fixed typo

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Josia Pietsch 2023-07-12 16:39:58 +02:00
parent 5013048fbf
commit 2b34e3b5fb
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@ -61,24 +61,24 @@ where $\mu = \bP X^{-1}$.
\begin{theorem} % Theorem 3
\label{thm:lec10_3}
Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
Let $\bP \in M_1(\R)$ such that $\phi_\bP \in L^1(\lambda)$.
Then $\bP$ has a continuous probability density given by
\[
f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) \dif t.
f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\bP}(t) \dif t.
\]
\end{theorem}
\begin{example}
\begin{itemize}
\item Let $\bP = \delta_{\{0\}}$.
\item Let $\bP = \delta_{0}$.
Then
\[
\phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
\phi_{\bP}(t) = \int e^{\i t x} \dif \delta_0(x) = e^{\i t 0 } = 1
\]
\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
Then
\[
\phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
\phi_{\bP}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
\]
\end{itemize}
\end{example}
@ -88,20 +88,21 @@ where $\mu = \bP X^{-1}$.
If $x_n \to x$, then $f(x_n) \to f(x)$.
\end{claim}
\begin{subproof}
If $e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x } \phi(t) $ for all $t$.
Then
Suppose that
$e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x} \phi(t)$
for all $t$.
Since
\[
|e^{-\i t x} \phi(t)| \le |\phi(t)|
\]
and $\phi \in L^1$, hence $f(x_n) \to f(x)$
and $\phi \in L^1$,
we get $f(x_n) \to f(x)$
by the dominated convergence theorem.
\end{subproof}
We'll show that for all $a < b$ we have
\[
\bP\left( (a,b] \right) = \int_a^b (x) \dif x.\label{thm10_3eq1}
\bP\left( (a,b] \right) = \int_a^b f(x) \dif x.\label{thm10_3eq1}
\]
Let $F$ be the distribution function of $\bP$.
It is enough to prove \autoref{thm10_3eq1}
@ -153,7 +154,7 @@ However, Fourier analysis is not only useful for continuous probability density
Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
Then
\begin{enumerate}[(a)]
\item $\phi(0) = 1$, $|\phi(t)| \le t$ and $\phi(\cdot )$ is continuous.
\item $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous.
\item $\phi$ is a \vocab{positive definite function},
i.e.~
\[\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