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 JosuaKugler
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\documentclass[uebung]{lecture}
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\title{Wtheo 0: Übungsblatt 10}
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\author{Josua Kugler, Christian Merten}
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\usepackage[]{mathrsfs}
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\newcommand{\E}{\mathbb{E}}
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\renewcommand{\P}{\mathbb{P}}
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\begin{document}
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\punkte[36]
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\begin{aufgabe}
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Sei $X \in \mathscr{A}^{n}$ die Flughöhe von $n$ Barock-Raketen und $Y \in \mathscr{A}^{m}$ die Flughöhe
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von $m$ Renaissance-Raketen. Laut
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Aufgabenstellung ist $(X,Y) \sim (N_{(\mu_B, \sigma^2)}^{n} \otimes N_{(\mu_R, \sigma^2)}^{m})$.
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Sei außerdem $\mathscr{H}_0\colon \mu_B \ge \mu_R$. Nach Satz 26.43 hält dann
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der linksseitige Test
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\[
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\varphi_c^{l} = \mathbbm{1}_{ \{ \overline{X}_n - \overline{Y}_{m} \le -c \frac{\sqrt{n + m} }{\sqrt{nm} } \hat{S}_{n,m}\}} = \mathbbm{1}_{\left\{ - \frac{\overline{X}_n - \overline{X}_m}{\hat{S}_{n,m}} \frac{\sqrt{nm} }{\sqrt{n+m} } \ge c\right\} }
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\] mit $c = t_{(n+m-2),(1-\alpha)}$ das Niveau $\alpha$ ein.
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Einsetzen aller Werte ergibt
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\[
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- \frac{\overline{X}_n - \overline{X}_m}{\hat{S}_{n,m}} \frac{\sqrt{nm} }{\sqrt{n+m} }
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\approx 0.941 < 1.734 = t_{18,0.95}
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.\] Also kann $\mathscr{H}_0$ nicht zum Signifikanzniveau $0.05$ abgelehnt werden.
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Für alle $\delta > 0$ gilt per Definition
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\begin{align*}
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\lim\limits_{n \to \infty} \lim\limits_{m \to \infty} \P(|Y_n - y| > \delta) &= 0\\
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\lim\limits_{n \to \infty} \lim\limits_{m \to \infty} \P(|Z_n - z| > \delta) &= 0
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\end{align*}
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Da $h$ eine stetige Funktion ist und $y$ und $z$ bereits feststehen gilt
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\begin{align*}
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\forall \epsilon > 0 \exists \delta > 0: \lVert(Y_n, Z_n) - (y,z)\rVert_1 \leq \delta &\implies |h(Y_n, Z_n) - h(y, z)| \leq \epsilon\\
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|Y_n - y| + |Z_n -z| \leq \delta &\implies |h(Y_n, Z_n) - h(y, z)| \leq \epsilon\\
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\{|h(Y_n, Z_n) - h(y, z)| \leq \epsilon\}&\supset \{|Y_n - y| + |Z_n -z| \leq \delta\}\\
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\{|h(Y_n, Z_n) - h(y, z)| > \epsilon\}&\subset \{|Y_n - y| + |Z_n -z| > \delta\}\\
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\P(|h(Y_n, Z_n) - h(y, z)| > \epsilon) &\leq P(|Y_n - y| + |Z_n -z| > \delta)\\
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\P(|h(Y_n, Z_n) - h(y, z)| > \epsilon) &\leq P(|Y_n - y| > \delta) + \P(|Z_n -z| > \delta)
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\intertext{$Y_n \xrightarrow{\P} y$,$Z_n \xrightarrow{\P} z$}
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\lim\limits_{n \to \infty} \P(|h(Y_n, Z_n) - h(y, z)| > \epsilon) &= 0.
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\end{align*}
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\item Auch $(a_n)_{n\in \N}$ kann als eine Folge von (konstanten) Zufallsvariablen aufgefasst werden.
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Weil $h(a,X) = aX$ eine stetige Funktion ist, gilt $a_nX_n \xrightarrow{\P} aX$.
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Weil $h(X, Y) = X + Y$ eine stetige Funktion ist, gilt $a_nX_n + Y_n \xrightarrow{\P} aX +Y$.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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Sei $X, X_n\colon \Omega \to \R$ für $n \in \N$.
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$X_n \xrightarrow{\mathbb{P}\text{ f.s.}} X \implies X_n \xrightarrow{\mathbb{P}} X$ nach VL.
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Sei also $X_n \xrightarrow{\mathbb{P}} X$. Sei weiter
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$\mathcal{X} \coloneqq \{ \omega \in \Omega \mid \mathbb{P}(\omega) > 0\} $. Dann
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ist $\mathbb{P}(\Omega \setminus \mathcal{X}) = 0$. Es genügt also
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zu zeigen, dass $\lim_{n \to \infty} |X_n(\omega) - X(\omega)| = 0$ für $\omega \in \mathcal{X}$.
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Sei dazu $\epsilon > 0$ und $\omega \in \mathcal{X}$.
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Da $\lim_{n \to \infty} \mathbb{P}(|X_n - X| > \epsilon) = 0$ ex.
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ein $n_0 \in \N$, s.d. $\forall n \ge n_0$:
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\[
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\mathbb{P}(|X_n - X| > \epsilon) < \mathbb{P}(\omega)
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.\] Damit folgt $w \not\in \{|X_n - X| > \epsilon \} $, also
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$|X_n(\omega) - X(\omega)| = 0$.
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\end{aufgabe}
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\begin{aufgabe}
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Zunächst berechne für $n \in \N$:
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\begin{salign*}
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\mathbb{P}^{U}([n, \infty)) &= 1 - \mathbb{P}^{U}((-\infty, n))
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= 1 - \int_{0}^{n} \exp(-v) \d{v} = \exp(-n) \\
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\mathbb{P}^{V}([n, \infty)) &= 1 - \mathbb{P}^{V}((-\infty, n))
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= 1 - \int_{1}^{n} \frac{1}{v^2} \d{v} = \frac{1}{n}
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.\end{salign*}
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\begin{enumerate}[(a)]
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\item Sei $\epsilon > 0$ und $n \in \N$ mit $n > \epsilon$. Dann gilt
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\begin{salign*}
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\mathbb{P}(|X_n| > \epsilon) = \mathbb{P}(n \mathbbm{1}_{[n, \infty)}(U) > \epsilon) \;
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\stackrel{n > \epsilon}{=} \; \mathbb{P}(\mathbbm{1}_{[n, \infty)}(U) > 0 )
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= \mathbb{P}^{U}([n, \infty)) = \exp(-n)
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.\end{salign*}
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Also folgt $\lim_{n \to \infty} \mathbb{P}(|X_n| > \epsilon) = 0$ also
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$X_n \xrightarrow{\mathbb{P}} 0$.
