Pearson residues applied to the binary model

Question

My question would be about Pearson residuals applied to the binary model.

When we build confidence intervals for a proportion, p: we have a sample, and we count the number of individuals who have a certain characteristic in n (sample size), and therefore, p ^ = x / n, where x is the number of individuals with this characteristic. In this case, X is a binomial (n, p), that is, X is the sum of n random variables with Bernoulli distribution. When n is large, we apply the Central Limit theorem and say that X converges to Normal because the sum of many independent random variables converges to Normal.

But if Y has bernoulli distribution (1 or 0), then E(Y)=p and Var(Y)=1-p, so can we say that Z=(Y - E(Y))/sqrt(Var(Y) converges to a standard Normal distribution? Can I apply the Central Limit theorem?

In my study, each individual follows a Bernoulli distribution, that is, a Binomial(1, p). How do we bring together a group of individuals to add several Bernoulli in such a way that we can apply the Central Limit theorem? Or would you have another theory?

Thanks

Answer 1

Random thoughts:

In the following I'm assuming that your logistic regression provides as good enough fit that $\hat{p}$ from the logistic regression is a good approximation of the true success probability $p$.

(1) Definitely $Z = (Y-p)/\sqrt{np(1-p)}$ won't converge in distribution to a normal distribution, because $Y$ is a single, Bernoulli random variable and does not represent a sequence of random variables. Central Limit Theorem does not apply at all in this case. ($Z$ has a shifted and scaled Bernoulli distribution with only two possible values, it's very much not normally distributed).

(2) As you mentioned, if your data values $Y$ represent a collection of $n$ observations averaged together all with the same (under the logistic model) success probability $\hat{p}$, then, yes, if $n$ is fairly large (usual rule of thumb is that $np > 7$ and $n(1-p) > 7$, though this varies), $Y$ will be approximately normally distributed and $(Y-\hat{p})/\sqrt{n\hat{p}(1-\hat{p})}$ will be roughly $N(0,1)$. However, this won't be very satisfying, because (i) you might not have any two observations with the same predicted success probability, so you would have to pool observations with similar success probabilities and (ii) this rather ruins the idea of looking for outliers, because pooling an outlier with a bunch of non-outliers will probably result in an innocent-looking residual. This works if the original experiment was based on groups with the same regressor values and you want do detect which groups are anomalous, but not helpful in determining which individual observations are anomalous.

(3) You actually can use Pearson residuals such as $Z=(Y-\hat{p})/\sqrt{\hat{p}(1-\hat{p})}$. They won't have a normal distribution, but you can still get a concept of outliers by using bounds based on the Chebyshef Inequality, so, for instance, $|Z| > 5$ would be flagged as an outlier (not adjusting for multiple comparisons).