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$$y = \alpha_0 + \alpha_1 x + (\beta_0 - \alpha_0)1_{x > x_0} + (\beta_1 - \alpha_1)x1_{x > x_0} + \epsilon$$ Let $\gamma_0 := \beta_0 - \alpha_0$ and $\gamma_1 := \beta_1 - \alpha_1$. Now define $X := \begin{bmatrix}1 && x && 1_{x > x_0} && x1_{x > x_0}\end{bmatrix}$ and $\beta := \begin{bmatrix}\alpha_0 && ... 2 There are many different ways to look at degrees of freedom. I wanted to provide a rigorous answer that starts from a concrete definition of degrees of freedom for a statistical estimator as this may be useful/satisfying to some readers: Definition: Given an observational model of the form $$y_i=r(x_i)+\xi_i,\ \ \ i=1,\dots,n$$ where ... 1 Since$\hat{y_i}$is determined from the linear regression, it has two degrees of freedom, corresponding to the fact that we specify a line by two points. When we consider the equation of a line in slope-intercept form, this becomes the slope value and the y-intercept value. When we subtract the mean response,$\overline{y}$, it cancels the y-intercept value ... 1 If you standardize both your$X$and your$Y$, then you are predicting the values of standardized$Y$, in which case you can convert it back to its original scale after prediction. You can also only standardize your predictor$X$while leaving your$Y$unstandardized (this is more common) in order to predict$Y$based on the way$X$deviates from its mean. ... 1 Since$\epsilon_i \sim N(0,9)$, it follows that (see the derivation bellow if you do not undestand why) $$\hat{b}-b \sim N\left(0,\frac{9}{\sum_i{(x_i-\bar{x})^2}}\right),$$ and, noticing that given the numbers above$\sum_i{(x_i-\bar{x})^2}=16, this leads to $$\sqrt{\frac{16}{9}} (\hat{b}-b) \sim N(0,1).$$ Finally, let ... 1 Optimizing a boolean function is NP-Hard in general, so I wouldn't say that there is one popular approach that would work every time. There are exact methods and heuristics. The exacts methods might not be quick enough depending on the complexity of the boolean function and there are no guarantees that the heuristics will ever find the optimal solution. A ... 1 Here is the proof my teacher gaved me today, shorter than the very good one from Daniel. \begin{align*} R^2&=\frac{ESS}{TSS}=\frac{\sum(\hat y -\bar y)^2}{\sum (y_i-\bar y)^2}\\ &=\frac{\sum (\hat\beta_1+\hat \beta_2x_i-\bar y)^2}{\sum(y_i-\bar y)^2}\\ &=\frac{\sum (\bar y- \hat \beta_2\bar x+\hat \beta_2x_i-\bar y)^2}{\sum(y_i-\bar y)^2}\\ ... 1 Here are a few good options, both free and commercial. Commercial: TableCurve It's been around for more than 20 years. It takes two columns of data and fits over a thousand functions to the data. You can choose what types of functions you want to fit, fit them all, or define your own functions. Free: Mathulus.com You input two columns of data. Then you ... 1 No, none of these assumptions imply the full rank condition onX$. It has to be assumed separately. The full rank condition on$X$does not mean independence of regressors from each other -- it just means that they are not related by a linear functional relationship. Independence of the error can be relaxed (to uncorrelatedness), and homoschedasticity and ... 1 No and also independence of the$X_i$'s (row vector, i-th row of design matrix) does not have anything to do with full rank either. For the design matrix you need additional assumptions. There are two ways in which people usually proceed: 1) Fixed (deterministic) design: Since it is deterministic, it either has full rank or it does not, so you just have to ... 1 1) The condition$\mathbb{E}[\tilde{\beta}]=\beta$is just the condition "the estimator is unbiased" in mathematical form. Let's say you are considering the least squares estimator, then$\begin{align} \mathbb{E}[\hat{\beta}] &= \mathbb{E}[(X^{\rm T}X)^{-1}X^{\rm T}Y]\\ &= \mathbb{E}[(X^{\rm T}X)^{-1}X^{\rm T}X\beta+\epsilon]\\ &= \beta, ... 1 The Gauss-Markov theorem states that, under the usual assumptions, the OLS estimator\beta_{OLS}$is BLUE (Best Linear Unbiased Estimator). To prove this, take an arbitrary linear, unbiased estimator$\bar{\beta}$of$\beta$. Since it is linear, we can write$\bar{\beta} = Cy$in the model$y = \beta X + \varepsilon\$. Furthermore, it is necessarily ...