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Sei nun $\sqrt{n} > \epsilon$. Dann gilt
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\begin{salign*}
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\mathbb{P}(|Y_n| > \epsilon) = \mathbb{P}(\sqrt{n} \mathbbm{1}_{[n, \infty)}(V) > \epsilon)
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\; \stackrel{\sqrt{n} > \epsilon}{=}
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\mathbb{P}^{V}([n, \infty)) = \frac{1}{n}
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.\end{salign*}
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Also folgt $\lim_{n \to \infty} \mathbb{P}(|Y_n| > \epsilon) = 0$ also
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$Y_n \xrightarrow{\mathbb{P}} 0$.
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\item Betrachte
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\begin{salign*}
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\E(|X_n|^2) = \E(n^2 \mathbbm{1}_{[n, \infty)}(U))
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= n^2 \int_{\R}^{} \mathbbm{1}_{[n, \infty)}(v) f^{U}(v) \d{v}
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= n^2 \mathbb{P}^{U}((n, \infty))
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= n^2 \exp(-n)
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.\end{salign*}
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Betrachte $f(x) \coloneqq x^2 \exp(-x) \in C^{\infty}(\R)$.
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Dann ist durch mehrfache Anwendung von de l'Hospital
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($*$):
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\[
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\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x^2 \exp(-x) = \lim_{x \to \infty} \frac{x^2}{\frac{1}{\exp(-x)}}
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\stackrel{(*)}{=} \lim_{x \to \infty} \frac{2x}{\exp(x)}
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\stackrel{(*)}{=} \lim_{x \to \infty} \frac{2}{\exp(x)} = 0
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.\]
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Mit der Folge $(a_n)_{n \in \N}$ mit $a_n \coloneqq n$ folgt also
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$\lim_{n \to \infty} n^2\exp(-n) = f(n) = \lim_{x \to \infty} f(x) = 0$.
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Also folgt insgesamt $\lim_{n \to \infty} \Vert X_n \Vert_{L^2} = 0$ und damit
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$X_n \xrightarrow{\mathscr{L}_2} 0$.
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Weiter folgt
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\begin{salign*}
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\E(|Y_n|^2) = \E(n\mathbbm{1}_{[n, \infty)}(V))
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= n \int_{\R}^{} \mathbbm{1}_{[n, \infty)}(v) f^{V}(v) \d{v}
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= n \mathbb{P}^{V}((n, \infty))
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= n \frac{1}{n} = 1
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.\end{salign*}
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Damit folgt $\lim_{n \to \infty} \Vert Y_n \Vert_{L^2} = \sqrt{1} = 1 \neq 0$. Da
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$Y_n \xrightarrow{\mathbb{P}} 0$ konvergiert $Y_n$ nicht in $\mathscr{L}_2$ gegen
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ein $Y \in \overline{\mathscr{A}}$ mit $Y \neq 0$ $\mathbb{P}$ f.s., da
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sonst auch $Y_n \xrightarrow{\mathbb{P}} Y \neq 0$ und
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stochastische Grenzwerte $\mathbb{P}$ f.s. übereinstimmen.
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Also konvergiert $Y_n$ nicht in $\mathscr{L}_2$.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Es gilt für $ 0 <\epsilon < 1$
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\begin{align*}
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\lim\limits_{n \to \infty} \P(X_n > \epsilon) &= \lim\limits_{n \to \infty} \P(\sqrt{n}\mathbbm{1}_{[0,\frac{1}{n}]}(U) > \epsilon)\\
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&= \lim\limits_{n \to \infty} \P(\mathbbm{1}_{[0,\frac{1}{n}]}(U))\\
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&= \lim\limits_{n \to \infty} \frac{1}{n}\\
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&= 0.
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\end{align*}
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Also gilt $X_n \xrightarrow{\P} 0$
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Gleichzeitig erhalten wir
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\begin{align*}
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\E(|X_n|^2) &= \lim\limits_{n \to \infty} \int_\R (X_n)^2 \P(\d{x}) \\
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&= \lim\limits_{n \to \infty} \int_\R n \mathbbm{1}_{[0,\frac{1}{n}]}(U)\P(\d{x})\\
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&= \lim\limits_{n \to \infty} n \frac{1}{n}\\
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&= 1\\
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&\neq 0.
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\end{align*}
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Daraus folgt $X_n \not \xrightarrow{L^2} 0$.
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\item Es gilt
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\begin{align*}
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\E(|X - X_n|^2) &= \E(|X- X_n|^2 \mathbbm{1}_{|X_n-X| > \epsilon}) + \E(|X- X_n|^2 \mathbbm{1}_{|X_n-X| \leq \epsilon})\\
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\intertext{Wir nutzen die Hölder-Ungleichung $\E(|X_nX|) \leq \sqrt{\E(|X|^2)\E(|X_n|^2)}$ und erhalten}
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&= \E(|X|^2\mathbbm{1}_{|X_n-X| > \epsilon}) + \E(|X_n|^2\mathbbm{1}_{|X_n-X| > \epsilon}) - 2\E(|XX_n|\mathbbm{1}_{|X_n-X| > \epsilon}) + \E(|X- X_n|^2 \mathbbm{1}_{|X_n-X| \leq \epsilon})
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\intertext{Wegen $X_n \xrightarrow{\P} X$ ist $\{|X_n - X| > \epsilon\}$ eine Nullmenge und es gilt}
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&= 0 + \E(|X- X_n|^2 \mathbbm{1}_{|X_n-X| \leq \epsilon})\\
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&= \epsilon^2 \E(\mathbbm{1}_{|X_n-X| \leq \epsilon})\\
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&= \epsilon^2 (1 - \P(|X_n - X| > \epsilon))\\
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&= \epsilon^2
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\end{align*}
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Für $\epsilon \to 0$ erhalten wir daraus die Behauptung.
