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Question 2: (1) $y$ is modelled as linear function of $x$. (2) $y$ is modelled as quadratic function in $x$ - note that this means that $y$ is (and will look) nonlinear in $x$, however it is linear in … 3-d space with the two variables $x$ and $x^2$. (3) Assuming that you mean the same thing by $X$ and $x$, $y$ is just linear in $x$, however the regression slope is $d=b^2$ rather than $b$. …

So you can do linear or nonlinear regression, and cross-validation will tell you how good your model is at prediction (at least within the range of the observed data) no matter whether the assumed model … On top of that, if you run a linear model and look at diagnostic residual plots, this may give you a better idea of whether linearity is a good assumption or what other relation could hold than looking …

Actually the Pearson correlation can be computed for any data, not just linear data. … distribution theory relies on linearity (which you need if you want to test the null hypothesis $\rho=0$ or compute a confidence interval) and interpretation may be a bit more tricky if the relationship is not linear …

Linear regression is a tool that is used to achieve a goal. So any answer will depend on the goal to be achieved. … If however you don't know $Z$, linear regression may work well for predicting $X$ from $Y$. …

The first thing to understand about issues like this one is that statistical models live in the land of mathematics rather than in real life, and in real life no statistical model is ever fulfilled (a …

I think a linear regression addresses the question "quantify how much suffering varies with each added unit score of cognitive distortion" straight away, so I'd probably run it. … Note that discrete data with limited value range can never be normally distributed, but then nothing is ever normally distributed ("all models are wrong but some are useful"), and we apply linear regression …

As commented by Nick Cox, I'd suspect that you understand the special case of standard linear regression properly, but the equation 1.5 involving $\mu(x)$ and $\sigma^2(x)$ regards a more general case, … of which the standard linear model is a special case with $\mu(x)=w^Tx$ and $ \sigma^2(x)=\sigma^2$ constant. …

Any values that these statistics can take can be achieved with linear as well as non-linear relationships (so in fact a non-linear relationship is indeed possible, but it's not possible to say with any … certainty that the relationship is non-linear). …

Let $Y$ be income, $X_1$ be the indicator for female and $X_2$ be experience. The indicator for male is $1-X_2$. I assume that there are only male and female (somewhat outdated to be honest). Then mod …

It is not true in general that an insignificant variable has no effect on the response. A variable can be insignificant because the sample size is too low or the random variation too large to find a c …

M- and other robust estimators are not linear in the observations. Note that this is a different issue from fitting a linear model; it concerns how the estimator is computed! … In fact being linear in the observations means that outliers cannot be down-weighted, which is what robust estimators aim to do. …

Some general issues: Model assumptions generally don't hold precisely in reality, and the idea that they have to hold is wrong (otherwise model-based methods could never be used). A model is an ideal …