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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 $B$ a 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.


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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 ...


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Regression Analysis by Example By Samprit Chatterjee and Ali S. Hadi http://www.amazon.in/Regression-Analysis-Example-Probability-Statistics/dp/0470905840


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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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