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\item Betrachte
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\begin{align*}
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\limsup\limits_{n \to \infty} \E(|X_n|^{2 + \alpha}) &= \limsup\limits_{n \to \infty} \int_\R \sqrt{n}^{2 + \alpha} \cdot \mathbbm{1}_{[0,1]}(U) \P^U(\d x)\\
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&= \limsup\limits_{n \to \infty} n \cdot n^{\frac{\alpha}{2}} \cdot \frac{1}{n}\\
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&= \limsup\limits_{n \to \infty} n^{\frac{\alpha}{2}}\\
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&= \infty
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\end{align*}
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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\documentclass[uebung]{../../../lecture}
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\documentclass[uebung]{lecture}
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\title{Wtheo 0: Übungsblatt 8}
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\title{Wtheo 0: Übungsblatt 8}
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\author{Josua Kugler, Christian Merten}
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\author{Josua Kugler, Christian Merten}
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\newcommand{\E}{\mathbb{E}}
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\newcommand{\E}{\mathbb{E}}
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\renewcommand{\P}{\mathbbm{P}}
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\usepackage[]{mathrsfs}
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\usepackage[]{mathrsfs}
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\newcommand{\cov}{\mathbb{C}\text{ov}}
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\newcommand{\cov}{\mathbb{C}\text{ov}}
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\newcommand{\var}{\mathbb{V}\text{ar}}
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\newcommand{\var}{\mathbb{V}\text{ar}}
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@@ -51,7 +52,42 @@
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\end{enumerate}
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\end{enumerate}
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\end{aufgabe}
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\end{aufgabe}
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\stepcounter{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Es gilt
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\begin{align*}
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\int_0^\infty \P(X > y) \d{y} &= \int_0^\infty \int_y^\infty \mathbbm{f}^X(x) \d{x} \d{y}\\
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&= \int_0^\infty \int_0^\infty \mathbbm{f}^X(x)\mathbbm{1}_{x>y} \d{x} \d{y}\\
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\intertext{Fubini}
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&= \int_0^\infty \int_0^\infty \mathbbm{f}^X(x)\mathbbm{1}_{x>y} \d{y} \d{x}\\
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&= \int_0^\infty \int_0^x \mathbbm{f}^X(x) \d{y} \d{x}\\
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&= \int_0^\infty x\mathbbm{f}^X(x) \d{x}\\
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&= \int_\Omega X(\omega) \mathbbm{f}(\omega) \d{\omega}\\
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&= \E(X)
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\end{align*}
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\item Es gilt
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\begin{align*}
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\E(X) &= \int_0^\infty \P(X > y) \d{y}\\
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&= \int_0^\infty \int_y^\infty \mathbbm{f}^X(\omega) \d{\omega}\d{y}\\
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&= \int_0^\infty \int_y^\infty \lambda e^{-\lambda x} \d{x} \d{y}\\
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&= \int_0^\infty e^{-\lambda y} \d{y}\\
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&= \frac{1}{\lambda}
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\end{align*}
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\item Es gilt
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\begin{align*}
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\E(X) &= \sum_{n = 1}^{\infty} \P(X \geq n)\\
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&= \sum_{n = 1}^{\infty} \sum_{k = n}^{\infty} \mathbbm{p}^X(k) \\
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&= \sum_{n = 1}^{\infty} \sum_{k = n}^{\infty} (1-p)^{k - 1}p\\
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&= \sum_{n = 1}^{\infty} p(1-p)^{n-1}\sum_{k = 0}^{\infty} (1-p)^k
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\intertext{geometrische Reihe}
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&= \sum_{n = 1}^{\infty} p(1-p)^{n-1} \frac{1}{1-(1-p)}\\
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&= \sum_{n = 1}^{\infty} (1-p)^{n-1}\\
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\intertext{geometrische Reihe}
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&= \frac{1}{1 - (1-p)}\\
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&= \frac{1}{p}
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\end{align*}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
|
\begin{aufgabe}
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\begin{enumerate}[(a)]
|
\begin{enumerate}[(a)]
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@@ -105,4 +141,52 @@
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\end{enumerate}
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\end{enumerate}
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\end{aufgabe}
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\end{aufgabe}
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|
\begin{aufgabe}
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|
\begin{enumerate}[(a)]
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||||||
|
\item Wir definieren den Wahrscheinlichkeitsraum $\Omega = \{(i_1, \dots, i_n)|i_j \in \{1,\dots, m\}\}$ als die Menge aller $n$-Tupel mit Werten zwischen 1 und $m$, wobei das $j$-te Element eines Tupels angibt, welche Ente der $j$-te Jäger gewählt hat. Dann enthält das Ereignis
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|
\begin{align*}
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|
A_i \coloneqq \{(i_1, \dots, i_n)\in \Omega, i\neq i_l \forall l \in \{1,\dots, n\}\}
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\end{align*}
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|
alle Elementarereignisse, in denen die $i$-te Ente nicht getroffen wird.
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Die Zufallsvariable $X_i \colon \Omega \to \{0,1\}, \omega \mapsto \mathbbm{1}_{A_i}$ gibt an, ob die $i$-te Ente überlebt (1) oder nicht (0). Dann ist durch $X \coloneqq \sum_{i = 1}^{m} X_i$ gerade die Anzahl der überlebenden Enten gegeben.
