I've read here that

... (Bayesian linear regression) is most similar to Bayesian inference in logistic regression, but in some ways logistic regression is even simpler, because there is no variance term to estimate, only the regression parameters.

Why is it the case, why no variance term in Bayesian logistic regression?


Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.

  • $\begingroup$ @patrick for linear regression $y = mx + c + \epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c). $\endgroup$
    – seanv507
    Apr 3 '19 at 21:02
  • $\begingroup$ @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+\epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty? $\endgroup$
    – Patrick
    Apr 3 '19 at 21:37
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    $\begingroup$ @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used? $\endgroup$
    – Tim
    Apr 3 '19 at 21:39
  • $\begingroup$ @Patrick When you describe the conditional expectation in terms of a distribution there's no error term. $\endgroup$
    – Firebug
    Apr 3 '19 at 22:12
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    $\begingroup$ @Patrick linear regression is $Y \sim \mathcal{N}(X\beta, \sigma)$ while logistic regression is $Y \sim \mathcal{B}(h^{-1}(X\beta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant. $\endgroup$
    – Tim
    Apr 4 '19 at 19:12

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