fixed lemma about orthogonal projection
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@ -101,7 +101,7 @@ and then do the harder proof.
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i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$
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making $H$ a complete metric space.
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For any $x \in H$ and $K \subseteq H$ closed,
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For any $x \in H$ and $K \subseteq H$ closed and convex,
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there exists a unique $z \in K$ such that the following equivalent conditions hold:
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\begin{enumerate}[(a)]
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\item $\forall y \in K : \langle x-z, y\rangle_H = 0$,
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@ -148,6 +148,7 @@ However, some subsets can be easily described, e.g.
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\end{proof}
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\begin{theorem}
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\label{thm:l1iffuip}
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Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$.
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Then the following are equivalent:
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\begin{enumerate}[(1)]
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@ -189,7 +190,7 @@ However, some subsets can be easily described, e.g.
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(1) $\implies$ (2)
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$X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$
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by Markov's inequality.
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by Markov's inequality (see \autoref{claim:convimpll1p}).
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Fix $\epsilon > 0$.
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We have
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@ -108,7 +108,8 @@ from the lecture on stochastic.
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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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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
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\label{claim:convimpll1p}
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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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Suppose $\bE[|X_n - X|] \to 0$.
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@ -151,7 +152,9 @@ from the lecture on stochastic.
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\end{subproof}
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\begin{claim}
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\label{claim:convimplpl1}
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$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
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$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$.%
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\footnote{Note that the implication holds under certain assumptions,
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see \autoref{thm:l1iffuip}.}
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\end{claim}
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\begin{subproof}
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Take $([0,1], \cB([0,1 ]), \lambda)$
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