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## Hot answers tagged regression

4

Consider: The Pearson Product Moment Correlation Coefficient $r$ is an estimate of $\rho$, the population correlation coefficient, which measures the strength of a linear relationship between the two variables $x$ and $y$ ($x$ independent and $y$ dependent): $r$ $=$ $\dfrac{\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2 \cdot ... 2 The matrix $$Y:=X^TX, \quad X:=[X_1,X_2],$$ (which is generally positive semidefinite) is invertible iff$[X_1,X_2]$has full column rank. So necessarily,$X_1$must have full column rank. However, full rank of$X_2$is not sufficient for the nonsingularity of$Y$. From the block inversion formulas it follows that$X$is invertible iff$X_1$has ... 2 This is the incorrect expression for$E$. You're neglecting the noncommuting nature of matrices. The actual expression is $$E = (Y+XB)(Y+XB) = Y^2 + YXB + XBY + XBXB.$$ You have to multiply the matrices from left-to-right. In the event that$Y$and$B$are meant to be vectors and$Ba matrix, this really should be written more as E = ... 1 Note the Following. Let \bar{e} denote the mean of e. Then I can re-write your model as y_i = (\beta_1 + \bar{e}) + \beta_2x_i + (e_i - \bar{e}). This model is identical to yours and now has a mean-zero error term (though, as you point out, it is still not normally distributed), but the intercept will be "biased" by the mean of the original error. 1 Perhaps calculating this variance will help \begin{align} var(\hat{\alpha} + x_i \hat{\beta})&= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) + 2cov(\hat{\alpha},\hat{\beta}x_i)\\ &= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) + 2x_icov(\hat{\alpha},\hat{\beta})\\ &= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) - 2x_i \frac{\sigma^2\bar{x}}{S_{xx}} ... 1 One approach is to consider transformed variables. If you define e.g. X^*=W^{1/2}X\\ Y^*=W^{1/2}Y $$and apply this transformation, you can write your estimator as$$ \hat{\beta}=(X'WX)^{-1}X'WY=({X^*}'X^*)^{-1}{X^*}'Y^*$which is regular OLS, but it is applied to a transformed regression. Can you see what such a transformation means in terms of the ... 1 The most flexible thing you can do is assign binary variables for each individual category and include them all in your model. For example, let$Age_1 = 1$if age was between 20-30 and$0$otherwise. Let$Age_2 = 1$if age was between 31-40 and$0$otherwise..., etc. The cost of this approach is that it significantly increases the number of parameters you ... 1 Regression Analysis by Example By Samprit Chatterjee and Ali S. Hadi http://www.amazon.in/Regression-Analysis-Example-Probability-Statistics/dp/0470905840 1 Sure: For every random variables$x$,$y$,$z$, with$y$integrable, one has$E(E(y\mid x,z)\mid x)=E(y\mid x)$. This is called the tower property and is mentioned in every decent introductory chapter on conditional expectations. In your case,$E(y\mid x,z)=z+x'\beta$hence$E(y\mid x)=E(z+x'\beta\mid x)=E(z\mid x)+x'\beta$. 1 No, you are wrong it have nothing common with log likelihood function. Moreover the model has defects, because you do not guarantee that$y_i>0\$ otherwise you cannot make a log.

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