387 lines
13 KiB
TeX
387 lines
13 KiB
TeX
\documentclass[12pt]{article}
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\usepackage{natbib}
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\usepackage{url}
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\usepackage{mathtools}%
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\usepackage{booktabs}%
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\title{\color{report_main}{Assignment Econometrics 2024}} % Title
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\author{Hendrik Marcel W Tillemans} % Author
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\date{\today} % Date
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\makeatletter
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% This is where the actual document starts
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%
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% This section details the group information
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{titlepage}
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\centering
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\vspace*{0.5 cm}
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\includegraphics[scale = 0.95]{../figures/vub.png}\\[1.0 cm] % University Logo
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\textsc{\LARGE \newline\newline Free University Brussels}\\[2.0 cm] % University Name
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\textsc{\Large \color{report_main}{Class: Econometrics}}\\[0.5 cm] % Course Code
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\rule{\linewidth}{0.2 mm} \\[0.4 cm]
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{ \huge \bfseries \thetitle}\\
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\rule{\linewidth}{0.2 mm} \\[1.5 cm]
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\begin{minipage}{0.5\textwidth}
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\begin{flushleft} \large
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\emph{Professor:}\\
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Jeroen Kerkhof\\
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Faculty of Economic Sciences\\
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\end{flushleft}
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\end{minipage}~
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\begin{minipage}{0.4\textwidth}
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\begin{flushright} \large
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\emph{Group:} \\
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Hendrik Marcel W Tillemans\\
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\end{flushright}
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\end{minipage}\\[2 cm]
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% takes the current date
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\thedate
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\end{titlepage}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% This details the inclusion (or not) of the table of contents
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% and list of figures and tables.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\tableofcontents
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\pagebreak
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\listoffigures
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\listoftables
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\pagebreak
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% This is the start of the actual document content
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% You can just write text in here as you would in any other word processor.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Simulation Study}
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\subsection{Question 1.1}
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We investigate a linear model with noise
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\[y=\beta_0 + \beta_1 x1 + \beta_2 x2 + u\]
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where
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\[x1 \sim \mathcal{N}(3,\,36)\]
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\[x2 \sim \mathcal{N}(2,\,25)\]
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\[u \sim \mathcal{N}(0,\,9)\]
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In figure \ref{fig::plot_1_1} we have a 3D representation of the generated model.
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\begin{figure}[hb]
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\includegraphics[width=0.6\paperwidth]{../figures/question_1_1}
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\caption{Generated points for Question 1.1.}
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\label{fig::plot_1_1}
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\end{figure}
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\subsection{Question 1.2}
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We now estimate the parameters of $\beta_0$, $\beta_1$ and $\beta_2$ using the \textbf{Ordinary Least Squares} (OLS) method. With model:
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\[y_i=\beta_0 + \beta_1 x_1 + \beta_2 x2 + u_i\]
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\begin{table}[ht]
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\centering
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\input{table_1_2}
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\caption{Linear Fit on Generated Data}
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\label{tab::table_1_2}
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\end{table}
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In table 1 we can see that those estimates are are close to their true values within 2\%. Because our estimation model is the same as the model used to generate the data, we have a sufficient number of points and the assumptions of OLS are satisfied. In this situation we can expect good results of OLS estimation.
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\subsection{Question 1.3}
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If we compare the estimates with those of question 1.2. We see that the estimate of the intersect is not close to the true value with a difference of 4.
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We can explain the bias of $\beta_0$ because this new model has a new error term: $\beta_2x_{2i} + u_i$. This error term no longer has an expacted value of 0, but in fact $\beta_2E(x_{2i}) + E(u_i)= 4$ wich is very close to the bias we find. For $\beta_1$, there is little to no bias. This can we explained because $x_2$ and $u$ are stochasticaly independent from $x_1$. The standard error is bigger because $\beta_2E(x_{2i}) + E(u_i)$ has a bigger variance than just $u$.
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Wich model would you choose? If I have sufficient calculation power, I would choose model 1.2 as it is much more accurate. However for a very resource constraint situation model 1.3 might give acceptable estimates.
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\begin{table}[ht]
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\centering
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\input{table_1_3}
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\caption{Linear Fit with 1 Variable}
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\label{tab::table_1_3}
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\end{table}
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\subsection{Question 1.4}
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In figure \ref{fig::plot_1_4} we have a 3D representation of the generated model.