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Es gilt aufgrund der Linearität des Erwartungswerts
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\begin{align*}
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|
\E(X) &= \E\left(\sum_{i = 1}^{m} X_i\right)\\
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&= \sum_{i = 1}^{m} \E(X_i)\\
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&= \sum_{i = 1}^{m} \E(\mathbb{1}_{A_i})\\
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&= \sum_{i = 1}^{m} \P(A_i)\\
|
||||||
|
&= \sum_{i = 1}^{m} \frac{\# A_i}{\# \Omega}\\
|
||||||
|
&= \sum_{i = 1}^{m} \frac{(m-1)^n}{m^n}\\
|
||||||
|
&= m \cdot \left(\frac{m-1}{m}\right)^n
|
||||||
|
\end{align*}
|
||||||
|
\item Wir bestimmen zunächst
|
||||||
|
\begin{align*}
|
||||||
|
\E(X_iX_j) &= \E(\mathbbm{1}_{A_i} \cdot \mathbbm{1}_{A_j})\\
|
||||||
|
&= \E(\mathbbm{1}_{A_i \cap A_j})\\
|
||||||
|
&= \P(A_i \cap A_j)\\
|
||||||
|
\intertext{Für $i = j$ gilt $\P(A_i \cap A_j) = \P(A_i) = m\left(\frac{m-1}{m}\right)^n$. Sei also $i\neq j$}
|
||||||
|
&= \frac{\# A_i \cap A_j}{\# \Omega}\\
|
||||||
|
&= \left(\frac{m-2}{m}\right)^2
|
||||||
|
\end{align*}
|
||||||
|
Es gilt daher
|
||||||
|
\begin{align*}
|
||||||
|
\var(X) &= \E(X^2) - \E(X)^2\\
|
||||||
|
&= \E\left(\sum_{i = 1}^{m} X_i \sum_{j = 1}^{m} X_j\right)- \E(X)^2\\
|
||||||
|
&= \E\left(\sum_{i, j = 1}^m X_iX_j\right)- \E(X)^2\\
|
||||||
|
&= \sum_{i,j = 1}^{m} \E(X_iX_j)- \E(X)^2\\
|
||||||
|
&= \sum_{i = 1}^{m} \E(X_iX_i) + \sum_{i \neq j, 1\leq i, j \leq m} \E(X_iX_j)- \E(X)^2\\
|
||||||
|
&= m \cdot \left(\frac{m-1}{m}\right)^n + (m^2 - m) \left(\frac{m-2}{m}\right)^2 - m^2 \cdot \left(\frac{m-1}{m}\right)^{2n}
|
||||||
|
\end{align*}
|
||||||
|
\item Für $n = m = 50$ gilt $7^{-2}\var(X) \approx 0.0996$ und $\E(X) \approx 18.2$. Für $m_1 = 11, m_2 = 26$ erhalten wir
|
||||||
|
\begin{align*}
|
||||||
|
\P(X \in [m_1, m_2]) &\geq \P(|X - \E(X)| \leq 7)\\
|
||||||
|
&= 1 - \P(|X - \E(X)| > 7)\\
|
||||||
|
\intertext{Ungleichung von Tschebycheff}
|
||||||
|
&\geq 1 - 7^{-2}\var(X)\\
|
||||||
|
&\geq 1 - 0.0996\\
|
||||||
|
&\geq 0.9
|
||||||
|
\end{align*}
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
\end{document}
|
\end{document}
|
||||||
|
|||||||
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|
|||||||
|
\documentclass[uebung]{lecture}
|
||||||
|
|
||||||
|
\title{Wtheo 0: Übungsblatt 9}
|
||||||
|
\author{Josua Kugler, Christian Merten}
|
||||||
|
\newcommand{\E}{\mathbb{E}}
|
||||||
|
\renewcommand{\P}{\mathbb{P}}
|
||||||
|
\usepackage[]{mathrsfs}
|
||||||
|
\newcommand{\cov}{\mathbb{C}\text{ov}}
|
||||||
|
\newcommand{\var}{\mathbb{V}\text{ar}}
|
||||||
|
\newcommand{\tageq}{\stepcounter{equation}\tag{\theequation}}
|
||||||
|
\newcommand{\indep}{\perp \!\!\! \perp}
|
||||||
|
|
||||||
|
\begin{document}
|
||||||
|
|
||||||
|
\punkte[33]
|
||||||
|
|
||||||
|
\begin{aufgabe}
|
||||||
|
\begin{enumerate}[(a)]
|
||||||
|
\item Anwenden der Formel der VL ergibt sofort für $x \in \R$:
|
||||||
|
\begin{salign*}
|
||||||
|
f^{X}(x) &= \int_{\R}^{} f^{X,Y}(x,y) \d{y} \\
|
||||||
|
&= \int_{\R}^{} \frac{1}{\pi} \mathbbm{1}_{\{(x,y) \in E\} } \d{y} \\
|
||||||
|
&= \mathbbm{1}_{\{|x| \le 1\}} \int_{-\sqrt{1-x^2} }^{\sqrt{1-x^2} } \frac{1}{\pi} \d{y} \\
|
||||||
|
&= \frac{2}{\pi} \sqrt{1-x^2} \mathbbm{1}_{\{|x| \le 1\}}
|
||||||
|
\intertext{
|
||||||
|
Ganz analog folgt für $y \in \R$:}
|
||||||
|
f^{Y}(y) &= \frac{2}{\pi} \sqrt{1-y^2} \mathbbm{1}_{\{|y| \le 1\} }
|
||||||
|
.\end{salign*}
|
||||||
|
\item Anwenden der Formel der VL ergibt zunächst mit Anwendung des Transformationssatzes
|
||||||
|
\begin{salign*}
|
||||||
|
\E(X) &= \int_{\R}^{} x f^{X}(x) \d{x} \\
|
||||||
|
&= \frac{1}{\pi} \int_{-1}^{1} 2x \sqrt{1-x^2} \d{x} \\
|
||||||
|
&= \frac{1}{\pi} \left[ \int_{-1}^{0} 2x \sqrt{1-x^2} \d{x}
|
||||||
|
+ \int_{0}^{1} 2x \sqrt{1-x^2} \d{x} \right] \\
|
||||||
|
&\stackrel{z = 1-x^2}{=} \frac{1}{\pi} \left[ \int_{0}^{1} - \sqrt{z} \d{z}
|
||||||
|
+ \int_{1}^{0} - \sqrt{z} \d{z} \right] \\
|
||||||
|
&= 0
|
||||||
|
\intertext{Unter erneuter Formelanwendung folgt}
|
||||||
|
\E(X^2) &= \int_{\R}^{} x^2 f^{X}(x) \d{x} \\
|
||||||
|
&= \frac{2}{\pi} \int_{-1}^{1} x^2 \sqrt{1-x^2} \d{x} \\
|
||||||
|
&\stackrel{x = \sin(\varphi)}{=} \frac{2}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
|
||||||
|
\sin^2(\varphi) \cos^2(\varphi)\d{\varphi} \\
|
||||||
|
