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I get that for simple linear regression, say y=b0+b1*x + error, the errors should be normally distributed, have no autocorrelation, constant variance. I also understand that there should be no collinearity, and that the relationship between x and y should be linear.

Those are the standard assumptions in an intro to regression. However, as I'm learning about power transforms such as the Box Cox transformation, it's said that the Box Cox attempts to transform either the predictor x or response y to normally distributed if they are skewed in the first place. By this I mean the marginal distribution of x and y, not the distribution of y conditioned on x (this is related to residues).

But other sources say it's not important for x and y to be normally distributed. So I'm thinking, if both x and y are uniformly distributed rather than normally distributed, and then there is a roughly linear relationship between them, wouldn't linear regression also work? You can still have the error be normally distributed in this case right?

For example, say you have a cylindrical bar of chocolate. You want to regress the weight against the length of chocolate that you cut. Since the cross section is the same, the weight is linear with the length. But let's say you cut it using a machine that randomly decides on a length from a uniform distribution (1-10 inches). Then the weight should be uniformly distributed too. You should still have normally distributed residues due to manufacturing tolerance, accuracy of the machine, accuracy of measurement, etc...

A linear regression clearly is suitable in this case, but neither x and y are normally distributed. So I'm guessing, it's not required for x and y to be normally distributed. Please advise.

user173565
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    This is possibly a duplicate of [How does linear regression use the normal distribution?](https://stats.stackexchange.com/q/148803/22228); also related are [Why is the normality of residuals “barely important at all” for the purpose of estimating the regression line?](https://stats.stackexchange.com/q/152674/22228) and [What if residuals are normally distributed, but y is not?](https://stats.stackexchange.com/q/12262/22228) – Silverfish Aug 14 '17 at 14:53
  • You seem to confuse "normally distributed" with "multinormally distributed." The distinction is this: the values of $y$ are a set of numbers. Their distribution depends both on the distribution of $x$ and the errors. The regression assumptions are only that the errors are multinormally distributed and their expectations depend linearly on $x$. When $x$ does not have a normal distribution (and having a normal distribution of regressors is *not* a standard regression assumption), then the $y$ will not have a normal distribution, either. – whuber Aug 14 '17 at 14:58
  • Please correct me if i'm wrong but $x$ doesn't have a distribution. If you write the model out in terms of distributions you get $y_i \sim N( X_i\beta, \sigma^2)$. That is the $y_i$ is a normally distributed variable with mean $X_i\beta$. The assumption then is a linear increase in $x$ will cause a linear expected increase in $y$. The idea of the box transformation is that we attempt to "repair" a non-linear relationship between $y$ and $x$. Happy to delete this comment if I am mistaken in this ! – gowerc Aug 14 '17 at 17:36
  • @C_G I don't think you are mistaken, but your view might be too narrow. It expresses what is sometimes called the "fixed effect" model of regression. Economists (and many others) who rely on observational data often feel a need to model their regressors as random variables. Thus arises a subtle conceptual distinction between regression as estimating properties of a response distribution as parameterized by the regressors (on the one hand) and regression as estimating properties of the *conditional* distribution of the response (on the other hand). Your causal interpretation is wrong, however. – whuber Aug 14 '17 at 17:53

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