Is the model below an example of nonlinear regression? $$y = \beta_1 + \beta_2^2x$$ Some books say, it is, but I'm not convinced as $\beta_2^2$ can itself considered a constant denoted by (say) $\beta_3$.    I have studied other examples of nonlinear regression and been convinced about them.  The question differs from an  suggested in that in my question, power is over $\beta$, the regression coefficient, while on the answer, power is over x, the independent variable.
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I was reading the following link on non linear regression . My understanding from reading the first section "Nonlinear Regression vs. Linear Regression" was that the equation below is actually a linear regression, is that correct? If so why?  $$y = b_1x^3 + b_2x^2 + b_3x + c$$  Am I also to understand that in non linear regression multicollinearity isn't an issue? I know that multicollinearity can be an issue in linear regression so surely if the model above is in fact a linear regression there would be multicollinearity?
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Like, I don't get it. In regression, we say  $$Y_i = a + bX_i + \epsilon$$ where $X_i$ is just a real number for each observation $i$,  and in analysis of variance with two groups, we say $$Y_i = a + bX_i + \epsilon$$ where $X_i$ is again just a real number, but this time either 0 (group 1) or 1 (group 2), so a is mean of group 1, and a+b is mean of group 2.  But despite the interpretation, the model is exactly the same. So why do people split it up into regression and anova? the model is the same?
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One of the assumption of logistic regression is the linearity in the logit. So once I got my model up and running I test for nonlinearity using Box-Tidwell test. One of my continuous predictors (X) has tested positive for nonlinearity. What am I suppose to do next?   As this is a violation of the assumptions shall I get rid of the predictor (X) or include the nonlinear transformation (X*X). Or transform the variable into a categorical? If you have a reference could you please point me to that too?
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Is it necessary to exponentiate the predicted values in a log-log regression model?  For example my model is:  $\log(y) = \log(x)$  $\log(y) = -0.5141 + 0.5377 \log(x)$  if I wanted to make a prediction from this model do I need to exponentiate the result or log the x value? What is the correct way of evaluating the equation?  if $x = 100$ then y=   $y = -0.5141 + 0.5377 \times 100 = 53.2559$ $y = -0.5141 + 0.5377 \times \log(100) = 1.9621$ $y = -0.5141 + 0.5377 \times \log(100) = 1.9621 = \exp(1.9621) = 7.114251$
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Imagine   You run a linear regression with four numeric predictors (IV1, ..., IV4) When only IV1 is included as a predictor the standardised beta is +.20 When you also include IV2 to IV4 the sign of the standardised regression coefficient of IV1 flips to -.25 (i.e., it's become negative).   This gives rise to a few questions:   With regards to terminology, do you call this a "suppressor effect"? What strategies would you use to explain and understand this effect? Do you have any examples of such effects in practice and how did you explain and understand these effects?
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The regression model y= b0+ b1 x + b2 x^2 + b3 x^3  and the second regression model y = b0 +b1 (x-u) + b2 (x-u)^2 + b3 (x-u)^3 where u is the mean of x  These two models lead to the same curves, or says the same fitted values. I understand centering the variables for polynomial or interaction regressors will lower the correlations of regressors and therefore reduce the multicolinearity, variance inflation, prediction interval... But they are still the same model, does the difference between the variances of two identical models really mean something important?  R code example:  y &lt;- rnorm(100) x &lt;- runif(100) u &lt;- mean(x)  model1&lt;- lm(y~x+I(x^2)+I(x^3)) model2 &lt;- lm(y~x+I((x-u)^2)+I((x-u)^3))  summary(model1) summary(model2)   They have the same sd error, R squared.... Their fitted values are the same as well.