&= \frac{2}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4} \sin^2(2\varphi) \d{\varphi} \\
|
||||||
|
&\stackrel{\psi = 2 \varphi}{=}
|
||||||
|
\frac{1}{2\pi} \int_{-\pi}^{\pi} \sin^2(\psi) \frac{1}{2} \d{\psi} \\
|
||||||
|
&= \frac{1}{4 \pi} \int_{-\pi}^{\pi} \frac{1}{2}(\sin^2(\psi) + \cos^2(\psi)) \d{\psi} \\
|
||||||
|
&= \frac{1}{8 \pi} 2 \pi \\
|
||||||
|
&= \frac{1}{4}
|
||||||
|
\intertext{Damit folgt}
|
||||||
|
\var(X) &= \frac{1}{4}
|
||||||
|
\intertext{Ganz analog}
|
||||||
|
\var(Y) &= \frac{1}{4}
|
||||||
|
\intertext{Betrachte nun zunächst}
|
||||||
|
\int_{0}^{2\pi} \sin(\varphi) \cos(\varphi)\d{\varphi}
|
||||||
|
&\stackrel{\text{part. Integrat.}}{=} \underbrace{\sin^2\varphi \Big|_{0}^{2\pi}}_{= 0}
|
||||||
|
- \int_{0}^{2\pi} \sin\varphi \cos\varphi \d{\varphi} \\
|
||||||
|
\intertext{Damit folgt}
|
||||||
|
\int_{0}^{2\pi} \sin\varphi \cos\varphi \d{\varphi} &= 0
|
||||||
|
\tageq \label{eq:1}
|
||||||
|
\intertext{Es ist $\int_{\R^2}^{} \left|\frac{xy}{\pi} \right| \mathbbm{1}_{\{(x,y) \in E\}}
|
||||||
|
\d{(x,y)} \le \frac{1}{\pi}\mathscr{L}^{2}(E) = 1 < \infty$,
|
||||||
|
d.h. Fubini ist anwendbar. Damit folgt}
|
||||||
|
\E(XY) &= \int_{\R^2}^{} xy f^{X,Y}(x,y) \d{(x,y)} \\
|
||||||
|
&= \int_{\R}^{} \int_{\R}^{} \frac{xy}{\pi} \mathbbm{1}_{\{(x,y) \in E\} } \d{x} \d{y} \\
|
||||||
|
&\stackrel{\text{Trafosatz}}{=}
|
||||||
|
\frac{1}{\pi}\int_{0}^{1} \d{r} \int_{0}^{2\pi} \d{\varphi} r^{3} \cos(\varphi) \sin(\varphi) \\
|
||||||
|
&= \frac{1}{4\pi} \int_{0}^{2\pi} \cos(\varphi) \sin(\varphi)\d{\varphi} \\
|
||||||
|
&\stackrel{\text{(\ref{eq:1})}}{=} 0
|
||||||
|
\intertext{Da $\E(X) = \E(Y) = 0$ folgt also}
|
||||||
|
\cov(X,Y) &= 0
|
||||||
|
\intertext{Und damit}
|
||||||
|
\rho(X,Y) &= 0
|
||||||
|
.\end{salign*}
|
||||||
|
\item Es gilt nach VL: $X \indep Y \iff f^{X,Y}(x,y) = f^{X}(x) f^{Y}(y)$ $\mathscr{L}$-f.ü. Nun
|
||||||
|
betrachte $A \coloneqq (-1,1)^2 \setminus E$. Dann ist
|
||||||
|
\[
|
||||||
|
\mathscr{L}^2(A) = \mathscr{L}^2(A) - \mathscr{L}^2(E) = 2^2 - \pi = 4 - \pi > 0
|
||||||
|
.\] Also ist $A$ keine $\mathscr{L}$-Nullmenge. Jedoch gilt $\forall (x,y) \in A$:
|
||||||
|
\[
|
||||||
|
f^{X,Y}(x,y) = 0 \neq \frac{4}{\pi^2} \underbrace{\sqrt{1-x^2}}_{> 0} \underbrace{\sqrt{1-y^2} }_{> 0}
|
||||||
|
.\] Also folgt $X$ und $Y$ nicht unabhängig.
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
|
\begin{aufgabe}
|
||||||
|
\begin{enumerate}[(a)]
|
||||||
|
\item Es gilt
|
||||||
|
\begin{salign*}
|
||||||
|
\mathbb{F}^{Z_p}(x) &= \P^{Z_p}((-\infty, x])\\
|
||||||
|
&= \P^{(-1)^{V_p}\cdot Y}((-\infty, x] \cap \{V_p = 0\}) + \P^{(-1)^{V_p}\cdot Y}((-\infty, x] \cap \{V_p = 1\})\\
|
||||||
|
&= \P^Y((-\infty, x] \cap \{V_p = 0\}) + \P^{-Y}((-\infty, x] \cap \{V_p = 1\})\\
|
||||||
|
&\stackrel{Y \indep V_p}{=} \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}) + \P(\{-Y \leq x\}) \cdot \P(\{V_p = 1\})\\
|
||||||
|
&\stackrel{\text{Symmetrie } N_{(0,1)}}{=} \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}) + \P(\{Y \leq x\}) \cdot \P(\{V_p = 1\})\\
|
||||||
|
&= \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}\cup \{V_p = 1\})\\
|
||||||
|
&= \P(\{Y \leq x\})\\
|
||||||
|
&= \mathbb{F}^{Y}(x)\\
|
||||||
|
\end{salign*}
|
||||||
|
Daher gilt $Z_p \sim Y \sim N_{(0,1)}$.
|
||||||
|
\item Es gilt
|
||||||
|
\begin{salign*}
|
||||||
|
\P(\{Y < -1, Z_p < -1\}) &= \P(\{Y < -1\} \cap \{V_p = 0\})
|
||||||
|
&\stackrel{Y \indep V_p}{=} \P(\{Y < -1\}) \cdot \P(\{V_p = 0\})\\
|
||||||
|
&= (1-p) \cdot \P(\{Y < -1\})
|
||||||
|
\end{salign*} und völlig analog
|
||||||
|
\begin{salign*}
|
||||||
|
\P(\{Y < -1, Z_p > 1\}) &= \P(\{Y < -1\} \cap \{V_p = 1\})
|
||||||
|
&\stackrel{Y \indep V_p}{=} \P(\{Y < -1\}) \cdot \P(\{V_p = 1\})\\
|
||||||
|
&= p \cdot \P(\{Y < -1\})
|
||||||
|
\end{salign*}
|
||||||
|
Angenommen, $Y \indep Z_p$. Dann gilt
|
||||||
|
\begin{salign*}
|
||||||
|
\P(\{Y < -1, Z_p < -1\}) &= \P(\{Y< -1\})\P(\{Z_p < -1\})\\
|
||||||
|
(1-p) \cdot \P(\{Y < -1\}) &\stackrel{\text{(a)}}{=} \P(\{Y < -1\})^2\\
|
||||||
|
(1-p) &= \P(\{Y < -1\})
|
||||||
|
\end{salign*} und völlig analog
|
||||||
|
\begin{salign*}
|
||||||
|
\P(\{Y < -1, Z_p > 1\}) &= \P(\{Y< -1\})\P(\{Z_p > 1\})\\
|
||||||
|
p \cdot \P(\{Y < -1\}) &\stackrel{\text{(a), Symmetrie } N_{(0,1)}}{=} \P(\{Y < -1\})^2\\
|
||||||
|
p &= \P(\{Y < -1\})
|
||||||
|
\end{salign*}
|
||||||
|
Nun führen wir eine Fallunterscheidung durch.