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\begin{figure}[ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_1_4}
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\caption{Generated points for Question 1.4.}
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\label{fig::plot_1_4}
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\end{figure}
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The estimation results compared to the results in question 1.2 are similar, there is very little bias. It appears that $x_2^{new}$ is sufficiently independent from $x_1$. We expected very little bias because $x2_{new}$ has a large independent part compared to $x_1$. The standard errors of the estimates of $\beta_1$ and $\beta_2$ are about 25\% higher wich can be explained partly bij the lower standard deviation in $x_2^{new}$.
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\begin{table}[ht]
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\centering
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\input{table_1_4}
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\caption{New Linear Fit on Generated Data}
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\label{tab::table_1_4}
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\end{table}
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\subsection{Question 1.5}
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Similar as in question 1.3 we estimated the parameter with a single independent variable.
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\begin{table}[ht]
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\centering
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\input{table_1_5}
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\caption{Linear Fit with 1 Variable}
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\label{tab::table_1_5}
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\end{table}
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The OLS estimators for the slope coefficients are biased. We see that $\beta_1$ is $-3$ instead of the true value of $-4$. We can explain this bias in the following way, lets start from the model.
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\[y^{new}=\beta_0 + \beta_1 x_1 + \beta_2 x_2^{new} + u_i\]
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We now have:
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\[x_2^{new} = 0.5 * x_1 + x_2^{'}\]
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Where:
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\[x_2^{'}\sim \mathcal{N}(5,\,16)\]
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Substituting in the model:
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\[\Longrightarrow y^{new}=\beta_0 + \beta_1 x_1 + \beta_2(0.5 * x_1 + x_2^{'}) + u_i\]
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Lets fill in the betas with the actual values:
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\[\Longrightarrow y^{new}= 3 + -4 x_1 + 2(0.5 * x_1 + x_2^{'}) + u_i\]
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\[\Leftrightarrow y^{new}= 3 - 4 x_1 + x_1 + 2x_2^{'}) + u_i\]
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\[\Leftrightarrow [y^{new}= 3 - 3 x_1 + 2x_2^{'}) + u_i\]
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Here we can see in table \ref{tab::table_1_5} easily that the OLS estimator will find -3 as the estimate for $\beta_1$.
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Similarly as in question 1.3 we can explain the bias on the intercept.
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\subsection{Question 1.6}
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Now we replace $x_1$ in the original model with
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\[x_1 \sim \mathcal{N}(3,\,1)\]
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If we now estimate the parameters we find:
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\begin{table}[ht]
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\centering
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\input{table_1_6}
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\caption{Generate Data with Small Variance on x1}
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\label{tab::table_1_6}
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\end{table}
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We find in table \ref{tab::table_1_6} that the parameters are essentially unbiased but have a bigger standard error for the intersect and $\beta_1$. The standard error of $\beta_1$ is 6 times bigger (from 0.016 to 0.10). We see no difference of the estimates $\beta_2$. Because nothing has changed in $x_2$.
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\begin{figure}[ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_1_6}
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\caption{Generated points for Question 1.6.}
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\label{fig::plot_1_6}
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\end{figure}
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We expected a similar estimation result as in 1.2 because there are no changes except of the standard deviation of $x_1$. This means that the OLS assumptions are equally valid and we expect unbiased estimates.
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We can explain the difference in standard error of the estimates of $\beta_1$ using the formula of $Var(\beta_1)$.
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\[Var(\beta_1) = \sigma^2(X^tX)_{11}^{-1}\]
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We can write this as
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\[Var(\beta_1) = \sigma^2/Var(x_1)\]
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This means that $Var(\beta_1) \sim 1/Var(x_1)$.
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Because $Var(x_1)$ changed from 36 in to 1, we expect the standard error to be $/sqrd(36) = 6$ times bigger. Which is exactly what we found.
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If the standard deviation from $x_1$ changes to 0, $\beta_1$ cannot we calculated. As we have seen with the no multicollinearity assumption.
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\section{Empirical Investigation}
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\subsection{Question 2.1}
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We retain 2510 observations.
