242 lines
8.2 KiB
TeX
242 lines
8.2 KiB
TeX
\lecture{18}{2023-06-20}{}
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Recall our key lemma \ref{lec17l3} for supermartingales from last time:
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\[
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(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].
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\]
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What happens for submartingales?
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If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
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Hence the same holds for submartingales, i.e.
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\begin{lemma}
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A (sub-/super-) martingale bounded in $L^1$ converges
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a.s.~to a finite limit, which is a.s.~finite.
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\end{lemma}
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\subsection{Doob's $L^p$ Inequality}
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\begin{question}
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What about $L^p$ convergence of martingales?
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\end{question}
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\begin{example}[A martingale not converging in $L^1$ ]
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Fix $u > 1$ and let $p = \frac{1}{1+u}$.
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Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
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$\bP[Z_n = 1] = p$.
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Let $X_0 = x > 0$ and
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define $X_{n+1} \coloneqq u^{Z_{n+1}} X_{n}$.
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Then $(X_n)_n$ is a martingale,
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since
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\begin{IEEEeqnarray*}{rCl}
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\bE[X_{n+1} |\cF_n] &=& X_n \bE[u^{Z_{n+1}}]\\
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&=& X_n \left(p \cdot u + (1-p)\cdot\frac{1}{u}\right)\\
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&=& X_n \left(\frac{p (u^2-1) + 1}{u}\right)\\
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&=& X_n.
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\end{IEEEeqnarray*}
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By \autoref{doobmartingaleconvergence},
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there exists an a.s.~limit $X_\infty$.
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By the SLLN, we have almost surely
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\[
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\frac{1}{n} \sum_{k=1}^{n} Z_k \xrightarrow{a.s.} \bE[Z_1] = 2p - 1.
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\]
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Hence
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\[
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\left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k}
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\xrightarrow{\text{a.s.}} u^{2p -1}.
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\]
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Since $(X_n)_{n \ge 0}$ is a martingale, we have $\bE[u^{Z_1}] = 1$.
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Hence $2p - 1 < 0$, because $u > 1$.
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Choose $\epsilon > 0$ small enough such that $u^{2p - 1}(1 + \epsilon) < 1$.
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Then there exists $N_0(\epsilon)$ (possibly random)
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such that for all $n > N_0(\epsilon)$ almost
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\[
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\left( \frac{X_n}{x} \right)^{\frac{1}{n}} \overset{\text{a.s.}}{\le}
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u^{2p - 1}(1 + \epsilon) %
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\implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow[n \to \infty]{\bP} 0.
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\]
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However, $X_n$ cannot converge to $0$ in $L^1$,
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as $\bE[X_n] = \bE[X_0] = x > 0$.
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\end{example}
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$L^2$ is nice, since it is a Hilbert space. So we will first
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consider $L^2$.
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\begin{fact}[Martingale increments are orthogonal in $L^2$ ]
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\label{martingaleincrementsorthogonal}
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Let $(X_n)_n$ be a martingale with $X_n \in L^2$ for all $n$
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and let $Y_n \coloneqq X_n - X_{n-1}$
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denote the \vocab{martingale increments}.
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Then for all $m \neq n$ we have that
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\[
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\langle Y_m | Y_n\rangle_{L^2} = \bE[Y_n Y_m] = 0.
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\]
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\end{fact}
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\begin{proof}
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As $\bE[Y_n^2] = \bE[X_n^2] - 2\bE[X_nX_{n-1}] + \bE[X_{n-1}^2] < \infty$,
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we have $Y_n \in L^2$.
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Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s.,
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by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le n$.
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In particular $\bE[Y_n | \cF_k] = 0$ for $k < n$.
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Suppose that $m < n$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\bE[Y_n Y_m] &=& \bE[\bE[Y_n Y_m | \cF_m]]\\
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&=& \bE[Y_m \bE[Y_n | \cF_m]]\\
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&=& 0
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{fact}[\vocab{Parallelogram identity}]
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Let $X, Y \in L^2$.
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Then
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\[
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2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
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\]
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\end{fact}
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\begin{theorem}\label{martingaleconvergencel2}
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Suppose that $(X_n)_n$ is a martingale bounded in
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$L^2$,\\
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i.e.~$\sup_n \bE[X_n^2] < \infty$.
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Then there is a random variable $X_\infty$ such that
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\[
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X_n \xrightarrow{L^2} X_\infty.
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\]
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\end{theorem}
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\begin{proof}
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Let $Y_n \coloneqq X_n - X_{n-1}$ and write
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\[
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X_n = \sum_{j=1}^{n} Y_j.
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\]
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We have
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\[
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\bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
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\]
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by \autoref{martingaleincrementsorthogonal}.
