diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index dd99575..dc9e110 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -19,8 +19,8 @@ Hence the same holds for submartingales, i.e. What about $L^p$ convergence of martingales? \end{question} -\begin{example}[\vocab{Branching process}] - Fix $u > 0$ and let $p = \frac{1}{1+u}$. +\begin{example}[A martingale not converging in $L^1$ ] + Fix $u > 1$ and let $p = \frac{1}{1+u}$. Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with $\bP[Z_n = 1] = p$. @@ -38,28 +38,27 @@ Hence the same holds for submartingales, i.e. By \autoref{doobmartingaleconvergence}, there exists an a.s.~limit $X_\infty$. - By the SLLN, we have + By the SLLN, we have almost surely \[ \frac{1}{n} \sum_{k=1}^{n} Z_k \xrightarrow{a.s.} \bE[Z_1] = 2p - 1. \] Hence \[ \left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k} - \xrightarrow{a.s.} u^{zp -1}. + \xrightarrow{\text{a.s.}} u^{2p -1}. \] - Since $(X_n)_{n \ge 0}$ is a martingale, we must have $\bE[u^{Z_1}] = 1$. + Since $(X_n)_{n \ge 0}$ is a martingale, we have $\bE[u^{Z_1}] = 1$. Hence $2p - 1 < 0$, because $u > 1$. - - Hence, if $\epsilon > 0$ is small, there exists - $N_0(\epsilon)$ (possibly random) - such that for all $n > N_0(\epsilon)$ + Choose $\epsilon > 0$ small enough such that $u^{2p - 1}(1 + \epsilon) < 1$. + Then there exists $N_0(\epsilon)$ (possibly random) + such that for all $n > N_0(\epsilon)$ almost \[ - \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) % - \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0. + \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \overset{\text{a.s.}}{\le} + u^{2p - 1}(1 + \epsilon) % + \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow[n \to \infty]{\bP} 0. \] - Thus it can not converge in $L^1$. - % TODO Make this less confusing - + However, $X_n$ cannot converge to $0$ in $L^1$, + as $\bE[X_n] = \bE[X_0] = x > 0$. \end{example} $L^2$ is nice, since it is a Hilbert space. So we will first @@ -67,7 +66,7 @@ consider $L^2$. \begin{fact}[Martingale increments are orthogonal in $L^2$ ] \label{martingaleincrementsorthogonal} - Let $(X_n)_n$ be a martingale + Let $(X_n)_n$ be a martingale with $X_n \in L^2$ for all $n$ and let $Y_n \coloneqq X_n - X_{n-1}$ denote the \vocab{martingale increments}. Then for all $m \neq n$ we have that @@ -76,10 +75,18 @@ consider $L^2$. \] \end{fact} \begin{proof} + As $\bE[Y_n^2] = \bE[X_n^2] - 2\bE[X_nX_{n-1}] + \bE[X_{n-1}^2] < \infty$, + we have $Y_n \in L^2$. Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s., by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le n$. - Play with conditional expectation. - \todo{Exercise} + In particular $\bE[Y_n | \cF_k] = 0$ for $k < n$. + Suppose that $m < n$. + Then + \begin{IEEEeqnarray*}{rCl} + \bE[Y_n Y_m] &=& \bE[\bE[Y_n Y_m | \cF_m]]\\ + &=& \bE[Y_m \bE[Y_n | \cF_m]]\\ + &=& 0 + \end{IEEEeqnarray*} \end{proof} \begin{fact}[\vocab{Parallelogram identity}] @@ -108,8 +115,7 @@ consider $L^2$. \[ \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2] \] - by \autoref{martingaleincrementsorthogonal} - % (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here? + by \autoref{martingaleincrementsorthogonal}. In particular, \[ \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty. @@ -128,7 +134,7 @@ consider $L^2$. Since $\bE[(X_\infty - X_n)^2]$ converges to the increasing limit \[ - \sum_{j \ge n + 1} \xrightarrow{n\to \infty} 0 + \sum_{j \ge n + 1} \bE[Y_j^2] \xrightarrow{n\to \infty} 0 \] we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$. \end{proof} @@ -156,7 +162,8 @@ In order to prove \autoref{dooblp}, we first need Let $p > 1$ and $X,Y$ non-negative random variable such that \[ - \forall \ell > 0 .~ \bP[Y \ge \ell] \le \frac{1}{\ell} \int_{\{Y \ge \ell\} } x \dif \bP + \forall \ell > 0 .~ \bP[Y \ge \ell] \le + \frac{1}{\ell} \int_{\{Y \ge \ell\} } X \dif \bP \] Then \[ @@ -168,18 +175,25 @@ In order to prove \autoref{dooblp}, we first need Then \begin{IEEEeqnarray}{rCl} - \|Y\|_{L^p}^p &=& \bE[Y^p]\\ - &=& \int Y(\omega)^p \dif \bP(\omega)\\ - &=&k \int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell \right) \dif \bP(\omega)\\ - &\overset{\text{Fubini}}{=}& \int_0^\infty \int_\Omega \underbrace{\One_{Y \ge \ell}\dif \bP\dif}_{\bP[Y \ge \ell]} \ell. \label{l18star}\\ + \|Y\|_{L^p}^p + &=& \bE[Y^p]\\ + &=& \int Y(\omega)^p \dif \bP(\omega)\\ + &=&\int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell + \right) \dif \bP(\omega)\\ + &\overset{\text{Fubini}}{=}& + \int_0^\infty p \ell^{p-1}\underbrace{\int_\Omega \One_{Y \ge \ell}\dif \bP}_% + {\bP[Y \ge \ell]} \dif\ell. \label{l18star} \end{IEEEeqnarray} - We have + By the assumption it follows that \begin{IEEEeqnarray*}{rCl} - \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}, + \eqref{l18star} + &\le& \int_0^\infty p \ell^{p-2} + \int_{\{Y(\omega) \ge \ell\}} X(\omega) \bP(\dif \omega)\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_{\omega} X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\ + &\overset{\text{Hölder}}{\le}& \frac{p}{p-1} \|X\|_{L^p} \|Y\|_{p}^{p-1}, \end{IEEEeqnarray*} where the assumption was used to apply Hölder. @@ -225,3 +239,4 @@ In order to prove \autoref{dooblp}, we first need For the second part, we apply the first part and \autoref{dooplplemma} (choose $Y \coloneqq X_n^\ast$). \end{refproof} +\todo{Branching process}