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