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Beta regression (i.e. GLM with beta distribution and usually the logit link function) is often recommended to deal with response aka dependent variable taking values between 0 and 1, such as fractions, ratios, or probabilities: .  However, it is always claimed that beta regression cannot be used as soon as the response variable equals 0 or 1 at least once. If it does, one needs to either use zero/one-inflated beta model, or make some transformation of the response, etc.: .  My question is: which property of beta distribution prevents beta regression from dealing with exact 0s and 1s, and why?  I am guessing it is that $0$ and $1$ are not in the support of beta distribution. But for all shape parameters $\alpha&gt;1$ and $\beta&gt;1$, both zero and one are in the support of beta distribution, it's only for smaller shape parameters that the distribution goes to infinity at one or both sides. And perhaps the sample data are such that $\alpha$ and $\beta$ providing best fit would both turn out to be above $1$.  Does it mean that in some cases one could in fact use beta regression even with zeros/ones?   Of course even when 0 and 1 are in the support of beta distribution, probability of observing exactly 0 or 1 is zero. But so is the probability to observe any other given countable set of values, so this cannot be an issue, can it? (Cf.  by @Glen_b).  $\hskip{8em}$  In the context of beta regression, beta distribution is parameterized differently, but with $\phi=\alpha+\beta&gt;2$ it should still be well-defined on $[0,1]$ for all $\mu$.
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Does the following qualify as a univariate regression?  $$y=b_0+b_1x+b_2x^2+\epsilon$$  I fully comprehend the implications of adding regressors and need no background information - a "yes" or "no" will do :-)
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Consider the  function     $$ r(x) = \mathbb{E}(Y \mid X = x) $$   This has been called the regression function in a textbook I'm using. I'm trying to figure out the relationship between this function and the classical linear regression model.  So, I know that it is a theorem* that we may write  $$ Y = r(X) + \epsilon $$  for some random variable $\epsilon$ s.t. $\mathbb{E}(\epsilon) = 0$.  Now suppose that we have     $$ Y = \beta_0 + \beta_1 X + \epsilon $$   This is the classical 1-dimensional regression function (assuming the $\beta_0$ and $\beta_1$ minimize the residual sum of squares).  Question: Is it then a mathematical theorem that if $Y$ is defined as above, that   $$ r(X) = \mathbb{E}(Y \mid X) = (\beta_0 + \beta_1 X)? $$  And is this why the function $\mathbb{E}(Y \mid X)$ is called the "regression function"?  EDIT: The theorem that I am making use of is as follows (from All of Statistics pg. 89):     Regression models are sometimes written as      $$ Y = r(X) + \epsilon $$      where $\mathbb{E}(\epsilon) = 0$. We can always rewrite a regression model this way. To see this, define $\epsilon = Y - r(X)$ and hence $Y = Y + r(X) - r(X) = r(X) + \epsilon$. Moreover, $\mathbb{E}(\epsilon) = \mathbb{E}\mathbb{E}(\epsilon \mid X) = \mathbb{E}(\mathbb{E}(Y - r(X)) \mid X) = \mathbb{E}(\mathbb{E} ( Y \mid X) - r(X)) = \mathbb{E}(r(X) - r(X)) = 0$.
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I want to run a regression with 4 to 5 explanatory variables, but I have only 15 observations. Not being able to assume these variables are normally distributed, is there a non-parametric or any other valid regression method?
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I am trying to understand the confidence interval for linear regression parameters. At this link  an answer is provided. However, I did not understand in the derivation fully. Do we consider the regression parameters as random variables?   That is what are the random variables in the following regression: $$ y=\beta_1 x+\beta_0 + \epsilon $$ From what I understand is that just the $\epsilon$ is random variable with mean 0 and variance $\sigma^2$. Considering $\beta_1, x, \beta_0$ as constants $y$ also becomes random variable with mean $\beta_1 x+\beta_0$ and variance  $\sigma^2$, from which confidence interval can be calculated.  However,for regression coefficients confidence intervals are also calculated. Can you please clarify the random variables in regression please?  Thanks.
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Consider a simple regression (normality not assumed): $$Y_i = a + b X_i + e_i,$$ where $e_i$ is with mean $0$ and standard deviation $\sigma$. Are the Least Square Estimates of $a$ and $b$ uncorrelated?
