fixed typo
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@ -141,7 +141,7 @@ We have
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Then
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Then
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\begin{itemize}
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\begin{itemize}
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\item $\phi_{a X + b}(t) = e^{\i t b} \phi_X(\frac{t}{a})$,
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\item $\phi_{a X + b}(t) = e^{\i t b} \phi_X(\frac{t}{a})$,
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\item $\phi_{X + Y}(t) = \phi_X(t) + \phi_Y(t)$.
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\item $\phi_{X + Y}(t) = \phi_X(t) \cdot \phi_Y(t)$.
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\end{itemize}
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\end{itemize}
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\end{fact}
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\end{fact}
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\begin{proof}
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\begin{proof}
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@ -48,6 +48,7 @@ from the lecture on stochastic.
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\pagebreak
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\pagebreak
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\begin{theorem}+
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\begin{theorem}+
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% \footnote{see exercise 3.4}
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\label{thm:convergenceimplications}
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\label{thm:convergenceimplications}
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\vspace{10pt}
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\vspace{10pt}
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Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
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Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
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@ -247,7 +248,20 @@ from the lecture on stochastic.
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We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
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We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
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\end{proof}
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\end{proof}
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\begin{theorem}[Riemann-Lebesgue]
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\begin{theorem}+[Change of variables formula]
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Let $X$ be a random variable with a continuous density $f$,
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and let $g: \R \to \R$ be continuous such that
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$g(X)$ is integrable.
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Then
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\[
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\bE[g(X)] = \int g \circ X \dif \bP
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= \int_{-\infty}^\infty g(y) f(y) \lambda(\dif y)
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= \int_{-\infty}^\infty g(y) f(y) \dif y.
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\]
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\end{theorem}
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\begin{theorem}+[Riemann-Lebesgue]
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%\footnote{see exercise 3.3}
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\label{riemann-lebesgue}
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\label{riemann-lebesgue}
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Let $f: \R \to \R$ be integrable.
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Let $f: \R \to \R$ be integrable.
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Then
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Then
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@ -256,7 +270,27 @@ from the lecture on stochastic.
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\]
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\]
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\end{theorem}
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\end{theorem}
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\begin{theorem}+[Fubini-Tonelli]
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%\footnote{exercise sheet 1}
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Let $(\Omega_{i}, \cF_i, \bP_i), i \in \{0,1\}$
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be probability spaces and $\Omega \coloneqq \Omega_0 \otimes \Omega_1$,
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$\cF \coloneqq \cF_1 \otimes\cF_2$,
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$\bP \coloneqq \bP_0 \otimes \bP_1$.
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Let $f \ge 0$ be $(\Omega, \cF)$-measurable,
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then
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\[
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\Omega_0 \ni x \mapsto \int_{\Omega_{2}} f(x,y) \bP_2(\dif y)
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\]
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and
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\[
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\Omega_1 \ni y \mapsto \int_{\Omega_1} f(x,y) \bP_1(\dif x)
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\]
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are measurable, and
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\[
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\int f \dif \bP = \int_{\Omega_1} \int_{\Omega_2} f(x,y) \bP_2(\dif y) \bP_1(\dif x)(\dif x)
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= \int_{\Omega_2} \int_{\Omega_1} f(x,y) \bP_1(\dif x) \bP_2(\dif y).
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\]
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\end{theorem}
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\subsection{Inequalities}
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\subsection{Inequalities}
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This is taken from section 6.1 of the notes on Stochastik.
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This is taken from section 6.1 of the notes on Stochastik.
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