typo
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@ -133,7 +133,7 @@ However, Fourier analysis is not only useful for continuous probability density
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RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \bP(\dif y) \\
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&\overset{\text{Fubini}}{=}& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} \dif t\\
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&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} \bP(\dif y) \int_{-T}^T \cos(t(y - x)) \dif t\\
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&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \bP(\dif y)\\
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&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x))}{T (y-x)} \bP(\dif y)\\
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\end{IEEEeqnarray*}
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Furthermore
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\[
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@ -141,10 +141,11 @@ However, Fourier analysis is not only useful for continuous probability density
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1, &y = x,\\
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0, &y \neq x.
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\end{cases}
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% TODO TODO TODO
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\]
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
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\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x))}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
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\end{IEEEeqnarray*}
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% TODO by dominated convergence?
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\end{refproof}
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@ -154,7 +155,7 @@ However, Fourier analysis is not only useful for continuous probability density
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Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
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Then
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\begin{enumerate}[(a)]
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\item $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous.
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\item $\phi(0) = 1$, $|\phi(t)| \le 1$ and $\phi(\cdot )$ is continuous.
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\item $\phi$ is a \vocab{positive definite function},
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i.e.~
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\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
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@ -164,8 +165,6 @@ However, Fourier analysis is not only useful for continuous probability density
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\end{theorem}
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\begin{refproof}{thm:lec_10thm5}
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Part (a) is obvious.
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% TODO
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For part (b) we have:
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\begin{IEEEeqnarray*}{rCl}
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