Mat3 Probeklausur SoSe 2022 2

1

Bruchstelle $X\in [0,1]$
$a=[0,X)$
$b=(X,1]$

a)

$\mathbb{P}\left(X<\frac{1}{3}\vee X>\frac{2}{3}\right)=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

b)

$\frac{\frac{5}{8}}{2}=\frac{5}{16}$

$$1-\frac{5}{8}=\frac{3}{8}$$ $$f_X(x)=\{\begin{array}{ll}\frac{5}{16}& x\le \frac{1}{3}\vee \frac{2}{3}\le x\\ \frac{3}{8}& \frac{1}{3}<x\vee x<\frac{2}{3}\\ 0& \text{sonst}\end{array}$$

2

a)

$$\begin{array}{rl}1& ={\int }_{-\infty }^{\infty }{f}_X(x)dx\\ & ={\int }_0^{1}a+b{x}^{2}dx\\ & ={\left[ax+\frac{1}{3}b{x}^3\right]}_0^1\\ & =a+\frac{1}{3}b\end{array}$$ $$\begin{array}{rl}\mathbb{E}[X]& ={\int }_{-\infty }^{\infty }x\cdot f_X(x)dx\\ \frac{2}{3}& ={\int }_0^{1}ax+b{x}^{3}dx\\ & ={\left[\frac{1}{2}a{x}^2+\frac{1}{4}b{x}^4\right]}_0^1\\ & =\frac{1}{2}a+\frac{1}{4}b\end{array}$$ $$\begin{array}{rl}1& =a+\frac{1}{3}b\\ a& =1-\frac{1}{3}b\end{array}$$ $$\begin{array}{rl}\frac{2}{3}& =\frac{1}{2}a+\frac{1}{4}b\\ \frac{2}{3}& =\frac{1}{2}-\frac{1}{6}b+\frac{1}{4}b\\ \frac{24}{3}& =\frac{12}{2}-\frac{12}{6}b+\frac{12}{4}b\\ 4& =6-2b+3b\\ b& =2\end{array}$$ $$\begin{array}{rl}a& =1-\frac{1}{3}b\\ a& =1-\frac{1}{3}2\\ a& =\frac{3}{3}-\frac{2}{3}\\ a& =\frac{1}{3}\end{array}$$

3

a

$$Z(y,\theta )=\prod _{i=1}^{n}p(y_i,\theta )$$ $$L(y,\theta )=\sum _{i=1}^n\log (p(y_i,\theta ))$$ $$\begin{array}{rl}L(y,\theta )& =\sum _{i=1}^n\ln (2\theta y_i\exp (-\theta y_i^2))\\ & =n\cdot \ln (2)+n\cdot \ln (\theta )+\sum _{i=1}^n\ln (y_i)-\theta \sum _{i=1}^n{y}_i^2\end{array}$$ $$\begin{array}{rl}\frac{d}{d\theta }L(y,\theta )& =\frac{n}{\theta }-\sum _{i=1}^n{y}_i^2\\ 0& =\frac{n}{\theta }-\sum _{i=1}^n{y}_i^2\\ \frac{n}{\theta }& =\sum _{i=1}^n{y}_i^2\\ \theta & =\frac{n}{\sum _{i=1}^n{y}_i^2}\end{array}$$