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% Document Info
\author{dsjac3}
\pdfinfo{
  /Title (econometrics.pdf)
  /Creator (Cheatography)
  /Author (dsjac3)
  /Subject (Econometrics Cheat Sheet)
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Econometrics Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{dsjac3} via \textcolor{DarkBackground}{\uline{cheatography.com/31611/cs/9635/}}}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}dsjac3 \\
  \uline{cheatography.com/dsjac3} \\
  \end{tabulary}
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\begin{tabulary}{5.8cm}{L}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 31st October, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{properties of OLS Matrix}}  \tn
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Sum of Squared Residuals & (y − Xβˆ)′(y − Xβˆ) \tn 
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 & y′y − βˆ′X′y − y′Xβˆ + βˆ′X′Xβˆ \tn 
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 & y′y − 2βˆ′X′y + βˆ′X′Xβˆ \tn 
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Minimise the SSR & ∂(SSR)/∂βˆ = −2X′y + 2X′Xβˆ = 0 \tn 
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from the minimum we get: "normal equation" & (X′X)βˆ = X′y \tn 
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Solve for OLS estimator βˆ; by pre multiplying both sides by (X′X) & \seqsplit{(X′X)−1(X′X)βˆ} = (X′X)−1X′y \tn 
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by definition, (X′X)−1(X′X) = I & Iβˆ = (X′X)−1X′y \tn 
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 & βˆ = (X ′ X )−1 X ′ y \tn 
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\mymulticolumn{2}{x{8.4cm}}{{\bf{Properties}}} \tn 
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The observed values of X are uncorrelated with the residuals. & X′e = 0 implies that for every column xk of X, x′ke = 0. \tn 
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substitute in y = Xβˆ + e into normal equation & (X′X)βˆ = X′(Xβˆ + e) \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{properties of OLS Matrix (cont)}}  \tn
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 & (X′X)βˆ = (X′X)βˆ + X′e \tn 
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 & X′e = 0 \tn 
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The sum of the residuals is zero. & If there is a constant, then the first column in X (i.e. X1) will be a column of ones. This means that for the first element in the X′e vector (i.e. X11 ×e1 +X12 ×e2 +...+X1n ×en) to be zero, it must be the case that  ei = 0. \tn 
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The sample mean of the residuals is zero. & e= ∑e i/n  = 0. \tn 
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The regression hyperplane passes through the means of the observed values (X    and y). & This follows from the fact that e = 0. Recall that e = y − Xβˆ. Dividing by the number of observations, we get e = y − xβˆ = 0. This implies that y = xβˆ. This shows that the regression hyperplane goes through the point of means of the data. \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{properties of OLS Matrix (cont)}}  \tn
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The predicted values of y are uncorrelated with the residuals. & ˆ′e = (Xβˆ)′e = b′X′e = 0 \tn 
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\mymulticolumn{2}{x{8.4cm}}{The mean of the predicted Y's for the sample will equal the mean of the observed Y's : y\textasciicircum{}-=y-} \tn 
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{\bf{The Gauss-Markov Theorem:}} Proof that βˆ is an unbiased estimator of β & βˆ = (X′X)\textasciicircum{}−1\textasciicircum{}X′y=(X′X)\textasciicircum{}−1\textasciicircum{}X′(Xβ + ε) \tn 
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 & β + (X′X)\textasciicircum{}−1\textasciicircum{}X′ε \tn 
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given (X′X)\textasciicircum{}−1\textasciicircum{}X′X = I & E{[}βˆ{]} = E{[}β{]} + E{[}(X′X)\textasciicircum{}−1\textasciicircum{}X′ε{]} = β + (X′X)\textasciicircum{}−1\textasciicircum{}X′E{[}ε{]} \tn 
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where E{[}X′ε{]}=0 & E{[}βˆ{]}=β \tn 
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{\bf{Proof that βˆ is a linear estimator of β.}} & βˆ = β + (X′X)\textasciicircum{}−1\textasciicircum{}X′ε; where (X′X)\textasciicircum{}−1\textasciicircum{}X′= A \tn 
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 & βˆ = β + Aε =\textgreater{} linear equation \tn 
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\begin{tabularx}{8.4cm}{x{3.12 cm} x{4.88 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Hetroskedasticity}}  \tn
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consequence: & the statistics used to test hypotheses under Gauss-Markov assumptions are not valid in the presence of hetroskedasticity. \tn 
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Valid estimator (any form) & ∑{[}(x1- x-)\textasciicircum{}2\textasciicircum{} uˆi\textasciicircum{}2\textasciicircum{}{]}/{[}SST\textasciicircum{}2\textasciicircum{}x{]} \tn 
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 & SSTx=∑(x1- x-)\textasciicircum{}2\textasciicircum{} \tn 
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Robust Standard error & Varˆ(βˆj)=∑{[}rˆij\textasciicircum{}2\textasciicircum{}û\textasciicircum{}2\textasciicircum{}i{]}/{[}SSR\textasciicircum{}2\textasciicircum{}j{]} \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Inference}}  \tn
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Normality Assumption: & zero mean and Variance \tn 
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 & Var(u)= σ\textasciicircum{}2\textasciicircum{} \tn 
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{\bf{T-test:}} & (βˆ j- β j)/se(βˆ j)\textasciitilde{} t n-k-1 =t df \tn 
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H0 : βj = 0 & used in testing hypotheses about a single population parameter as in . \tn 
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Test statistic & t βˆ j=(βˆ j)/se(βˆ j)\textasciitilde{} t n-k-1 \tn 
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 & t = (estimate − hypothesised value)/  standard error \tn 
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\mymulticolumn{2}{x{8.4cm}}{{\bf{Alternative Hypothesis/one sided}}} \tn 
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H1: βj \textgreater{} 0 & t βˆj \textgreater{} c  {[}c @5\%{]} \tn 
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H1: βj \textless{} 0 & t βˆj \textless{}- c {[}c @5\%{]} \tn 
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\mymulticolumn{2}{x{8.4cm}}{{\bf{Two sided}}} \tn 
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H1: βj =/= 0 & |tβˆj | \textgreater{} c   {[}c @2.5\%{]} \tn 
