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In regression, it is abundantly clear that $Y$ can be non-normal while the residuals $\epsilon = Y - \beta_0 - \beta_1 X$ are normal. But can $Y$ be binary when the $\epsilon$ are normally distributed? This question is motivated by students' regression projects where their $Y$ was quite discrete, and obviously non-normal for that reason, yet their residual histogram and q-q plots "looked normal." Here is a simulation to illustrate:

par(mfrow=c(1,2)); set.seed(12345) 
X = rnorm(1000,3,1); Y = round(X + rnorm(1000,0,1)) 
table(Y) # Y is highly discrete and obviously non-normal 
model = lm(Y ~ X) ## But the diagnostic plots and tests suggest normality is reasonable 

qqnorm(model$residuals, main= "Residual q-q Plot"); qqline(model$residuals) 
shapiro.test(model$residuals); plot(X,Y, main = "Raw Scatterplot")

Despite the large sample size ($n=1000$), the Shapiro test "accepts" normality ($p = 0.1228$), and the normal q-q plot looks fine. However, the raw scatterplot shows obviousness discreteness, and hence, obvious nonnormality that is not discernable from the analysis of the residual distribution.

Discrete Y

I am pretty sure that the actual distribution of $\epsilon$ is not precisely normal in my example, but could it be normal?

How bad can this problem be? In the most extreme case, suppose $Y$ is binary (0 or 1), and a linear predictor is $\beta_0 + \beta_1 X$. Is it possible that $\epsilon = Y - \beta_0 - \beta_1 X$ has precisely a normal distribution in this case?

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  • $\begingroup$ $Y$ in my problem statement is either 0 or 1. The predictor which gives the residual is linear, but the problem could also be stated in terms of a nonlinear predictor. But that does not change the essential issue, and it makes the math harder. $\endgroup$ – BigBendRegion Sep 15 '20 at 21:55
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    $\begingroup$ Of course $Y$ can be binary! Simply begin with arbitrary binary $Y$ data $(y_i)$ and the Normal errors $\epsilon_i,$ pick $(\beta_0,\beta_1),$ and set $x_i = (y_i - \epsilon_i - \beta_0)/\beta_1.$ Done. $\endgroup$ – whuber Sep 15 '20 at 22:27
  • $\begingroup$ Very nice. But I should revise my question though for better realism. To correspond with the intended application of the naïve student who performs the regression and concludes "approximately normal $Y$" based on the residual plots, the $\beta_0$ and $\beta_1$ should correspond to a (theoretical) line of best fit. I don't think they need to be identified explicitly, though. $\endgroup$ – BigBendRegion Sep 16 '20 at 11:10
  • $\begingroup$ In my comment, the $\beta_i$ define a "theoretical line of best fit," so I don't understand the distinction you are trying to make. It is true that if you were to use OLS to fit these data, the residuals wouldn't look at all Normal -- but the reason is that a fundamental assumption of OLS (and most regression models) is strongly violated: the $x_i$ and $\epsilon_i$ will have a strong correlation; the $\epsilon_i$ will not be independent. If you insist the $\epsilon_i$ be independent, then they cannot be Normally distributed. $\endgroup$ – whuber Sep 16 '20 at 12:28
  • $\begingroup$ Please run: y=rbinom(1000,1,.5); eps=rnorm(1000); b0=2; b1=3; x=(y-eps-b0)/b1; plot(x,y); abline(b0,b1); abline(lsfit(x,y), lty=2) . The least squares fit is very different from your $\beta_0+\beta_1x$ line. For the naïve student problem, I think the $\beta$'s should correspond to some "least-squares like" theoretical quantities. $\endgroup$ – BigBendRegion Sep 16 '20 at 21:13

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