From f775ff03c3a906916a719950a18b93d769c99e61 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 22 Jun 2023 17:50:06 +0200 Subject: [PATCH] lecture 19 --- inputs/lecture_18.tex | 16 +-- inputs/lecture_19.tex | 238 +++++++++++++++++++++++++++++++++++++++++ probability_theory.tex | 1 + 3 files changed, 247 insertions(+), 8 deletions(-) create mode 100644 inputs/lecture_19.tex diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index 138d532..9a3d5f4 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -51,8 +51,8 @@ Hence the same holds for submartingales, i.e. $N_0(\epsilon)$ (possibly random) such that for all $n > N_0(\epsilon)$ \[ - \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) \implies x [\underbrace_{< 1}{u^{2p - 1} (1+\epsilon)}]^n \xrightarrow{n \to \infty} 0. - \] + \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 @@ -132,27 +132,27 @@ First, we need a very important inequality: Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$ denote the \vocab{running maximum}. \begin{enumerate}[(1)] - \item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \] + \item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \] (Doob's $L^1$ inequality). \item Fix $p > 1$. Then \[ \bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p]. - \] + \] (Doob's $L^p$ inequality). \end{enumerate} \end{theorem} -We first need +We first need \begin{lemma} \label{dooplplemma} Let $p > 1$ and $X,Y$ non-negative random variable such that \[ \forall \ell > 0 .~ \bP[Y \ge \ell] \le \frac{1}{\ell} \int_{\{Y \ge \ell\} } x \dif \bP - \] + \] Then \[ \bE[Y^p] \le \left( \frac{p}{p-1} \right)^p \bE[X^p]. - \] + \] \end{lemma} \begin{proof} First, assume $Y \in L^p$. @@ -167,7 +167,7 @@ We first need We have \begin{IEEEeqnarray*}{rCl} - \eqref{l18star} &\le & \int_0^\infty \frac{1}{\ell} \int_{\{Y(\omega) \ge \ell\} \ell^p \dif \ell\\ + \eqref{l18star} &\le & \int_0^\infty \frac{1}{\ell} \int_{\{Y(\omega) \ge \ell\}} \ell^p \dif \ell\\ &\overset{\text{Fubini}}{=}& \int_\Omega X(\omega) \int_{0}^{Y(\omega)} p \ell^{p-2} \dif \ell\bP(\dif \omega)\\ &=& \frac{p}{p-1} \int X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\ &\overset{\text{Hölder}}{\le}& \frac{p}{p-1} \|X\|_{L^p} \|Y\|_{p}^{p-1}, diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex new file mode 100644 index 0000000..6766371 --- /dev/null +++ b/inputs/lecture_19.tex @@ -0,0 +1,238 @@ +\lecture{19}{2023-06-22}{} + +\subsection{Uniform integrability} + +\begin{example} + Let $\Omega = [0,1]$, $\cF = \cB$ + and $\bP = \lambda \defon{[0,1]}$. + Consider $X_n \coloneqq n \One_{(0,\frac{1}{n}))}$. + We know that $X_n \xrightarrow{n \to \infty} 0$ a.s., + however $\bE[X_n] = \bE[|X_n|] = 1$, + hence $X_n$ does not converge in $L^1(\bP)$. + + Let $\mu_n(\cdot ) = \bP[X_n \in \cdot ]$. + + Intuitively, for a series that converges in probability, + for $L^1$-convergence to hold we somehow need to make sure + that probability measures don't assign mass far away from $0$. + This will be made precise in the notion of uniform integrability. +\end{example} +\begin{goal} + We want to show that uniform integrability and convergence in probability + is equivalent to convergence in $L^1$. +\end{goal} + +\begin{definition} + A sequence of random variables $(X_n)_n$ is called \vocab{uniformly integrable} (UI), + if + \[\forall \epsilon > 0 .~\exists k > 0 .~ \forall n.~\bE[|X_n| \One_{\{|X_n > k\} }] < \epsilon.