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