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started App. Stats cousework
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\usepackage{mathoperators}
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\title{\textbf{Applied statistics (spring term 2019)}}
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\author{readers: \person{Nikolai Bode} and \person{Ksenia Shalonova}}
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\author{readers: Dr \person{Nikolai Bode} and Dr \person{ Ksenia Shalonova}}
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\date{written by \person{Henry Haustein}}
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\begin{document}
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Erasmus/Applied statistics/Coursework 1.pdf
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Erasmus/Applied statistics/Coursework 1.pdf
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Erasmus/Applied statistics/Coursework 1.tex
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Erasmus/Applied statistics/Coursework 1.tex
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\documentclass[british,a4paper,order=firstname]{mathscript}
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\usepackage{mathoperators}
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\title{\textbf{Applied statistics: Coursework 1}}
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\author{\person{Henry Haustein}}
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\begin{document}
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\pagenumbering{roman}
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\pagestyle{plain}
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\maketitle
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\hypertarget{tocpage}{}
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\tableofcontents
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\bookmark[dest=tocpage,level=1]{Table of contents}
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\pagebreak
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\pagenumbering{arabic}
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\pagestyle{fancy}
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\section{Task 1}
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\subsection{Part (1)}
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\subsection{Part (2)}
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\subsection{Part (3)}
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\subsection{Part (4)}
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\pagebreak
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\section{Task 2}
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\subsection{Part (1)}
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\subsection{Part (2)}
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\subsection{Part (3)}
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\pagebreak
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\section{Task 3}
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\subsection{Part (1)}
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\subsection{Part (2)}
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\pagebreak
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\section{Task 4}
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\subsection{Part (1)}
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The probability density function $f(t)$ is
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\begin{align}
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f(t) = \frac{2t\cdot\frac{\exp(-t^2)}{100}}{100} = \frac{t\cdot\exp(-t^2)}{5000}\notag
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\end{align}
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(x*exp(-x^2))/5000};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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The cumulative distribution function $F(t)$ is then
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\begin{align}
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F(t) &= \int_0^t f(\xi)\,\diff\xi \notag \\
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&= \int_0^t \frac{\xi\cdot\exp(-\xi^2)}{5000}\,\diff\xi\notag \\
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&= \frac{\exp(-t^2)\Big(\exp(t^2)-1\Big)}{10000} \notag
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\end{align}
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(exp(-x^2)*(exp(x^2)-1))/10000};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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For the survival function we get
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\begin{align}
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R(t) &= 1 - F(t) \notag \\
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&= \frac{\exp(-t^2)+9999}{10000} \notag
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\end{align}
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(exp(-x^2)+9999)/10000};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0.9999, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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y tick label style={
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/pgf/number format/.cd,
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precision=5,
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/tikz/.cd
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},
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]
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\addplot+[mark=none] {(exp(-x^2)+9999)/10000};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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To get the reliability of the component at $t=7$ we simply evaluate $R(7)$ which is 0.9999.
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The hazard function is defined as
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\begin{align}
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h(t) &= \frac{f(t)}{1-F(t)} \notag \\
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&= \frac{2t}{9999\cdot \exp(t^2)+1} \notag
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\end{align}
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(2*x)/(9999*exp(x^2)+1)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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The hazard function describes how an item ages where $t$ affects the risk of failure. It is the frequency with which the item fails, expressed in failures per unit of time.
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\subsection{Part (2)}
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Given $h(x)\sim(\sqrt{x})^{-1}$ we will try to find out the $shape$-parameter of the \person{Weibull} distribution first.
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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yticklabels={,,},
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xticklabels={,,}
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]
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\addplot+[mark=none] {1/sqrt(x)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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Comparing this graph to graphs of the hazard function with different $shape$-parameters we see that $shape=0.5$ fits best.
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\begin{center}
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\begin{tabular}{p{5cm}|p{5cm}|p{5cm}}
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$shape = 0.5$ & $shape = 1$ & $shape = 2$ \\
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\hline
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\multicolumn{3}{c}{\cellcolor{gray!50}\textbf{Hazard function} $\left(h = \frac{\text{PDF}}{1-\text{CDF}}\right)$} \\
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\hline
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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]
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\addplot+[mark=none] {(0.5*x^(-0.5)*exp(-x^0.5))/(1-(1-exp(-x^0.5)))};
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\end{axis}
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\end{tikzpicture} &
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(1/exp(x))/(1-(1-1/exp(x)))};
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\draw[blue] (axis cs: 0,1) -- (axis cs: 2,1);
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\end{axis}
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\end{tikzpicture} &
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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]
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\addplot+[mark=none] {(2*x^(1)*exp(-x^2))/(1-(1-exp(-x^2)))};
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\end{axis}
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\end{tikzpicture} \\
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\end{tabular}
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\end{center}
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To get the $scale$-parameter of the distribution we use the other provided information:
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\begin{align}
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5 &= \mu \notag \\
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&= scale\cdot\Gamma\left(1+\frac{1}{shape}\right) \notag \\
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&= scale\cdot\Gamma(3) \notag \\
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\Rightarrow scale &= \frac{5}{2} \notag
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\end{align}
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Let's build the survival function:
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\begin{align}
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R(t) &= 1-\Bigg(1-\exp\left(-\sqrt{\frac{x}{\nicefrac{5}{2}}}\right)\Bigg) \notag \\
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&= \exp(-\sqrt{x}\cdot\sqrt{2.5})\notag
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\end{align}
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That mean that the probability of surviving 6 years (30 years) is $R(6) = 0.0208$ ($R(30) = 0.0002$).
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\end{document}
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