251 lines
8.8 KiB
TeX
251 lines
8.8 KiB
TeX
\lecture{19}{2023-06-22}{}
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\subsection{Uniform Integrability}
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\begin{example}
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Let $\Omega = [0,1]$, $\cF = \cB$
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and $\bP = \lambda \defon{[0,1]}$.
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Consider $X_n \coloneqq n \One_{(0,\frac{1}{n}))}$.
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We know that $X_n \xrightarrow{n \to \infty} 0$ a.s.,
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however $\bE[X_n] = \bE[|X_n|] = 1$,
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hence $X_n$ does not converge in $L^1(\bP)$.
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Let $\mu_n(\cdot ) = \bP[X_n \in \cdot ]$.
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Intuitively, for a series that converges in probability,
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for $L^1$-convergence to hold we somehow need to make sure
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that probability measures don't assign mass far away from $0$.
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This will be made precise in the notion of uniform integrability.
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\end{example}
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\begin{goal}
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We want to show that uniform integrability and convergence in probability
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is equivalent to convergence in $L^1$.
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\end{goal}
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\begin{definition}
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\label{def:ui}
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A sequence of random variables $(X_n)_n$ is called \vocab{uniformly integrable} (UI),
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if
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\[\forall \epsilon > 0 .~\exists K > 0 .~ \forall n.~
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\bE[|X_n| \One_{\{|X_n| > K\} }] < \epsilon.\]
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Similarly, we define uniformly integrable for sets of random variables.
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\end{definition}
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\begin{example}
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$X_n \coloneqq n \One_{(0,\frac{1}{n})}$ is not uniformly integrable.
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\end{example}
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There is no nice description of uniform integrability.
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However, some subsets can be easily described, e.g.
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\begin{fact}\label{lec19f1}
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If $(X_n)_{n \ge 1}$ is a sequence bounded in $L^{1 + \delta}(\bP)$
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for some $\delta > 0$ (i.e.~$\sup_n \bE[|X_n|^{1+\delta}] < \infty$),
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then $(X_n)_n$ is uniformly integrable.
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\end{fact}
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\begin{proof}
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Let $\epsilon > 0$.
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Let $p \coloneqq 1 + \delta > 1$.
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Choose $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_n| \One_{|X_n| > K}]
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&\le& \bE[|X_n|^p]^{\frac{1}{p}} \bP[|X_n| > k]^{\frac{1}{q}},\\
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\end{IEEEeqnarray*}
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i.e.
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\begin{IEEEeqnarray*}{rCl}
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\sup_n\bE[|X_n| \One_{|X_n| > k}]
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&\le& \underbrace{\sup_n\bE[|X_n|^p]^{\frac{1}{p}}}_{< \infty}
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\sup_n \underbrace{\bP[|X_n| > K]^{\frac{1}{q}}}_%
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{\le K^{-\frac{1}{q}} \bE[|X_n|]^{\frac{1}{q}}}\\
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\end{IEEEeqnarray*}
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where we have applied \yaref{thm:markov}.
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Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
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we have that $\sup_n \bE[|X_n|] < \infty$ by \yaref{jensen}.
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Hence for $K$ large enough the relevant term is less than $\epsilon$.
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\end{proof}
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\begin{fact}\label{lec19f2}
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If $(X_n)_n$ is uniformly integrable,
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then $(X_n)_n$ is bounded in $L^1$.
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\end{fact}
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\begin{proof}
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Take some $\epsilon > 0$ and $K$ such that
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$\sup_n\bE[|X_n| \One_{|X_n| > K}] < \epsilon$.
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Then $\sup_n\|X_n\|_{L^1} \le K + \epsilon$.
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\end{proof}
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\begin{fact}\label{lec19f3}
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Suppose $Y \in L^1(\bP)$ and $\sup_n |X_n(\cdot )| \le Y(\cdot )$.
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Then $(X_n)_n$ is uniformly integrable.
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\end{fact}
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\begin{fact}\label{lec19f4}
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Let $X \in L^1(\bP)$.
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\begin{enumerate}[(a)]
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\item $\forall \epsilon > 0 .~ \exists \delta > 0 .~\forall F \in \cF .~ \bP(F) < \delta \implies\int_F |X| \dif \bP < \epsilon$.
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\item $\forall \epsilon > 0 .~ \exists k \in (0,\infty) .~ \int_{|X| > k} | X| \dif \bP < \epsilon$.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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\begin{enumerate}[(a)]
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\item Suppose not. Then for $\delta = 1, \frac{1}{2}, \frac{1}{2^2}, \ldots$
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there exists $F_n$ such that $\bP(F_n) <\frac{1}{2^n}$
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but $\int_{F_n} |X| \dif \bP \ge \epsilon$.
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Since $\sum_{n} \bP(F_n) < \infty$,
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by \yaref{thm:borelcantelli},
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\[\bP[\underbrace{\limsup_n F_n}_{\text{\reflectbox{$\coloneqq$}}F}] = 0.\]
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We have
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\begin{IEEEeqnarray*}{rCl}
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\int_F | X| \dif \bP &=& \int |X| \One_F \dif \bP\\
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&=& \int \limsup_n (|X| \One_{F_n}) \dif \bP\\
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&\overset{\text{Reverse Fatou}}{\ge }&
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\limsup_n \int |X| \One_{F_n} \dif \bP\\
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&\ge & \epsilon
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\end{IEEEeqnarray*}
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where the assumption that $X$ is in $L^1$ was used to apply
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the reverse of Fatou's lemma. % TODO reverse fatou
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This yields a contradiction since $\bP(F) = 0$.
