From e516317a4ef65963436d5a3cb4bf4a0bf27da654 Mon Sep 17 00:00:00 2001 From: Andreas Tsouchlos Date: Thu, 11 Dec 2025 16:22:26 +0100 Subject: [PATCH] tut4: Add solutions for exercise 1 --- src/2025-12-09/presentation.tex | 148 ++++++++++++++++++++++++++++++++ 1 file changed, 148 insertions(+) diff --git a/src/2025-12-09/presentation.tex b/src/2025-12-09/presentation.tex index a7fa370..8b0655f 100644 --- a/src/2025-12-09/presentation.tex +++ b/src/2025-12-09/presentation.tex @@ -118,7 +118,155 @@ $\{\omega : 1 < X(\omega) \le 2\}$? \end{enumerate} % tex-fmt: on +\end{frame} +\begin{frame} + \frametitle{Aufgabe 1: Stetige Verteilungen} + + \vspace*{-15mm} + + Die Zufallsvariable X besitze die Dichte + + % tex-fmt: off + \begin{align*} + f_X (x) = \left\{ + \begin{array}{ll} + C \cdot x e^{-ax^2}, & x \ge 0 \\ + 0, &\text{sonst} + \end{array} + \right. + \end{align*} + % tex-fmt: on + + mit dem Parameter $a > 0$. + + % tex-fmt: off + \begin{enumerate}[a{)}] + \item Bestimmen Sie den Koeffizienten $C$, sodass $f_X(x)$ eine + Wahrscheinlichkeitsdichte ist. Welche Eigenschaften muss eine + \textbf{Wahrscheinlichkeitsdichte} erfüllen? Skizzieren Sie + $f_X (x)$ für $a = 0{,}5$. + \pause\begin{columns} + \column{\kitthreecolumns} + \begin{align*} + \text{Eigenschaften:} \hspace{5mm} + \left\{ + \begin{array}{rl} + f_X(x) &\ge 0 \\[3mm] + \displaystyle\int_{-\infty}^{\infty} f_X(x) dx &= 1 + \end{array} + \right. + \end{align*} + \pause\begin{gather*} + \int_{-\infty}^{\infty} f_X(x) dx + = \int_{-\infty}^{\infty} C\cdot x e^{-ax^2} dx + = \frac{C}{-2a} \int_{-\infty}^{\infty} (-2ax) e^{-ax^2} dx \\ + = \frac{C}{-2a} \int_{-\infty}^{\infty} (e^{-ax^2})' dx + = \frac{C}{-2a} \mleft[ e^{-ax^2} \mright]_0^{\infty} \overset{!}{=} 1 \hspace{10mm} \Rightarrow C = 2a + \end{gather*} + \centering + \column{\kitthreecolumns} + \pause \begin{align*} + f_X(x) = + \left\{ + \begin{array}{ll} + 2ax \cdot e^{-ax^2}, & x\ge 0\\ + 0, & \text{sonst} + \end{array} + \right. + \end{align*} + \begin{figure}[H] + \centering + \begin{tikzpicture} + \begin{axis}[ + domain=0:5, + width=12cm, + height=5cm, + samples=100, + xlabel={$x$}, + ylabel={$f_X(x)$}, + ] + \addplot+[mark=none, line width=1pt] + {x * exp(-0.5*x*x)}; + % {x *exp(-a*x*x)}; + \end{axis} + \end{tikzpicture} + \end{figure} + \end{columns} + \end{enumerate} + % tex-fmt: on +\end{frame} + +\begin{frame} + \frametitle{Aufgabe 1: Stetige Verteilungen} + + \vspace*{-20mm} + + % tex-fmt: off + \begin{enumerate}[a{)}] + \item Welche Eigenschaften muss eine \textbf{Verteilungsfunktion} + erfüllen? + \pause\vspace{-10mm}\begin{columns}[t] + \column{\kitonecolumn} + \column{\kittwocolumns} + \centering + \begin{gather*} + \lim_{x\rightarrow -\infty} F_X(x) = 0\\ + \lim_{x\rightarrow +\infty} F_X(x) = 1 + \end{gather*} + \column{\kittwocolumns} + \centering + \begin{gather*} + x_1 \le x_2 \Rightarrow F_X(x_1) \le F_X(x_2) \\ + F_X(x+) = \lim_{h\rightarrow 0^+} F_X (x+h) = F_X(x) + \hspace{5mm}\forall x\in \mathbb{R} + \end{gather*} + \column{\kitonecolumn} + \end{columns} + \pause\item Berechnen und skizzieren Sie die Verteilungsfunktion $F_X (x)$. + \begin{gather*} + f_X(x) = 2ax\cdot e^{-ax^2}, \hspace{5mm} x\ge 0 + \end{gather*} + \pause \vspace*{-6mm}\begin{gather*} + F_X(x) = \int_{-\infty}^{x} f_X(u) du + = \left\{ \begin{array}{ll} + \displaystyle\int_{0}^{x} 2au\cdot e^{-au^2} du, & x\ge 0 \\ + 0, & x < 0 + \end{array} \right. + \hspace{5mm} = \left\{ \begin{array}{ll} + \mleft[ -e^{-au^2} \mright]_0^{x}, & x\ge 0 \\ + 0, & x < 0 + \end{array} \right. + \hspace{5mm} = \left\{ \begin{array}{ll} + 1 - e^{-ax^2}, & x\ge 0\\ + 0, & x < 0 + \end{array} \right. + \end{gather*} + \pause\begin{figure}[H] + \centering + + \begin{tikzpicture} + \begin{axis}[ + domain=0:5, + width=14cm, + height=5cm, + xlabel={$x$}, + ylabel={$F_X(x)$}, + ] + \addplot+[mark=none, line width=1pt] + {1 - exp(-0.5 * x*x)}; + \end{axis} + \end{tikzpicture} + \end{figure} + \vspace*{-3mm} + \pause\item Welche Wahrscheinlichkeit hat das Ereignis + $\{\omega : 1 < X(\omega) \le 2\}$? + \pause \begin{gather*} + P(\mleft\{ \omega: 1 < X(\omega) \le 2 \mright\}) + = P(1 < X \le 2) = F_X(2) - F_X(1) = e^{-a} - e^{-4a} + \end{gather*} + \end{enumerate} + % tex-fmt: off \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%