|
||||||
|
Für $p = \frac{1}{2}$ folgt $\P(\{Y < -1\}) < \P(\{Y < 0\}) \leq \frac{1}{2}$, Widerspruch zu $p = \P(\{Y < -1\})$.
|
||||||
|
Für $p \neq \frac{1}{2}$ erhalten wir aus $p = \P(\{Y < -1\}) = (1-p)$ ebenfalls einen Widerspruch.
|
||||||
|
Daher ist $Y \not \indep Z_p$.
|
||||||
|
\item Es gilt
|
||||||
|
\begin{salign*}
|
||||||
|
\E(YZ_p) &= \int_\R \int_\R yz \mathbbm{f}^{Y, Z_p}(y, z) \d{y}\d{z}\\
|
||||||
|
&\stackrel{Z_p = (-1)^{V_p}Y}{=} \int_{0,1} \int_\R y^2 (-1)^v \mathbbm{f}^{Y, V_p}(y, v) \d{y}\d{v}\\
|
||||||
|
&\stackrel{Y \indep V_p}{=} (1-p) \cdot \int_\R y^2 \mathbbm{f}^Y(y) \d{y} + p\cdot \int_\R -y^2 \mathbbm{f}^Y(y) \d{y}\\
|
||||||
|
&= (1 - 2p) \int_\R y^2 \mathbbm{f}^Y(y) \d{y}
|
||||||
|
\end{salign*}
|
||||||
|
Außerdem gilt
|
||||||
|
\begin{salign*}
|
||||||
|
\E(y)\E(Z_p) &= \int_\R y\mathbbm{f}^Y(y) \d{y} \int_\R z\mathbbm{f}^{Z_p}(z) \d{z}\\
|
||||||
|
&\stackrel{Z_p = (-1)^{V_p}Y}{=} \int_\R y\mathbbm{f}^Y(y) \d{y} \cdot \int_{0,1} \int_\R y (-1)^v \mathbbm{f}^{Y, V_p}(y, v) \d{y}\d{v}\\
|
||||||
|
&\stackrel{Y \indep V_p}{=} (1-p) \cdot \left(\int_\R y \mathbbm{f}^Y(y) \d{y}\right)^2 - p\cdot \left(\int_\R y^2 \mathbbm{f}^Y(y) \d{y}\right)^2\\
|
||||||
|
&= (1 - 2p) \left(\int_\R y \mathbbm{f}^Y(y) \d{y}\right)^2
|
||||||
|
\end{salign*}
|
||||||
|
Für $p = \frac{1}{2}$ erhalten wir daher
|
||||||
|
\begin{align*}
|
||||||
|
\cov(Y, Z_p) = \E(YZ_p) - \E(Y)\E(Z_p) = 0 - 0 = 0.
|
||||||
|
\end{align*}
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
|
\begin{aufgabe}
|
||||||
|
\begin{enumerate}[(a)]
|
||||||
|
\item Rechnen ergibt für $z \in \R$
|
||||||
|
\begin{salign*}
|
||||||
|
\mathbb{F}^{M_1}(z) &= \mathbb{P}\left( \{M_1 \le z\} \right) \\
|
||||||
|
&= \mathbb{P}\left( \left\{ \min_{i \in \{1, \ldots, n\} }{X_i} \le z\right\} \right) \\
|
||||||
|
&= 1 - \mathbb{P}\left( \bigcap_{i=1}^{n} \{X_i > z\} \right) \\
|
||||||
|
&\stackrel{\text{unabh.}}{=}
|
||||||
|
1 - \prod_{i=1}^{n} \mathbb{P}(\{X_i > z\}) \\
|
||||||
|
&\stackrel{\text{idv}}{=} 1 - (1 - \mathbb{F}^{X}(z))^{n} \\
|
||||||
|
\mathbb{F}^{M_2}(z) &= \mathbb{P}\left( \{M_2 \le z\} \right) \\
|
||||||
|
&= \mathbb{P}\left( \prod_{i=1}^{n} \{X_i \le z\} \right) \\
|
||||||
|
&\stackrel{\text{unabh.}}{=} \prod_{i=1}^{n} \mathbb{P}(\{X_i \le z\}) \\
|
||||||
|
&\stackrel{\text{idv}}{=} \mathbb{F}^{X}(z)^{n}
|
||||||
|
.\end{salign*}
|
||||||
|
\item Für $X_1 \sim \text{Exp}_{\lambda}$ ist $\mathbb{F}^{X_1} = (1 - \exp(- \lambda z))\mathbbm{1}_{\R^{+}}$. Damit folgt
|
||||||
|
\begin{salign*}
|
||||||
|
\mathbb{F}^{M_1}(z) &= 1 - (1 - \mathbb{F}_{\text{Exp}_{\lambda}}(z))^{n} \\
|
||||||
|
&= 1 - \left[ 1 - \left( 1 - \exp(-\lambda z) \right) \mathbbm{1}_{\R^{+}(z)} \right]^{n} \\
|
||||||
|
&= 1 - \left[ 1 - \mathbbm{1}_{\R^{+}}(z) + \exp(- \lambda z) \mathbbm{1}_{\R^{+}}(z)\right]^{n}\\
|
||||||
|
&= 1 - \begin{cases}
|
||||||
|
\exp(- \lambda n z) & z \in \R^{+} \\
|
||||||
|
1 & z \not\in \R^{+}
|
||||||
|
\end{cases}\\
|
||||||
|
&= (1 - \exp(- \lambda n z)) \mathbbm{1}_{\R^{+}} \\
|
||||||
|
&= \mathbb{F}_{\text{Exp}_{n\lambda}}
|
||||||
|
.\end{salign*}
|
||||||
|
Da die Verteilungsfunktion das W-Maß eindeutig festlegt, folgt
|
||||||
|
$M_1 \sim \text{Exp}_{n\lambda}$.
