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fixed some typos in lecture 18

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Josia Pietsch 2023-07-17 18:39:05 +02:00
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inputs/lecture_18.tex
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@ -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
\[
@ -168,17 +175,24 @@ In order to prove \autoref{dooblp}, we first need
Then
\begin{IEEEeqnarray}{rCl}
\|Y\|_{L^p}^p &=& \bE[Y^p]\\
\|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}\\
&=&\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)\\
\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}