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tut4: Add solutions for exercise 1

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Andreas Tsouchlos 2025-12-11 16:22:26 +01:00
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src/2025-12-09/presentation.tex
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@ -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}
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