|
||||||
|
\item Für $X_1 \sim U_{[0, \theta]}$ ist die Dichte $f(x) = \frac{1}{\theta} \mathbbm{1}_{[0, \theta]}$
|
||||||
|
gegeben. Damit folgt
|
||||||
|
\begin{salign*}
|
||||||
|
\E_{\theta}(X_1) &= \int_{\R}^{} \frac{x}{\theta} \mathbbm{1}_{[0, \theta]} \d{x} \\
|
||||||
|
&= \frac{1}{\theta} \int_{0}^{\theta} x \d{x} \\
|
||||||
|
&= \frac{\theta}{2} \\
|
||||||
|
\E_{\theta}(X_1^2) &= \int_{\R}^{} \frac{x^2}{\theta} \mathbbm{1}_{[0, \theta]} \d{x} \\
|
||||||
|
&= \frac{1}{\theta} \int_{0}^{\theta} x^2 \d{x} \\
|
||||||
|
&= \frac{\theta^2}{3} \\
|
||||||
|
\var_{\theta}(X_1) &= \E_{\theta}(X_1^2) - \E_{\theta}(X_1)^2 \\
|
||||||
|
&= \frac{\theta^2}{3} - \frac{\theta^2}{4} \\
|
||||||
|
&= \frac{\theta^2}{12}
|
||||||
|
.\end{salign*}
|
||||||
|
Es gilt $f^{M_{2}} = (\mathbb{F}^{M_{2}})'$. Damit folgt für $z \in \R$:
|
||||||
|
\begin{salign*}
|
||||||
|
f^{M_2}(z) &= (\mathbb{F}^{M_2})'(z) \\
|
||||||
|
&= (\mathbb{F}^{X}(z)^{n})' \\
|
||||||
|
&= n (\mathbb{F}^{X}(z)^{n-1}) f^{X}(z) \\
|
||||||
|
&= n \left( \frac{(z \land \theta) \lor \theta}{\theta} \right)^{n-1}
|
||||||
|
\frac{1}{\theta} \mathbbm{1}_{[0, \theta]}(z) \\
|
||||||
|
&= \frac{n}{\theta^{n}} z^{n-1} \mathbbm{1}_{[0, \theta]}(z)
|
||||||
|
\intertext{Damit folgt}
|
||||||
|
\E_{\theta}(M_2) &= \int_{\R}^{} x \frac{n x^{n-1}}{\theta ^{n}} \mathbbm{1}_{[0, \theta]}
|
||||||
|
\d{x} \\
|
||||||
|
&= \frac{n}{\theta^{n}} \int_{0}^{\theta} x^{n} \d{x} \\
|
||||||
|
&= \frac{n}{n+1} \theta
|
||||||
|
.\end{salign*}
|
||||||
|
\item Rechnen ergibt
|
||||||
|
\begin{salign*}
|
||||||
|
\E_{\theta}(\overline{X}_n) &= \E_{\theta}\left( \frac{1}{n} \sum_{k=1}^{n} X_k \right) \\
|
||||||
|
&\stackrel{\text{lin.}}{=} \frac{1}{n} \sum_{k=1}^{n} \E_{\theta}(X_k) \\
|
||||||
|
&\stackrel{\text{idv}}{=} \E_{\theta}(X_1) \\
|
||||||
|
&= \frac{\theta}{2}
|
||||||
|
\intertext{Damit folgt}
|
||||||
|
\text{Bias}_{\theta}(\hat{\theta}_1) &= \E_{\theta}(\hat{\theta}_1 - \theta) \\
|
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|
&= 2 \E_{\theta}(\overline{X}_n) - \theta \\
|
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|
&= 2 \frac{\theta}{2} - \theta \\
|
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|
&= 0 \\
|
||||||
|
\text{Bias}_{\theta}(\hat{\theta}_2)
|
||||||
|
&= \E_{\theta}(\hat{\theta}_2 - \theta) \\
|
||||||
|
&= \E_{\theta}(M_2) - \theta \\
|
||||||
|
&= \frac{n}{n+1} \theta - \theta \\
|
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|
&= - \frac{\theta}{n+1}
|
||||||
|
\intertext{Nun rechne}
|
||||||
|
\E_{\theta}(M_2^2) &= \int_{\R}^{} x^2 \frac{n x^{n-1}}{\theta^{n}} \mathbbm{1}_{[0, \theta]}(x) \d{x} \\
|
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|
&= \frac{n}{\theta ^{n}} \int_{0}^{\theta} x^{n+1} \d{x} \\
|
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|
&= \frac{n}{n+2} \theta^2 \tageq \label{eq:2}
|
||||||
|
\intertext{
|
||||||
|
Es ist offensichtlich $\hat{\theta}_3$ nun erwartungstreu. Damit folgt für $n > 1$}
|
||||||
|
\var_{\theta}(\hat{\theta}_1) &= 4 \var_{\theta}(\overline{X}_n) \\
|
||||||
|
&\stackrel{\text{unabh.}}{=} \frac{4}{n^2} \sum_{k=1}^{n} \var_{\theta}(X_k) \\
|
||||||
|
&\stackrel{\text{idv}}{=} \frac{4}{n} \var_{\theta}(X_1) \\
|
||||||
|
&= \frac{\theta^2}{3n} \\
|
||||||
|
&> \frac{\theta^2}{n(n+2)} \\
|
||||||
|
&= \left( \frac{n+1}{n} \right)^2 \frac{n}{n+2} \theta^2 - \theta^2 \\
|
||||||
|
&\stackrel{\text{(\ref{eq:2})}}{=} \left( \frac{n+1}{n} \right)^2 \E_{\theta}(M_2^2) - \E_{\theta}\left( \frac{n+1}{n} M_2 \right)^2 \\
|
||||||
|
&= \E_{\theta} (\hat{\theta}_3^2) - \E_{\theta}(\hat{\theta}_3)^2 \\
|
||||||
|
&= \var_{\theta}(\hat{\theta}_3)
|
||||||
|
\intertext{Schlussendlich ergibt sich}
|
||||||
|
