I recently received a great question in a comment about whether the assumptions of normality, constant variance, and independence in linear models are about the residuals or the response variable.

The asker had a situation where Y, the response, was not normally distributed, but the residuals were.

**Quick Answer**: It’s just the residuals.

In fact, if you look at any (good) statistics textbook on linear models, you’ll see below the model, stating the assumptions:

ε~ i.i.d. N(0, σ²)

That *ε* is the residual term (and it ought to have an i subscript–one for each individual). The i.i.d. means every residual is independent and identically distributed. They all have the same distribution, which is defined right afterward.

You’ll notice there is nothing similar about Y. ε’s distribution is influenced by Y’s, which is why Y has to be continuous, unbounded, and measured on an interval or ratio scale.

But Y’s distribution is also influenced by the X’s. ε’s isn’t. That’s why you can get a normal distribution for ε, but lopsided, chunky, or just plain weird-looking Y.

{ 3 comments… read them below or add one }

Hello Karen. Thanks for this nice post on an issue that often confuses people. I have two minor comments. First, I think you meant to say interval OR ratio scale in the second to last paragraph. Second, I think it is useful (at least for more advanced users of statistics) to point out the important distinction between errors and residuals, as in this Wikipedia page:

http://en.wikipedia.org/wiki/Errors_and_residuals_in_statistics

The i.i.d. N(0, σ²) assumption applies to the errors, not the residuals. For example, if you give me n-1 of the residuals from your regression model, I can work out the last one, because they must sum to 0. So the residuals are not truly independent. The unobservable errors, on the other hand, can be truly independent.

Once again, thanks for a great post.

Cheers,

Bruce

Hi Karen,

Since Y = E(Y) + ε, and E(Y) is a constant (function of X’s and betas), this should imply that the variance, independence and distributional assumptions on ε applies to Y as well. Am I right to say this?

Hi Kevin,

One small change that makes all the difference: Y=E(Y|X) + e. If every individual had the same value of X, then yes, the distribution of Y would match that of e. Since they generally differ, the Y’s are affected by the X’s but the residuals aren’t.

The distribution of Y|X is the same as the distribution of e, but the distribution of Y isn’t necessarily. I’ve seen many data sets where Y is skewed, but e is normal.

Karen