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Consider the model $y_i = \beta x_i + \varepsilon_i$ (simple linear regression without an intercept). … Consider the fitted values that result from performing linear regression without an intercept. …

So I stared at the question long enough. Turns out it's just a matter of playing around with the summation's indices. If we substitute $\hat{\beta}$ into $\hat{y}_i$ $$\hat{y}_i = x_i \frac{\sum_{i=1} …

I have a dataset with multiple rows per individual. Each row represents the number of items sold by an employee on a given day. Suppose I have thousands of employees, and each of them has roughly five …

Let's say I have the following model: $$\ln\Big(\frac{\mathbb{P}(Y_i = 1 | X_i)}{\mathbb{P}(Y_i = 0 | X_i)}\Big) = \beta_0 + \beta_1 X_{1,i} + \beta_2 X_{2,i}$$ Let $\rho = \frac{\mathbb{P}(Y_i = 1 | …

In R, I turned the state variable into a factor and added it to the regression. …

I want to fit fit a cubic regression spline and select the optimal number of knots via grid search. … # Fit step-functions on grid spline_tune <- tune_grid(object = spline_wf, resamples = cv, grid = spline_grid) It is my understanding that a regression spline has $(d + 1) + k$ degrees of freedom, where …

My regression model had six parameters. … In order for the regression coefficients to match the differences in conditional means, I had to add two additional interaction terms: $X_1 \times X_2$ and $X_1 \times X_2 \times T$. …

I am posting this question following this other question I posted earlier today. Suppose we have the following model: $$\ln Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + ... + \varepsilon $$ Why is a marg …

As far as I'm concerned, all the regression result tables I've seen in Python and R return $t$ statistics for each estimated coefficient. …

In a nutshell, I want the regression coefficients of a model to match several differences in conditional means. You can download the data from this repo. … How come the linear regression results do not match the differences in conditional means? …

I have multiple observations per year and I was asked to estimate an individual regression line for each year within a single model. … In essence, this model fits one regression line per year to each group of observations. …

I am struggling to see why a one percent change in $X$ is associated with a $\frac{\beta_1}{100}$ change in $Y$ in the following model: $Y = \beta_0 + \beta_1 \ln X + \beta_2 W + ... + u$. It is clear …

I gave it some thought (sometimes posting here helps me structure my thought process) and this is what I came up with: $\frac{\delta z_i}{\delta X_1} = \frac{\delta \frac{y_i - \bar{y}}{\sigma_y}}{\de …

That is: $$d_1 = E(Y|x_1=1, x_2=0) - E(Y|x_1=0, x_2=0)$$ $$d_2 = E(Y|x_1=1, x_2=1) - E(Y|x_1=0, x_2=1)$$ How does this approach compare to the following regression model? …

Let's say we have the following regression model: $$z_i = \beta_0 + \beta_1 X_{1,i} + \beta_2 X_{2,1} + u_i$$ Where $z_i = \frac{y_i - \bar{y}}{\sigma_y}$ is the (standardized) dependent variable. …