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Unless I'm mistaken, in a linear model, the distribution of the response is assumed to have a systematic component and a random component. The error term captures the random component. Therefore, if we assume that the error term is Normally distributed, doesn't that imply that the response is also Normally distributed? I think it does, but then statements such as the one below seem rather confusing:

And you can see clearly that the only assumption of "normality" in this model is that the residuals (or "errors" $\epsilon_i$) should be normally distributed. There is no assumption about the distribution of the predictor $x_i$ or the response variable $y_i$.

Source: Predictors, responses and residuals: What really needs to be normally distributed?

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If the $x$'s are non-stochastic the normality of $\epsilon$ implies normality of the dependent variable. For stochastic independent variables this will not hold in general, it then depends on the distribution of the independent variables. – fcop Mar 28 at 11:52
up vote 15 down vote accepted

The standard OLS model is $Y = X \beta + \varepsilon$ with $\varepsilon \sim \mathcal N(\vec 0, \sigma^2 I_n)$ for a fixed $X \in \mathbb R^{n \times p}$.

This does indeed mean that $Y|\{X, \beta, \sigma^2\} \sim \mathcal N(X\beta, \sigma^2 I_n)$, although this is a consequence of our assumption on the distribution of $\varepsilon$, rather than actually being the assumption. Also keep in mind that I'm talking about the conditional distribution of $Y$, not the marginal distribution of $Y$. I'm focusing on the conditional distribution because I think that's what you're really asking about.

I think the part that is confusing is that this doesn't mean that a histogram of $Y$ will look normal. We are saying that the entire vector $Y$ is a single draw from a multivariate normal distribution where each element has a potentially different mean $E(Y_i|X_i) = X_i^T\beta$. This is not the same as being an iid normal sample. The errors $\varepsilon$ actually are an iid sample so a histogram of them would look normal (and that's why we do a QQ plot of the residuals, not the response).

Here's an example: suppose we are measuring height $H$ for a sample of 6th graders and 12th graders. Our model is $H_i = \beta_0 + \beta_1I(\text{12th grader}) + \varepsilon_i$ with $\varepsilon_i \sim \ \text{iid} \ \mathcal N(0, \sigma^2)$. If we look at a histogram of the $H_i$ we'll probably see a bimodal distribution, with one peak for 6th graders and one peak for 12th graders, but that doesn't represent a violation of our assumptions.

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Maybe a better statement would be $(Y | X) \sim \mathcal{N}(X \beta, \sigma^2 I_n)$? – Adrian Mar 28 at 13:11
Fair point. I've updated. – Chaconne Mar 28 at 13:14
Could someone please clarify notation $\sigma^2 I_n$? – snoram Mar 28 at 13:15
It means the $n \times n$ identity matrix multiplied by a scalar $\sigma^2$. – Chaconne Mar 28 at 13:16
The $n\times n$ identity matrix times some variance. – General Abrial Mar 28 at 13:16

Therefore, if we assume that the error term is Normally distributed, doesn't that imply that the response is also Normally distributed?

Not even remotely. The way I remember this is that the residuals are normal conditional on the deterministic portion of the model. Here's a demonstration of what that looks like in practice.

I start by randomly generating some data. Then I define an outcome which is a linear function of the predictors and estimate a model.

N <- 100

x1 <- rbeta(N, shape1=2, shape2=10)
x2 <- rbeta(N, shape1=10, shape2=2)

x <- c(x1,x2)
plot(density(x, from=0, to=1))

y <- 1+10*x+rnorm(2*N, sd=1)


Let's take a look at what these residuals look like. I suspect that they should be normally distributed, since the outcome y had iid normal noise added to it. And indeed that is the case.

enter image description here

plot(density(model$residuals), main="Model residuals", lwd=2)
s <- seq(-5,20, len=1000)
lines(s, dnorm(s), col="red")

plot(density(y), main="KDE of y", lwd=2)
lines(s, dnorm(s, mean=mean(y), sd=sd(y)), col="red")

Checking the distribution of y, however, we can see that it's definitely not normal! I've overlaid the density function with the same mean and variance as y, but it's obviously a terrible fit!

Density of y

The reason that this happened in this case is that the input data is not even remotely normal. Nothing about this regression model requires normality except in the residuals -- not in the independent variable, and not in the dependent variable.

Denisty of x

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No, it doesn't. For example, suppose we have a model predicting the weight of Olympic athletes. While weight could well be normally distributed among athletes in each sport, it won't be among all athletes - it might not even be unimodal.

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