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\item We want to apply part (a) to $F = \{ |X| > k\}$.
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By \yaref{thm:markov}, $\bP(F) \le \frac{1}{k} \bE[|X|]$.
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Since $\bE[|X|] < \infty$, we can choose $k$ large enough
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to get $\bP(F) \le \delta$.
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\end{enumerate}
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\end{proof}
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\begin{refproof}{lec19f3}
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Fix $\epsilon > 0$.
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We have
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\[
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\bE[|X_n| \One_{|X_n| > k}] \le \bE[|Y| \One_{|Y| > k}] < \epsilon
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\]
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for $k$ large enough by \yaref{lec19f4} (b).
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\end{refproof}
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\begin{fact}\label{lec19f5}
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Let $X \in L^1(\bP)$.
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Then $\bF \coloneqq \{ \bE[X | \cG] : \cG \subseteq \cF \text{ sub-$\sigma$-algebra}\}$ is uniformly integrable.
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\end{fact}
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\begin{proof}
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Fix $\epsilon > 0$.
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Choose $\delta > 0$ such that
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\begin{equation}
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\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon.
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\label{lec19eqstar}
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\end{equation}
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Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
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Then, by \yaref{cjensen}, $|Y| \le \bE[ |X| | \cG]$.
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Hence $\bE[|Y|] \le \bE[|X|]$.
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By \yaref{thm:markov},
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it follows that $\bP[|Y| > k] < \delta$
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for $k > \frac{\bE[|X|]}{\delta}$.
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Note that $\{|Y| > k\} \in \cG$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
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\end{IEEEeqnarray*}
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by \eqref{lec19eqstar}, since $\bP[|Y| > k] < \delta$.
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\end{proof}
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\begin{theorem}
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\label{thm:l1iffuip}
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Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$.
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Then the following are equivalent:
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\begin{enumerate}[(1)]
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\item $X_n \to X$ in $L^1$.
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\item $(X_n)_n$ is uniformly integrable and $X_n \to X$ in probability.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(2) $\implies$ (1)
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Define
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\begin{IEEEeqnarray*}{rCl}
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\phi(x) &\coloneqq & \begin{cases}
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-k, & x \le -k\\
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x, & x \in (-k,k)\\
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k, & x \ge k.
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\end{cases}
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\end{IEEEeqnarray*}
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$\phi$ is $1$-Lipschitz. % TODO
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We have
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\begin{IEEEeqnarray*}{rCl}
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\int |X_n - X| \dif \bP
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&\le & \int |X_n - \phi(X_n)| \dif \bP
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+ \int |\phi(X) - X| \dif \bP
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+ \int |\phi(X_n) - \phi(X)| \dif \bP\\
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\end{IEEEeqnarray*}
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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
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\yaref{lec19f4} part (b).
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Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
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Since $\phi$ is Lipschitz,
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$ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
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By the \yaref{thm:boundedconvergence}
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$|\phi(X_n)| \le k \implies \int | \phi(X_n) - \phi(X)| \dif \bP \to 0$.
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(1) $\implies$ (2)
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$X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$
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by \yaref{thm:markov} (see \yaref{claim:convimpll1p}).
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Fix $\epsilon > 0$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_n|] &=& \bE[|X_n - X + X|]\\
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&\le & \epsilon + \bE[|X|]\\
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&<& \delta k
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\end{IEEEeqnarray*}
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for all $\delta > 0$ and suitable $k$.
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Hence $\bP[|X_n| > k] < \delta$ by \yaref{thm:markov}.
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Then by \yaref{lec19f4} part (a) it follows that
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\[
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\int_{|X_n| > k} |X_n| \dif \bP \le \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le 2 \epsilon.
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\]
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\end{proof}
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\subsection{Martingale Convergence Theorems in \texorpdfstring{$L^p, p \ge 1$}{$Lp, p >= 1$}}
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Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
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\begin{fact}\label{lec19f6}
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Suppose that $X \in L^p$ for some $p \ge 1$.
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Then $(\bE[X | \cF_n])_n$ is an $\cF_n$-martingale.
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\end{fact}
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\begin{proof}
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It is clear that $(\bE[X | \cF_n])_n$ is adapted to $(\cF_n)_n$.
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Let $X_n \coloneqq \bE[X | \cF_n]$.
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Consider
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\begin{IEEEeqnarray*}{rCl}
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\bE[X_n - X_{n-1} | \cF_{n-1}]
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&=& \bE[\bE[X | \cF_n] - \bE[X | \cF_{n-1}] | \cF_{n-1}]\\
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&=& \bE[X | \cF_{n-1}] - \bE[X | \cF_{n-1}]\\
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&=& 0.
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{theorem}
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\label{ceismartingale}
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Let $X \in L^p$ for some $p \ge 1$
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and $\bigcup_n \cF_n \to \cF \supseteq \sigma(X)$.
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Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges
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to $X$ in $L^p$.
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\end{theorem}
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% \todo{Proof ?}
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\begin{theorem}
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\label{martingaleisce}
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Let $p > 1$.
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Let $(X_n)_n$ be a martingale bounded in $L^p$.
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Then there exists a random variable $X \in L^p$, such that
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$X_n = \bE[X | \cF_n]$ for all $n$.
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In particular, $X_n \xrightarrow{L^p} X$.
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\end{theorem}
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