better proof of convergence in L^1 => convergence in measure
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@ -101,7 +101,7 @@ from the lecture on stochastic.
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
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\|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
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&\overset{\text{Hölder}}{\le }& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
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&\overset{\text{Hölder}}{\le}& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
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&=& \|X_n - X\|_{L^q}
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&=& \|X_n - X\|_{L^q}
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
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Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
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@ -114,7 +114,7 @@ from the lecture on stochastic.
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Then for every $\epsilon > 0$
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Then for every $\epsilon > 0$
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\bP[|X_n - X| \ge \epsilon]
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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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&\overset{\text{Markov}}{\ge}& \frac{\bE[|X_n - X|]}{\epsilon}\\
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&\xrightarrow{n \to \infty} & 0,
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&\xrightarrow{n \to \infty} & 0,
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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hence $X_n \xrightarrow{\bP} X$.
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hence $X_n \xrightarrow{\bP} X$.
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