better proof of convergence in L^1 => convergence in measure

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Josia Pietsch 2023-07-13 16:31:28 +02:00
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@ -46,6 +46,7 @@ from the lecture on stochastic.
\end{definition}
% TODO Connect to AnaIII
\pagebreak
\begin{theorem}+
\label{thm:convergenceimplications}
\vspace{10pt}
@ -109,20 +110,14 @@ from the lecture on stochastic.
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
\end{claim}
\begin{subproof}
Let $\bE[|X_n - X|] \to 0$.
Suppose there exists an $\epsilon > 0$ such that
$\limsup\limits_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
W.l.o.g.~$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c$,
otherwise choose an appropriate subsequence.
We have
\begin{IEEEeqnarray*}{rCl}
\bE[|X_n - X|] &=& \int_\Omega |X_n - X | \dif\bP\\
&=& \int_{|X_n - X| > \epsilon} |X_n - X| \dif\bP
+ \underbrace{\int_{|X_n - X| \le \epsilon}|X_n-X|\dif\bP}_{\ge 0}\\
&\ge& \epsilon \int_{|X_n -X | > \epsilon} \dif\bP\\
&=& \epsilon \cdot c > 0 \lightning
\end{IEEEeqnarray*}
\todo{Improve this with Markov}
Suppose $\bE[|X_n - X|] \to 0$.
Then for every $\epsilon > 0$
\begin{IEEEeqnarray*}{rCl}
\bP[|X_n - X| \ge \epsilon]
&\overset{\text{Markov}}{\ge}& \frac{\bE[|X_n - X|]}{\epsilon}
&\xrightarrow{n \to \infty} & 0,
\end{IEEEeqnarray*}
hence $X_n \xrightarrow{\bP} X$.
\end{subproof}
\begin{claim} %+
$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
@ -142,13 +137,15 @@ from the lecture on stochastic.
we have
$\bP[|X_n - X| > \delta] < \frac{\epsilon}{2}$.
It is
\[|F_n(t) - F(t)|
= |\bP[X_n \le t] - F(t)|
\le \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
|\bP[X \le t -\delta] - F(t)|)\\
\le \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|, |F(t-\delta) -F(t)|
\le \epsilon,
\]
\begin{IEEEeqnarray*}{rCl}
|F_n(t) - F(t)|
&=& |\bP[X_n \le t] - F(t)|\\
&\le& \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
|\bP[X \le t -\delta] - F(t)|)\\
&\le& \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|,
|F(t-\delta) -F(t)|)\\
&\le& \epsilon,
\end{IEEEeqnarray*}
hence $F_n(t) \to F(t)$.
\end{subproof}
\begin{claim}