\text{MSE}_{\theta}(\hat{\theta}_1) &= \E_{\theta}(|\hat{\theta}_1 - \theta|^2) \\
|
||||||
|
&= 4 \E_{\theta}(\overline{X}_n^2) - \theta^2 \\
|
||||||
|
&= 4 \var_{\theta}(\overline{X}_n) + 4 \E_{\theta}(\overline{X}_n)^2 - \theta^2 \\
|
||||||
|
&\stackrel{\text{iid}}{=} \frac{4}{n} \var_{\theta}(X_1) \\
|
||||||
|
&= \frac{\theta^2}{3n} \\
|
||||||
|
\text{MSE}_{\theta}(\hat{\theta}_2) &= \E_{\theta}(\hat{\theta}_2^2) - 2 \theta \E_{\theta}(M_2)
|
||||||
|
+ \theta^2 \\
|
||||||
|
&= \frac{n}{n+2} \theta^2 - 2 \frac{n}{n+1} \theta^2
|
||||||
|
+ \theta^2 \\
|
||||||
|
&= \frac{2 \theta^2}{(n+2)(n+1)} \\
|
||||||
|
\text{MSE}_{\theta}(\hat{\theta}_3) &= \E_{\theta}(\hat{\theta}_3^2) - 2 \theta \E_{\theta}
|
||||||
|
(\hat{\theta}_3) + \theta^2 \\
|
||||||
|
&= \left( \frac{n+1}{n} \right)^2 \frac{n}{n+2} \theta^2 - \theta^2 \\
|
||||||
|
&= \frac{\theta^2}{n(n+2)}
|
||||||
|
.\end{salign*}
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
|
\begin{aufgabe}
|
||||||
|
\begin{enumerate}[(a)]
|
||||||
|
\item Wir wählen die Hypothesen $H_0 \colon \mu \leq \mu_0$ und $H_1 \colon \mu > \mu_0$.
|
||||||
|
Der Student-$t$-Test ist dann gegeben durch
|
||||||
|
\[
|
||||||
|
\phi_c^r = \mathbbm{1}_{\sqrt{n}(\overline{X_n} - \mu_0) \geq c \hat{S}_n}
|
||||||
|
\]
|
||||||
|
mit $ c= t_{(n-1), (1-\alpha)}$.
|
||||||
|
Wir berechnen also zunächst
|
||||||
|
\[
|
||||||
|
\overline{X_n} = \frac{1}{n}\sum_{i = 1}^{n} X_i = 103.64,
|
||||||
|
\]
|
||||||
|
\[
|
||||||
|
\hat{S}_n = \sqrt{\frac{1}{n-1} \sum_{i = 1}^{n} (X_i - \overline{X}_n)^2} \approx 5.22
|
||||||
|
\]
|
||||||
|
und
|
||||||
|
\[
|
||||||
|
c = t_{(n-1, 1-\alpha)} = t_{9, 0.95} = 1.833.
|
||||||
|
\]
|
||||||
|
Daraus folgt
|
||||||
|
\[
|
||||||
|
\phi_c^r = \mathbbm{1}_{\sqrt{10}(103.64-100) \geq 1.833 \cdot 5.22} = \mathbbm{1}_{11.51 \geq 9.57} = 1,
|
||||||
|
\]
|
||||||
|
wir lehnen also ab.
|
||||||
|
\item Wir wollen als Partition in richtige und falsche Parameter $\mathcal{R}_\mu = \{\mu\}$ und $\mathcal{F}_\mu = \R\setminus\{\mu\}$.
|
||||||
|
Dann erhalten wir als assoziierte Familie von Partitionen und Null- und Alternativhypothesen
|
||||||
|
$\mathscr H_\mu^0 = \{\mu\}$ und $\mathscr H_\mu^1 = \R \setminus \{\mu\}$.
|
||||||
|
Da der beidseitige Student-$t$-Test ein $\alpha$-Test der Nullhypothese $H_0\colon \mathscr H_\mu^0$ gegen die
|
||||||
|
Alternative $H_1 \colon \mathscr H_\mu^1$ für jedes $\mu \in \R$ ist, muss die assoziierte Bereichsschätzfunktion
|
||||||
|
für $(\{\mathcal R_\mu, \mathcal F_\mu\})_{\mu \in \R}$ ein $(1-\alpha)$-Konfidenzbereich sein.
|
||||||
|
Die assoziierte Bereichsschätzfunktion zum beidseitigen Student-$t$-Test
|
||||||
|
$\phi^b_{t_{(n-1), (1 - \alpha/2)}, \mu}(X_1, \dots X_n)$ ist gegeben durch
|
||||||
|
\begin{align*}
|
||||||
|
B(X_1, \dots, X_n) &= \{\mu \in \R: \phi^b_{t_{(n-1), (1 - \alpha/2)}, \mu}(X_1, \dots X_n) = 0\}\\
|
||||||
|
&= \{\mu \in \R: \sqrt{n}|\overline{X_n} - \mu| \leq t_{(n-1), (1-\alpha/2)} S\}\\
|
||||||
|
&= \{\mu \in \R: |\overline{X_n} - \mu| \leq \frac{S}{\sqrt{n}} t_{(n-1), (1-\alpha/2)}\}\\
|
||||||
|
&= \left[\overline{X_n} - \frac{S}{\sqrt{n}} t_{(n-1), (1-\alpha/2)}, \overline{X_n} + \frac{S}{\sqrt{n}} t_{(n-1), (1-\alpha/2)}\right]
|
||||||
|
\end{align*}
|
||||||
|
Das war zu zeigen.
|
||||||
|
\item Mithilfe unserer numerischen Resultate aus der (a) sowie der Aussage von Teilaufgabe (b) folgern wir, dass
|
||||||
|
\begin{align*}
|
||||||
|
[103.64 - \frac{5.22}{\sqrt{10}}t_{9, 0.975}, 103.64 + \frac{5.22}{\sqrt{10}}t_{9, 0.975}] &\subset [99.90, 107.38]
|
||||||
|
\end{align*}
|
||||||
|
ein 95\%-Konfidenzintervall ist.
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
|
\end{document}
|
||||||
Reference in New Issue
Block a user