +39,7 @@ \input{inputs/lecture_13.tex} \input{inputs/lecture_14.tex} \input{inputs/lecture_15.tex} +\input{inputs/lecture_16.tex} \cleardoublepage diff --git a/temp_lenovo b/temp_lenovo new file mode 100644 index 0000000..182d966 --- /dev/null +++ b/temp_lenovo @@ -0,0 +1,28 @@ +coretemp-isa-0000 +Adapter: ISA adapter +Package id 0: +47.0°C (high = +86.0°C, crit = +100.0°C) +Core 0: +47.0°C (high = +86.0°C, crit = +100.0°C) +Core 1: +42.0°C (high = +86.0°C, crit = +100.0°C) + +BAT0-acpi-0 +Adapter: ACPI interface +in0: 11.88 V +curr1: 1.93 A + +thinkpad-isa-0000 +Adapter: ISA adapter +fan1: 1969 RPM +CPU: +42.0°C +GPU: +0.0°C +temp3: +0.0°C +temp4: +0.0°C +temp5: +0.0°C +temp6: +0.0°C +temp7: +0.0°C +temp8: +0.0°C + +acpitz-acpi-0 +Adapter: ACPI interface +temp1: +42.0°C (crit = +100.0°C) +temp2: +26.8°C (crit = +99.0°C) +