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In particular,
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\[
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\sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
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\]
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Since $(X_n)_n$ is bounded in $L^2$,
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there exists $X_\infty$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$
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by \autoref{doob}.
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It remains to show $X_n \xrightarrow{L^2} X_\infty$.
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For any $r \in \N$, consider
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\[\bE[(X_{n+r} - X_n)^2] = \sum_{j=n+1}^{n+r} \bE[Y_j^2] \xrightarrow{n \to \infty} 0\]
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as a tail of a convergent series.
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Hence $(X_n)_n$ is Cauchy, thus it converges in $L^2$.
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Since $\bE[(X_\infty - X_n)^2]$ converges to the increasing
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limit
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\[
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\sum_{j \ge n + 1} \bE[Y_j^2] \xrightarrow{n\to \infty} 0
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\]
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we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$.
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\end{proof}
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Now let $p \ge 1$ be not necessarily $2$.
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First, we need a very important inequality:
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\begin{theorem}[Doob's $L^p$ inequality]
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\label{dooblp}
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Suppose that $(X_n)_n$ is a sub-martingale.
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Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
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denote the \vocab{running maximum}.
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\begin{enumerate}[(1)]
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\item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ge \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \]
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(Doob's $L^1$ inequality).
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\item Fix $p > 1$. Then \[
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\bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].
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\]
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(Doob's $L^p$ inequality).
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\end{enumerate}
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\end{theorem}
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In order to prove \autoref{dooblp}, we first need
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\begin{lemma}
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\label{dooplplemma}
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Let $p > 1$ and $X,Y$ non-negative random variable
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such that
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\[
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\forall \ell > 0 .~ \bP[Y \ge \ell] \le
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\frac{1}{\ell} \int_{\{Y \ge \ell\} } X \dif \bP
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\]
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Then
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\[
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\bE[Y^p] \le \left( \frac{p}{p-1} \right)^p \bE[X^p].
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\]
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\end{lemma}
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\begin{proof}
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First, assume $Y \in L^p$.
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Then
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\begin{IEEEeqnarray}{rCl}
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\|Y\|_{L^p}^p
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&=& \bE[Y^p]\\
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&=& \int Y(\omega)^p \dif \bP(\omega)\\
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&=&\int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell
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\right) \dif \bP(\omega)\\
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&\overset{\text{Fubini}}{=}&
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\int_0^\infty p \ell^{p-1}\underbrace{\int_\Omega \One_{Y \ge \ell}\dif \bP}_%
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{\bP[Y \ge \ell]} \dif\ell. \label{l18star}
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\end{IEEEeqnarray}
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By the assumption it follows that
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\begin{IEEEeqnarray*}{rCl}
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\eqref{l18star}
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&\le& \int_0^\infty p \ell^{p-2}
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\int_{\{Y(\omega) \ge \ell\}} X(\omega) \bP(\dif \omega)\dif \ell\\
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&\overset{\text{Fubini}}{=}&
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\int_\Omega X(\omega) \int_{0}^{Y(\omega)} p \ell^{p-2} \dif \ell\bP(\dif \omega)\\
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&=& \frac{p}{p-1} \int_{\omega} X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\
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&\overset{\text{Hölder}}{\le}& \frac{p}{p-1} \|X\|_{L^p} \|Y\|_{p}^{p-1},
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\end{IEEEeqnarray*}
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where the assumption was used to apply Hölder.
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Suppose now $Y \not\in L^p$.
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Then look at $Y_M = Y \wedge M$.
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Apply the above to $Y_M \in L^p$ and use the monotone convergence theorem.
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\end{proof}
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\begin{refproof}{dooblp}
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Let $E \coloneqq \{X_n^\ast \ge \ell\} = E_1 \sqcup \ldots \sqcup E_n$
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where
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\[
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E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
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\]
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Then
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\begin{equation}
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\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
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\label{lec18eq2star}
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\end{equation}
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Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
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(by \autoref{cjensen}).
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]
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&=& \One_{E_j} \bE[(|X_n| - |X_{j}|)|\cF_j]\\
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&\overset{\text{a.s.}}{\ge }& 0.
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\end{IEEEeqnarray*}
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By the law of total expectation, \autoref{totalexpectation},
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it follows that
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\begin{equation}
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\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
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\end{equation}
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Now
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\begin{IEEEeqnarray*}{rCl}
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\bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
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&\overset{\eqref{lec18eq2star}, \eqref{lec18eq3star}}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
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&=& \frac{1}{\ell} \int_E |X_n| \dif \bP
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\end{IEEEeqnarray*}
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This proves the first part.
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For the second part, we apply the first part and
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\autoref{dooplplemma} (choose $Y \coloneqq X_n^\ast$).
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\end{refproof}
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\todo{Branching process}
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