s23-probability-theory/inputs/prerequisites.tex

183 lines
6.3 KiB
TeX
Raw Normal View History

2023-07-05 17:53:41 +02:00
This section provides a short recap of things that should be known
from the lecture on stochastic.
2023-04-27 17:07:50 +02:00
2023-07-05 17:53:41 +02:00
\subsection{Notions of Convergence}
2023-05-10 18:56:36 +02:00
\begin{definition}
2023-07-07 17:42:38 +02:00
\label{def:convergence}
2023-05-10 18:56:36 +02:00
Fix a probability space $(\Omega,\cF,\bP)$.
Let $X, X_1, X_2,\ldots$ be random variables.
\begin{itemize}
\item We say that $X_n$ converges to $X$
\vocab[Convergence!almost surely]{almost surely}
($X_n \xrightarrow{a.s.} X$)
iff
\[
\bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
\]
\item We say that $X_n$ converges to $X$
\vocab[Convergence!in probability]{in probability}
($X_n \xrightarrow{\bP} X$)
iff
\[
\lim_{n \to \infty}\bP[|X_n - X| > \epsilon] = 0
\]
for all $\epsilon > 0$.
\item We say that $X_n$ converges to $X$
\vocab[Convergence!in mean]{in the $p$-th mean}
($X_n \xrightarrow{L^p} X$ )
iff
\[
\bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
\]
\end{itemize}
\end{definition}
2023-07-07 17:42:38 +02:00
% TODO Connect to AnaIII
2023-05-10 18:56:36 +02:00
\begin{theorem}
2023-07-07 17:42:38 +02:00
\label{thm:convergenceimplications}
2023-05-10 18:56:36 +02:00
\vspace{10pt}
Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
Then
\begin{figure}[H]
\centering
\begin{tikzpicture}
\node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
\draw[double equal sign distance, -implies] (as) -- (p);
\draw[double equal sign distance, -implies] (L1) -- (p);
\end{tikzpicture}
\end{figure}
and none of the other implications hold.
2023-04-27 17:07:50 +02:00
\end{theorem}
2023-05-10 18:56:36 +02:00
\begin{proof}
\begin{claim}
$X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
\end{claim}
\begin{subproof}
$\Omega_0 \coloneqq \{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
Let $\epsilon > 0$ and consider $A_n \coloneqq \bigcup_{m \ge n} \{\omega \in \Omega: |X_m(\omega) - X(\Omega)| > \epsilon\}$.
Then $A_n \supseteq A_{n+1} \supseteq \ldots$
Define $A \coloneqq \bigcap_{n \in \N} A_n$.
Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
Since $X_n \xrightarrow{a.s.} X$ we have that
$\forall \omega \in \Omega_0 \exists n \in \N \forall m \ge n |X_m(\omega) - X(\omega)| < \epsilon$.
We have $A \subseteq \Omega_0^{c}$, hence $\bP[A_n] \to 0$.
Thus \[
\bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0.
\]
\end{subproof}
\begin{claim}
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
\end{claim}
\begin{subproof}
We have $\bE[|X_n - X|] \to 0$.
Suppose there exists an $\epsilon > 0$ such that
2023-07-10 12:20:52 +02:00
$\limsup\limits_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
W.l.o.g.~$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c$,
otherwise choose an appropriate subsequence.
2023-05-10 18:56:36 +02:00
We have
\begin{IEEEeqnarray*}{rCl}
\bE[|X_n - X|] &=& \int_\Omega |X_n - X | d\bP\\
&=& \int_{|X_n - X| > \epsilon} |X_n - X| d\bP + \underbrace{\int_{|X_n - X| \le \epsilon} |X_n - X | d\bP}_{\ge 0}\\
&\ge& \epsilon \int_{|X_n -X | > \epsilon} d\bP\\
&=& \epsilon \cdot c > 0 \lightning
\end{IEEEeqnarray*}
\todo{Improve this with Markov}
2023-07-07 17:42:38 +02:00
\todo{Counter examples}
\todo{weak convergence}
\todo{$L^p$ convergence}
2023-05-10 18:56:36 +02:00
\end{subproof}
\begin{claim}
$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
\end{claim}
\begin{subproof}
2023-07-10 12:20:52 +02:00
Take $([0,1], \cB([0,1 ]), \lambda)$
2023-05-10 18:56:36 +02:00
and define $X_n \coloneqq n \One_{[0, \frac{1}{n}]}$.
We have $\bP[|X_n| > \epsilon] = \frac{1}{n}$
for $n$ large enough.
2023-04-27 17:07:50 +02:00
2023-05-10 18:56:36 +02:00
However $\bE[|X_n|] = 1$.
\end{subproof}
2023-04-27 17:07:50 +02:00
2023-05-10 18:56:36 +02:00
\begin{claim}
$X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
\end{claim}
\begin{subproof}
We can use the same counterexample as in c).
2023-07-06 00:36:26 +02:00
2023-05-10 18:56:36 +02:00
$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
We have already seen, that $X_n$ does not converge in $L_1$.
\end{subproof}
2023-04-27 17:07:50 +02:00
2023-05-10 18:56:36 +02:00
\begin{claim}
$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
\end{claim}
\begin{subproof}
Take $\Omega = [0,1], \cF = \cB([0,1]), \bP = \lambda$.
Define $A_n \coloneqq [j 2^{-k}, (j+1) 2^{-k}]$ where $n = 2^k + j$.
We have
\[
\bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
\]
However $X_n$ does not converge a.s.~as for all $\omega \in [0,1]$
the sequence $X_n(\omega)$ takes the values $0$ and $1$ infinitely often.
\end{subproof}
\end{proof}
2023-04-27 17:07:50 +02:00
How do we prove that something happens almost surely?
2023-05-10 18:56:36 +02:00
The first thing that should come to mind is:
2023-04-27 17:07:50 +02:00
\begin{lemma}[Borel-Cantelli]
2023-06-29 22:18:23 +02:00
\label{borelcantelli}
2023-04-27 17:07:50 +02:00
If we have a sequence of events $(A_n)_{n \ge 1}$
such that $\sum_{n \ge 1} \bP(A_n) < \infty$,
then $\bP[ A_n \text{for infinitely many $n$}] = 0$
(more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$).
2023-05-10 18:56:36 +02:00
For independent events $A_n$ the converse holds as well.
2023-04-27 17:07:50 +02:00
\end{lemma}
2023-05-10 18:56:36 +02:00
\iffalse
\todo{Add more stuff here}
\subsection{Some inequalities}
% TODO: Markov
\begin{theorem}[Chebyshev's inequality] % TODO Proof
Let $X$ be a r.v.~with $\Var(x) < \infty$.
Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
\end{theorem}
We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and $\Var(X_1 +\ldots + X_n) = \Var(X_1) + \ldots + \Var(X_n)$ for independent $X_i$.
2023-07-05 17:53:41 +02:00
\fi
\subsection{Some Facts from Measure Theory}
\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
Let $\mu$ be a finite measure on $(\R, \cB(\R))$.
Then for all $\epsilon > 0$,
there exists a compact set $K \in \cB(\R)$ such that
$\mu(K) > \mu(\R) - \epsilon$.
\end{fact}
\begin{proof}
We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
\end{proof}
\begin{theorem}[Riemann-Lebesgue]
\label{riemann-lebesgue}
Let $f: \R \to \R$ be integrable.
Then
\[
\lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
2023-07-06 00:36:26 +02:00
\]
2023-07-05 17:53:41 +02:00
\end{theorem}
2023-05-10 18:56:36 +02:00