\] + + Similarly, we define uniformly integrable for sets of random variables. +\end{definition} +\begin{example} + $X_n \coloneqq n \One_{(0,\frac{1}{n})}$ is not uniformly integrable. +\end{example} +There is no nice description of uniform integrability. +However, some subsets can be easily described, e.g. +\begin{fact}\label{lec19f1} + If $(X_n)_{n \ge 1}$ is a sequence bounded in $L^{1 + \delta}(\bP)$ + for some $\delta > 0$ (i.e.~$\sup_n \bE[|X_n|^{1+\delta}] < \infty$), + then $(X_n)_n$ is uniformly integrable. +\end{fact} +\begin{proof} + Let $\epsilon > 0$. + Let $p \coloneqq 1 + \delta > 1$. + Choose $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$. + Then + \begin{IEEEeqnarray*}{rCl} + \bE[|X_n| \One_{|X_n| > k}] &\le& \bE[|X_n|^p]^{\frac{1}{p}} \bP[|X_n| > k]^{\frac{1}{q}}\\ + \end{IEEEeqnarray*} + i.e. + \begin{IEEEeqnarray*}{rCl} + \sup_n\bE[|X_n| \One_{|X_n| > k}] &\le& \underbrace{\sup_n\bE[|X_n|^p]^{\frac{1}{p}}}_{< \infty} \sup_n \underbrace{\bP[|X_n| > k]^{\frac{1}{q}}}_{\le k^{\frac{1}{q}} \bE[|X_n|]^{\frac{1}{q}}}\\ + \end{IEEEeqnarray*} + 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 + Hence, choose $k$ large enough to make the relevant + term less than $\epsilon$. +\end{proof} + +\begin{fact}\label{lec19f2} + If $(X_n)_n$ is uniformly integrable, + then $(X_n)_n$ is bounded in $L^1$.k:w +\end{fact} + +\begin{fact}\label{lec19f3} + Suppose $Y \in L^1(\bP)$ and $\sup_n |X_n(\cdot )| \le Y(\cdot )$. + Then $(X_n)_n$ is uniformly integrable. +\end{fact} + +\begin{fact}\label{lec19f4} + Let $X \in L^1(\bP)$. + \begin{enumerate}[(a)] + \item $\forall \epsilon > 0 .~ \exists \delta > 0 .~\forall F \in \cF .~ \bP(F) < \delta \implies\int_F |X| \dif \bP < \epsilon$. + \item $\forall \epsilon > 0 .~ \exists k \in (0,\infty) .~ \int_{|X| > k} | X| \dif \bP < \epsilon$. + \end{enumerate} +\end{fact} +\begin{proof} + \begin{enumerate}[(a)] + \item Suppose not. Then for $\delta = 1, \frac{1}{2}, \frac{1}{2^2}, \ldots$ + there exists $F_n$ such that $\bP(F_n) <\frac{1}{2^n}$ + but $\int_{F_n} |X| \dif \bP \ge \epsilon$. + + Since $\sum_{n} \bP(F_n) < \infty$, + by \autoref{borelcantelli}, + \[\bP[\underbrace{\limsup_n F_n}_{\text{\reflectbox{$\coloneqq$}}F}] = 0.\] + We have + \begin{IEEEeqnarray*}{rCl} + \int_F | X| \dif \bP &=& \int |X| \One_F \dif \bP\\ + &=& \int \limsup_n (|X| \One_{F_n}) \dif \bP\\ + &\overset{\text{Reverse Fatou}}{\ge }& + \limsup_n \int |X| \One_{F_n} \dif \bP\\ + &\ge & \epsilon + \end{IEEEeqnarray*} + where the assumption that $X$ is in $L^1$ was used to apply + the reverse of Fatou's lemma. + + This yields a contradiction since $\bP(F) = 0$. + \item We want to apply part (a) to $F = \{ |X| > k\}$. + By Markov, $\bP(F) \le \frac{1}{k} \bE[|X|]$. + Since $\bE[|X|] < \infty$, we can choose $k$ large enough + to get $\bP(F) \le \delta$. + \end{enumerate} +\end{proof} + +\begin{refproof}{lec19f3} + Fix $\epsilon > 0$. + We have + \[ + \bE[|X_n| \One_{|X_n| > k}] \le \bE[|Y| \One_{|Y| > k}] < \epsilon + \] + for $k$ large enough by \autoref{lec19f4} (b). +\end{refproof} + +\begin{fact}\label{lec19f5} + Let $X \in L^1(\bP)$. + Then $\bF \coloneqq \{ \bE[X | \cG] : \cG \subseteq \cF \text{ sub-$\sigma$-algebra}\}$ is uniformly integrable. +\end{fact} +\begin{proof} + Fix $\epsilon > 0$. + Choose $\delta > 0$ such that + \[\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon. (\ast)\] + Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$. + Then, by \autoref{condjensen}, $|Y| \le \bE[ |X| | \cG]$. + Hence $\bE[|Y|] \le \bE[|X|]$. + It follows that $\bP[|Y| > k] < \delta$ + for $k$ suitably large, + since $\bE[|X|] \le \infty$. + Note that $\{Y > k\} \in \cG$. + We have + \begin{IEEEeqnarray*}{rCl} + \bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon + \end{IEEEeqnarray*} + by $(\ast)$, since $\bP[|Y| > k] < \delta$. +\end{proof} + +\begin{theorem} + Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$. + Then the following are equivalent: + \begin{enumerate}[(1)] + \item $X_n \to X$ in $L^1$. + \item $(X_n)_n$ is uniformly integrable and $X_n \to X$ in probability. + \end{enumerate} +\end{theorem} +\begin{proof} + (2) $\implies$ (1) + + Define + \begin{IEEEeqnarray*}{rCl} + \phi(x) &\coloneqq & \begin{cases} + -k, & x \le -k\\ + x, & x \in (-k,k)\\ + k, & x \ge k. + \end{cases} + \end{IEEEeqnarray*} + + $\phi$ is $1$-Lipshitz. % TODO + + We have + \begin{IEEEeqnarray*}{rCl} + \int |X_n - X| \dif \bP + &\le & \int |X_n - \phi(X_n)| \dif \bP + + \int |\phi(X) - X| \dif \bP + + \int |\phi(X_n) - \phi(X)| \dif \bP\\ + \end{IEEEeqnarray*} + We have $\int_{|X_n| > k} \underbrace{|X_n - \phi(X_n)|}_{\le |X_n| + | \phi(X_n)| \le 2 |X_n|} \dif \bP\le \epsilon$ by uniform integrability and + \autoref{lec19f4} part (b). + Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$. + + Since $\phi$ is Lipshitz, + $ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$. + By the bounded convergence theorem % TODO + $|\phi(X_n)| \le k \implies \int | \phi(X_n) - \phi(X)| \dif \bP \to 0$. + + + (1) $\implies$ (2) + + $X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$ + by Markov's inequality. + + Fix $\epsilon > 0$. + We have + \begin{IEEEeqnarray*}{rCl} + \bE[|X_n|] &=& \bE[|X_n - X + X|]\\ + &\le & \epsilon + \bE[|X|]\\ + &<& \delta k + \end{IEEEeqnarray*} + for all $\delta > 0$ and suitable $k$. + + Hence $\bP[|X_n| < k] < \delta$ by Markov's inequality. + Then by \autoref{lec19f4} part (a) it follows that + \[ + \int_{|X_n| > k} |X_n| \dif \bP \le \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le 2 \epsilon. + \] +\end{proof} + +\subsection{Martingale convergence theorems in $L^p, p \ge 1$} + +Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration. + +\begin{fact}\label{lec19f6} + Suppose that $X \in L^p$ for some $p \ge 1$. + + Then $(\bE[X | \cF_n])_n$ is an $\cF_n$-martingale. +\end{fact} +\begin{proof} + It is clear that $(\bE[X | \cF_n])_n$ is adapted to $(\cF_n)_n$. + + Let $X_n \coloneqq \bE[X | \cF_n]$. + Consider + \begin{IEEEeqnarray*}{rCl} + \bE[X_n - X_{n-1} | \cF_{n-1}] + &=& \bE[\bE[X | \cF_n] - \bE[X | \cF_{n-1}] | \cF_{n-1}]\\ + &=& \bE[X | \cF_{n-1}] - \bE[X | \cF_{n-1}]\\ + &=& 0. + \end{IEEEeqnarray*} +\end{proof} + +\begin{theorem} + Let $X \in L^p$ for some $p \ge 1$. + Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges + to $X$ in $L^p$. +\end{theorem} +\begin{proof} + +\end{proof} + +\begin{theorem} + Let $p > 1$. + Let $(X_n)_n$ be a martingale bounded in $L^p$. + Then there exists a random variable $X \in L^p$, such that + $X_n = \bE[X | \cF_n]$ for all $n$. + +\end{theorem} + + + diff --git a/probability_theory.tex b/probability_theory.tex index 0e25050..8cd381b 100644 --- a/probability_theory.tex +++ b/probability_theory.tex @@ -42,6 +42,7 @@ \input{inputs/lecture_16.tex} \input{inputs/lecture_17.tex} \input{inputs/lecture_18.tex} +\input{inputs/lecture_19.tex} \cleardoublepage