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If H0, rejected & x j is statistically significant, (significantly different from zero), @ the 5\% level \tn 
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if H0, not rejected & x j is statistically insignificant @the 5\% level \tn 
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P-value & smallest significant level at which the null hypotheses would be rejected \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Inference (cont)}}  \tn
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{\bf{Confidence Interval}} & βˆj ±c·se(βˆj) \tn 
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 & where c is 97.5 percentile in a t n-k-1 distribution \tn 
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CI given; @ 5\% significant level & H0 :βj =aj is rejected against H1:βj = ̸=aj ; if aj is not in the 95\% confidence interval \tn 
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H0:β1\textless{}β2 ⇔ β1−β2\textless{}0 & t= (βˆ1−βˆ2) /se(βˆ1 − βˆ2) \tn 
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\mymulticolumn{2}{x{8.4cm}}{se(βˆ1 − βˆ2) = √Var(βˆ1 − βˆ2)} \tn 
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 & Var(βˆ1 − βˆ2) = Var(βˆ1) + Var(βˆ2) − 2Cov(βˆ1, βˆ2) \tn 
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{\emph{alternative to calculating se(βˆ1 − βˆ2)}} & Let θ = βˆ1 − βˆ2; β1 = θ + βˆ2 \tn 
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H0: θ=0, H1: θ\textless{}0 & Substituting β1 = θ + βˆ2 into the model we obtain \tn 
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\mymulticolumn{2}{x{8.4cm}}{β0 +θ x1 +β2(x1 +x2)+β3 x3 +u} \tn 
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{\bf{F Test}} & \{\{noshy\}\} F ={[}(SSRr-SSRur )/q{]} / {[}SSRur/(n-k-1){]} \tn 
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q & =number of restrictions \tn 
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n-k-1= df ur & = df r- df ur \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Inference (cont)}}  \tn
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{\bf{R\textasciicircum{}2\textasciicircum{} F stat}} & SSR= SST(1 - R\textasciicircum{}2\textasciicircum{} ) \tn 
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 & \{\{noshy\}\} F= {[}(R\textasciicircum{}2\textasciicircum{}ur-R\textasciicircum{}2\textasciicircum{}r)/q{]} / {[}1-R\textasciicircum{}2\textasciicircum{}ur)/(df ur){]} \tn 
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\mymulticolumn{2}{x{8.4cm}}{remember to not square the R value thats already been done} \tn 
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\mymulticolumn{2}{x{8.4cm}}{Overall significance of the regression} \tn 
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Testing  joint exclusion & {[}R\textasciicircum{}2\textasciicircum{}/R{]}/{[}(1-R\textasciicircum{}2\textasciicircum{})/(n-k-1){]} \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{x{3.52 cm} x{4.48 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Data Scaling}}  \tn
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\mymulticolumn{2}{x{8.4cm}}{Changes:} \tn 
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if Xj is * by c & Its coefficient is / by c \tn 
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If dependant variable is * by c & ALL OLS coefficients are * by c \tn 
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\mymulticolumn{2}{x{8.4cm}}{neither t nor F statistics are affected} \tn 
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Beta coefficients & obtained from an OLS regression after the dependant and independent variables have been transformed into z-scores \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Dummy Variables}}  \tn
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\{\{noshy\}\}Dummy/Binary Variables & = yes/no variables \tn 
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 & = take on the values 0 and 1 to identify the mutually exclusive classes of the explanatory variables. \tn 
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 & = leads to regression models where the parameters have very natural interpretations \tn 
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\mymulticolumn{2}{x{8.4cm}}{Given: wage= β0+ ∂0 female + β1 edu + u} \tn 
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 & To solve for ∂0: \tn 
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 & ∂0=E(wage|female,edu)-E(wage|male,edu) \tn 
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 & where level of education is the same \tn 
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Graphically ∂0 = & an intercept shift \tn 
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 & male intercept= β0 \tn 
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 & female intercept= β0+∂0 \tn 
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\{\{noshy\}\}dummy variable trap= & when both dummy variables (male \& female) are included; resulting in perfect collinearity \tn 
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\{\{noshy\}\}If a qualitative variable has m levels; & then (m−1) dummy variables are required and each of them takes value 0 and 1. \tn 
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Dummy Variables (cont)}}  \tn
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\mymulticolumn{2}{x{8.4cm}}{{\bf{Hypothesis test}}} \tn 
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\mymulticolumn{2}{x{8.4cm}}{Test whether the two regression models are identical:} \tn 
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 & H0 :β2 =β3 0 \tn 
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 & H1 :β2 ≠0 and/or β3 ≠0. \tn 
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\mymulticolumn{2}{x{8.4cm}}{Acceptance of H0 indicates that only single model is necessary to explain the relationship.} \tn 
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\mymulticolumn{2}{x{8.4cm}}{Test is two models differ with respect to intercepts only and they have same slopes} \tn 
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 & H0 :β3 =0 \tn 
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 & H1:β3 ≠0. \tn 
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\mymulticolumn{2}{x{8.4cm}}{Treating a {\bf{quantitative variable}} as qualitative variable increases the complexity of the model.} \tn 
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 & The degrees of freedom for error are reduced. \tn 
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 & Can effect the inferences if data set is small \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
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% That's all folks
\end{multicols*}

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