probability-theory/inputs/lecture_19.tex

\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}
    \label{def:ui}
    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 (\autoref{jensen}).
    Hence for $K$ large enough relevant term is less than $\epsilon$.
\end{proof}

\begin{fact}\label{lec19f2}
    If $(X_n)_n$ is uniformly integrable,
    then $(X_n)_n$ is bounded in $L^1$.
\end{fact}
\begin{proof}
    Take some $\epsilon > 0$ and $K$ such that
    $\sup_n\bE[|X_n| \One_{|X_n| > K}] < \epsilon$.
    Then $\sup_n\|X_n\|_{L^1} \le  K + \epsilon$.
\end{proof}

\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. % TODO reverse fatou

            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
    \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{cjensen}, $|Y| \le  \bE[ |X| | \cG]$.
    Hence $\bE[|Y|] \le  \bE[|X|]$.
    By Markov's inequality,
    it follows that $\bP[|Y| > k] < \delta$
    for $k > \frac{\bE[|X|]}{\delta}$.
    Note that $\{|Y| > k\}  \in \cG$.
    We have
    \begin{IEEEeqnarray*}{rCl}
        \bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
    \end{IEEEeqnarray*}
    by \eqref{lec19eqstar}, since $\bP[|Y| > k] < \delta$.
\end{proof}

\begin{theorem}
    \label{thm:l1iffuip}
    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 $1$-Lipschitz. % 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 Lipschitz,
    $ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
    By the bounded convergence theorem, \autoref{thm:boundedconvergence},
    $|\phi(X_n)| \le k \implies \int |  \phi(X_n) - \phi(X)| \dif \bP \to  0$.


    (1) $\implies$ (2)

    $X_n \xrightarrow{L^1}  X \implies X_n \xrightarrow{\bP} X$
    by Markov's inequality (see \autoref{claim:convimpll1p}).

    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 \texorpdfstring{$L^p, p \ge 1$}{$Lp, p >= 1$}}

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}
    \label{ceismartingale}
    Let $X \in L^p$ for some $p \ge 1$
    and $\bigcup_n \cF_n \to \cF \supseteq \sigma(X)$.
    Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
    to $X$ in $L^p$.
\end{theorem}
% \todo{Proof ?}

\begin{theorem}
    \label{martingaleisce}
    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 $n$.
    In particular, $X_n \xrightarrow{L^p} X$.
\end{theorem}