Distribution in logistic regression

Question

Suppose we have $n$ observations. For example, consider $n$ people who each have their blood pressure ($x_1$), pulse ($x_2$), and blood glucose ($x_3$) levels measured. So there are are $3$ explanatory variables measured for each person. The outcome variable is presence or absence of obesity ($Y$). In this case, does logistic regression assume that the data are distributed as $\text{Bernoulli}(p_i)$? For example, for the first person, we measure $x_1,x_2,x_3$ and compute $p_1$ (the probability of observing this)?

Answer 1

Yes: the model is $\operatorname{logit} p_i = \beta_0 +\beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i}$.

That's true for bog-standard logistic regression anyway - the term is sometimes used where there's an extra parameter for dispersion, or for an estimating equation approach for which the Bernoulli model isn't assumed.

Re your comment: $\sum_{i=1}^{m_j} Y_{ij}$ has a binomial distribution $\operatorname{Bin}(m_j,p_j)$ for groups of $m_j$ people (from the original $n$) who have the same covariate pattern—the same blood pressure, pulse rate & glucose levels—& therefore the same probability $p_j$ of obesity. If no-one has the same covariate pattern, then there are $n$ groups, each with $m_j=1$, i.e. $n$ different Bernoulli distributions. To be clear, for each individual person $Y_i\sim\operatorname{Bin}(1,p_i)\equiv\operatorname{Bern}(p_i)$, & as @Frank says, there's no real need to consider people grouped together by covariate pattern, though it's sometimes useful for diagnostics.

To be really clear, if your model says this:–

Tom: 90 mmHg, 80 /min, 6 mmol/l => 60% chance of obesity

Dick: 90 mmHg, 80 /min, 6 mmol/l => 60%

Harry: 60 mmHg, 60 /min, 5 mmol/l => 20%

you can write this:–

$$Y_{\mathrm{Tom}}+Y_{\mathrm{Dick}}\sim \operatorname{Bin}(2,60\%)$$ $$Y_{\mathrm{Harry}}\sim \operatorname{Bin}(1,20\%)\equiv\operatorname{Bern}(20\%)$$

or this:–

$$Y_{\mathrm{Tom}}\sim \operatorname{Bin}(1,60\%)\equiv\operatorname{Bern}(60\%)$$ $$Y_{\mathrm{Dick}}\sim \operatorname{Bin}(1,60\%)\equiv\operatorname{Bern}(60\%)$$ $$Y_{\mathrm{Harry}}\sim \operatorname{Bin}(1,20\%)\equiv\operatorname{Bern}(20\%)$$

Note that $Y_{\mathrm{Tom}}+Y_{\mathrm{Dick}}+Y_{\mathrm{Harry}}$ is not binomially distributed because there's not a common probability for each person.

Answer 2

As @Scortchi correctly notes, the answer is yes. However, I think this is not quite the right question.

I suspect what you are wondering about is the way that probability, $p_i$, is related to the explanatory variables. In generalized linear models, this is done via a link function. The default link function for binary GLiMs is the logit, however, if BMI is normally distributed, but was categorized as obese, not obese for the study, then your response variable depends on a hidden Gaussian variable, and a different link function is appropriate (namely the probit). For more on this topic, you may want to read my answer here: difference-between-logit-and-probit-models.