# ward/vub

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math stat oef -- +ch3 oef4

 @@ -26,6 +26,7 @@ \newcommand{\C}{{\mathbb C}} \newcommand{\HQ}{{\mathbb H}} +\newcommand{\Var}{\mathrm{Var}} \newcommand{\Prob}{{\mathbb P}} \newcommand{\E}{{\mathbb E}} @@ -449,6 +450,95 @@ \section{Hypotheses testing} \begin{oplossing} \end{oplossing} +\section{Linear Regression} + +\begin{opgave} +\end{opgave} +\begin{oplossing} +\end{oplossing} + +\begin{opgave} +\end{opgave} +\begin{oplossing} +\end{oplossing} + +\begin{opgave} +\end{opgave} +\begin{oplossing} +\end{oplossing} + +\begin{opgave}%TODO Change the assignment \mu into \mu_0 to differentiate? + Two laboraties each take $c$ measurements on the same standard $\mu$. + Consider the model + \begin{align*} + Y_{ij} &= \mu + \epsilon_{ij} &&i = 1, 2 \text{ and } j = 1, \dots, c + \end{align*} + ... %TODO complete +\end{opgave} +\begin{oplossing} + \begin{align*} + \Var(Y_{1j}) &= \sigma^2\\ + \Var(Y_{2j}) &= 4\sigma^2 + \end{align*} + + First make the variances equal. + $+ \begin{cases} + Y_{1j} = \mu + \epsilon_{1j}\\ + \frac{Y_{2j}}{2} = \frac{\mu}{2} + \frac{\epsilon_{2j}}{2} + \end{cases} +$ + + The Gaussian linear model. + \begin{align*} + \mathbf{Y} &= C\mathbf{\beta} + \mathbf{\epsilon}\\ + &= \begin{pmatrix}1\\\vdots\\1\\0.5\\\vdots\\0.5\end{pmatrix} + \begin{pmatrix}\mu\end{pmatrix} + + \begin{pmatrix} + \epsilon_{11}\\\vdots\\\epsilon_{1c}\\ + \epsilon_{21}\\\vdots\\\epsilon_{2c} + \end{pmatrix} + \end{align*} + with $C \in \R^{2c \times 1}$, $\mathbf{\beta} \in \R^{1 \times 1}$ and + $\mathbf{\epsilon} \in \R^{2c \times 1}$. The $\mu$ is the one from the + assignment, not the one that equals $C\mathbf{\beta}$ (afaik). + + In other words to estimate the $\mu$ that they ask for, we need to estimate + $\beta$ in our Gaussian linear model. + + \begin{align*} + \begin{pmatrix}\mu\end{pmatrix} &= \hat{\beta}\\ + &= (C^{t} C)^{-1} C^{t} \mathbf{Y}\\ + &= \left[ + \begin{pmatrix}1 &\dots &1 &0.5 &\dots &0.5\end{pmatrix} + \begin{pmatrix}1\\\vdots\\1\\0.5\\\vdots\\0.5\end{pmatrix} + \right]^{-1} + \begin{pmatrix}1 &\dots &1 &0.5 &\dots &0.5\end{pmatrix} + \mathbf{Y}\\ + &= \left[ + \begin{pmatrix}c + \frac{c}{4}\end{pmatrix} + \right]^{-1} + \begin{pmatrix}1 &\dots &1 &0.5 &\dots &0.5\end{pmatrix} + \mathbf{Y}\\ + &= \begin{pmatrix}\frac{4}{5c}\end{pmatrix} + \begin{pmatrix}1 &\dots &1 &0.5 &\dots &0.5\end{pmatrix} + \mathbf{Y}\\ + &= \begin{pmatrix}\frac{4}{5c} &\dots &\frac{4}{5c} &\frac{2}{5c} &\dots &\frac{2}{5c}\end{pmatrix} + \begin{pmatrix}Y_{11}\\\vdots\\Y_{1c}\\Y_{21}/2\\\vdots\\Y_{2c}/2\end{pmatrix} + \end{align*} + and we get what we needed to prove. +\end{oplossing} + +\begin{opgave} +\end{opgave} +\begin{oplossing} +\end{oplossing} + +\begin{opgave} +\end{opgave} +\begin{oplossing} +\end{oplossing} + %\newpage %\begin{thebibliography}{99} % \bibitem{VOORBEELD} Voorbeeld, Ward Muylaert.