I want to perform logistic regression with the following binomial response and with $X_1$ and $X_2$ as my predictors.
I can present the same data as Bernoulli responses in the following format.
The logistic regression outputs for these 2 data sets are mostly the same. The deviance residuals and AIC are different. (The difference between the null deviance and the residual deviance is the same in both cases - 0.228.)
The following are the regression outputs from R. The data sets are called binom.data and bern.data.
Here is the binomial output.
Call:
glm(formula = cbind(Successes, Trials - Successes) ~ X1 + X2,
family = binomial, data = binom.data)
Deviance Residuals:
[1] 0 0 0
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.9649 21.6072 -0.137 0.891
X1Yes -0.1897 2.5290 -0.075 0.940
X2 0.3596 1.9094 0.188 0.851
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 2.2846e-01 on 2 degrees of freedom
Residual deviance: -4.9328e-32 on 0 degrees of freedom
AIC: 11.473
Number of Fisher Scoring iterations: 4
Here is the Bernoulli output.
Call:
glm(formula = Success ~ X1 + X2, family = binomial, data = bern.data)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.6651 -1.3537 0.7585 0.9281 1.0108
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.9649 21.6072 -0.137 0.891
X1Yes -0.1897 2.5290 -0.075 0.940
X2 0.3596 1.9094 0.188 0.851
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 15.276 on 11 degrees of freedom
Residual deviance: 15.048 on 9 degrees of freedom
AIC: 21.048
Number of Fisher Scoring iterations: 4
My questions:
I can see that the point estimates and standard errors between the 2 approaches are equivalent in this particular case. Is this equivalence true in general?
How can the answer for Question #1 be justified mathematically?
Why are the deviance residuals and AIC different?