# In binary logistic regression, must the binary Y be interpreted as the dependent variable?

If I have a binary variable, say sex, and I want to test whether multiple other variables are associated with it. To do this, I run a logistic regression of the form

$$logit(probability(sex = male)) = \beta X1 + \beta X2 ... + \beta Xk$$

Once I do this, I calculate the pseudo R-squared, and it is 0.45 meaning that the regression explained 45% of the variance in sex.

My question is, is it also fair/correct to interpret this as sex explained 45% of the variance in the regressors?

Likewise, if the odds ratio for X1 is 2.0, can one claim that being male increased the odds of X1 occurring by 100%?

Basically, can the equal sign in the regression equation truly be treated as an equal sign (i.e., bi-directional equivalence) if I have no claim to directionality?

• The linear equation estimated by the model is an equality, but that doesn't apply to the inferences made by the model. – david25272 Aug 21 '17 at 0:58
• Is there any specific reason you tagged "causality"? – Carlos Cinelli Aug 21 '17 at 6:00
• Adding to other answers: we can talk about "explained variance" only when conducting linear regression. For GLMs $R^2$ has nothing to do with explained variance. – Tim Aug 21 '17 at 6:25
• @carloscinelli I tagged causality because, if interferences are directionally constrained, that is a sort of causal inference. For example, if we can claim that X has an effect on Y but not the inverse, we at least know that Y doesn't cause X. – JRF1111 Aug 21 '17 at 22:49
• @JRF1111 ok, that opens a whole lot of different answers. Causality is definitely directionally constrained, that is, for causal models the equal sign is not a literal equality, it's more like an assignment operator, and the relationship is asymmetric $y \leftarrow x$ . – Carlos Cinelli Aug 21 '17 at 23:19

$$y \mid x \sim \text{Bernoulli}(p = logit^{-1}(\beta_0 + \beta_1 x_1 + \beta x_2 ... + \beta x_k))$$
Here, $\sim$ means "is distributed as".
You are making a distributional statement about $y$ conditional on $x$, and conditioning is not symmetric. Because of the conditional, the $x$'s are not considered random by the model, so $y$ cannot be thought of as explaining variance in the $x$'s (at least from the point of view of the model).