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\begin{table}[ht]
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\centering
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\input{summary_stats}
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\caption{Generate Data with Small Variance on x1}
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\label{tab::summary_stats}
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\end{table}
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\subsection{Question 2.2}
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\begin{figure} [ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_2_2_wage}
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\caption{Histogram wage}
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\label{fig::question_2_2_wage}
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\end{figure}
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\begin{figure} [ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_2_2_lwage}
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\caption{Histogram lwage}
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\label{fig::question_2_2_lwage}
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\end{figure}
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The lwage histogram in fig \ref{fig::question_2_2_lwage} is nicely centered so there is no need to remove any outliners. This is also close to a normal distribution. The wage historgam in fig \ref{fig::question_2_2_wage} is not symmetrical but is leaning to the left. Clealy not normal distributed.
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\subsection{Question 2.3}
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We are going to investigate the correlation between the variables wage, age, school, man, malay, chinese and indian.
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\begin{table}[ht]
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\centering
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\input{table_2_3}
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\caption{Correlation matrix}
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\label{tab::table_2_3}
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\end{table}
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We can see in table \ref{tab::table_2_3} that there is a positive correlation between wage and school. It means that people who go longer to school will get a higher wage. There is a negative correlation between age and school. The younger generation is higher educated than older generation. Chinese citizens are better payed than malay, indian citizens have a negative correlation with wage.
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\subsection{Question 2.4}
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We estimate a regression for lwage using the variables chinese and indian. We can calculate malay influence from the results.
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\begin{table}[ht]
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\centering
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\input{results_24}
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\caption{Linear model lwage}
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\label{tab::results_24}
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\end{table}
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$R^{2} = 00.0255$ this means that there is a very weak correlation found. If we look at the coefficients in table \ref{tab::results_24}. We see a negative value of 0.17 for indian and a positive value of 0.14 for chinese. This gives a slightly positive value for the malay of $0.14 - 0.17 + 0.03 = 0$
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This results implicate that there is a wage gap based on ethnicity.
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\subsection{Question 2.5}
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We estimate a regression for lwage using the variables chinese, indian and school.
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\begin{table}[ht]
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\centering
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\input{results_25}
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\caption{Linear model lwage/school}
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\label{tab::results_25}
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\end{table}
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$R^{2} = 0.224$ this means that there is a weak correlation found. If we look at the coefficients in table \ref{tab::results_25}. We see a negative value of 0.07 for indian and a positive value of 0.18 for chinese. This gives a negative value for the malay of $0.18 - 0.07 - 0.11 = 0$
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To see if lwage vs years of schooling is not linear. We plot it:
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\begin{figure} [ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_2_5}
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\caption{lwage vs school}
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\label{fig::question_2_5}
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\end{figure}
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In figure \ref{fig::question_2_5} there is no obvious non-linearity.
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\subsection{Question 2.6}
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We estimate a regression for lwage using the variables chinese, indian and school.
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\begin{table}[ht]
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\centering
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\input{results_26}
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\caption{Linear model lwage/age}
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\label{tab::results_26}
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\end{table}
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$R^{2} = 0.370$ this means that there is a weak correlation found. If we look at the coefficients in table \ref{tab::results_26}.
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To see if lwage vs age of schooling is not linear. We plot it:
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\begin{figure} [ht]
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\includegraphics[width=0.6\paperwidth]{../figures/question_2_6}
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\caption{lwage vs age}
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\label{fig::question_2_6}
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\end{figure}
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In figure \ref{fig::question_2_6} there is a banana shaped model indicating an non-linear relationship with a peak earnings around 40.
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We can use \textbf{agesq} to do a parabolic fit. If we run this model we get:
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\begin{table}[ht]
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\centering
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\input{results_26b}
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\caption{parabolic model lwage/age}
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\label{tab::results_26b}
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\end{table}
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We find a $R^{2} = 0.429$ which is higher than without the agesq.
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In table \ref{tab::results_26b} a negative coefficient for agesq wich explains the parabolic distribution with a maximum.
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\subsection{Question 2.8}
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\begin{table}[ht]
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\centering
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\input{results_28}
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\caption{Linear model 2.8}
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\label{tab::results_28}
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\end{table}
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From the table \ref{tab::results_28} we can conclude that age does not differ substantially between the tables. For school we see a little difference.
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\end{document}
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