typo
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@ -125,7 +125,7 @@ for any $k \in \N$.
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this also means $\sigma(X_1,X_2,\ldots) \subseteq\sigma\left( \bigcup_{n \in \N} \cF_n \right)$.
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``$\subseteq$ '' Since $\cF_n = \sigma(X_1,\ldots,X_n)$,
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obviously $\cF_n \subseteq \sigma(X_1,\ldots,X_n)$
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obviously $\cF_n \subseteq \sigma(X_1,X_2\ldots)$
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for all $n$.
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It follows that $\bigcup_{n \in \N} \cF_n \subseteq \sigma(X_1,X_2,\ldots)$.
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Hence $\sigma\left( \bigcup_{n \in \N} \cF_n \right) \subseteq\sigma(X_1,X_2,\ldots)$.
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@ -29,7 +29,7 @@ First, we need to prove some properties of characteristic functions.
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\begin{refproof}{charfprops}
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\begin{enumerate}[(i)]
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\item $\phi_X(0) = \bE[e^{\i 0 X}] = \bE[1] = 1$.
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For $t \in \R$, we have $|\phi_X(t)| = |\bE[e^{\i t X}]| \overset{\text{Jensen}}{\le} \bE|e^{\i t X}|] = 1$.
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For $t \in \R$, we have $|\phi_X(t)| = |\bE[e^{\i t X}]| \overset{\text{Jensen}}{\le} \bE[|e^{\i t X}|] = 1$.
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\item Let $t, h \in \R$.
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Then
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