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Usually regression models are evaluated using $R^2$. I understand this metric can be misleading too at times but as far as I understand the first parameter we look at is $R^2$.   There is another parameter which is often used and it is $MAPE$.  Both are functions of errors between predicted and true value. I am just wondering if there are certain cases where one should be preferred above another?   $R^2$ can become negative if being used on test data (on which the model is not built). Is this a reason to not use $R^2$ in these cases?  Can anyone give a qualitative insight on where $MAPE$ should be used and where $R^2$ should be used.
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I think this is a really simple question so please bear with me - I just had my first class in regression and I'm a little confused about nomenclature/labeling.  Does anyone recommend some good weblinks that explain beginning linear regression really well?  There's a question I've been looking at for a while and I'm not sure how to do it (although I'm sure the solution is simple):  Show that the sample estimate $\hat{\mu}(X) = \frac{1}{n} \sum X_i$ is a least square estimator of $\mu$ for a variable $X$ given $X_1, \ldots, X_n$.  My first thought was,  $\mathrm{SSE} = \sum (\mu - \hat{\mu})^2$  But I'm not sure if thats right. I'm confused about what the beta is (is it n?) and I don't know if there are enough parameters to expand it.  Thanks so much for your patience and if this doesn't make sense, I can clarify more. Thanks!
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Suppose, I have performed multiple regression analysis on the following data set where X1 and X2 are independent variables and Y is the dependent variable.    And achieved the following multiple regression analysis table from where it is clear that X2 is not correlated with Y as the p-value is greater than 0.05 which is 0.108 for X2 variable     So, if p value is greater than 0.05 for a variable, then why it is considered in regression equation  Y = 2.709 + 0.763(X1) + 0.463(X2) which has been shown in any online multiple regression equation calculator.  My question is Why it is not like the following Y = 2.709 + 0.763(X1) omitting the [0.463(X2)] part from the equation, as X2 is not correlated with Y  Please help to clarify the conceptual idea....
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In An Introduction to Statistical Learning (James et al.), in section 3.7 exercise 5, it states that the formula for $\hat{\beta}_1$ assuming linear regression without an intercept is  $$\hat{\beta}_1 = \dfrac{\displaystyle\sum\limits_{i=1}^{n}x_iy_i}{\displaystyle \sum\limits_{i=1}^{n}x_i^2}\text{,}$$ where $\hat{\beta}_0 = \bar{y}-\hat{\beta}_1\bar{x}$ and $\hat{\beta}_1 = \dfrac{\displaystyle S_{xy}}{S_{xx}}$ are the usual estimates under OLS for simple linear regression ($S_{xy} = \displaystyle\sum\limits_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})$).  This is not the actual exercise; I am merely wondering how to derive the equation. Without using matrix algebra, how do I derive it?   My attempt: with $\hat{\beta}_0 = 0$, we have $\hat{\beta}_1 = \dfrac{\bar{y}}{\bar{x}} = \dfrac{S_{xy}}{S_{xx}}$.  After some algebra, it can be shown that $\displaystyle S_{xy} = \sum\limits_{i=1}^{n}x_i^2 - n\bar{x}\bar{y}$ and $\displaystyle S_{xx} = \sum\limits_{i=1}^{n}x_i^2 - n\bar{x}^2$. From here, I'm stuck.
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If granted omniscience and we know that $\beta_1$ in a multiple linear regression model is truly 0, what does that mean in words (and math notation)?  The model is: $Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \epsilon$  Does $\beta_1 = 0$ mean:   Empirically: "If we make infinite, perfect measurements of our data, $\hat\beta_1$ approaches 0 at the limit?" [Math: ??] Formally: "In the true, finite population of our data, $X_1$ has 0 association with $Y$ and their vectors of values are perfectly linearly independent, uncorrelated, and orthogonal?" [Math: ??]   What does $\beta_1 = 0$ mean, precisely?