A log-likelihood function takes on the form of: logL$=\Sigma^{n}_{i=1}(y_ilog(\frac{e^{\beta' x_i}}{1+e^{\beta'x_i}} )+(1-y_i)(\frac{e^{\beta' x_i}}{1+e^{\beta'x_i}})$

My logit model is estimated as follows:


Estimating $y_i$


x's as a matrix of the dependent variable


Taking the estimates of $\beta$ from my model


Putting the these together:


Estimating the logistic form:


Putting everything together in the form of a log-likelihood model:


Issue with the fact that you can't take take the log of a negative value, so I only get about half of my coefficients in the output.

  • $\begingroup$ The default link function for the binomial family is logit. It looks like you want log. Try binomial("log") as the family. $\endgroup$ Commented Nov 6, 2020 at 22:06
  • $\begingroup$ what exactly do you want to do? Extract the predicted probabilities for each observation? Extract the log-likelihood for the model? Compute the log-likelihood for the model manually? (see ?predict, ?logLik ...) This might be more appropriate for StackOverflow, as it looks more like "how do I compute ... ?" rather than "what should I compute?" or "what does this mean?" $\endgroup$
    – Ben Bolker
    Commented Nov 6, 2020 at 22:37
  • $\begingroup$ @BenBolker I want to compute the log-likelihood model manually without using logLik. thanks $\endgroup$ Commented Nov 6, 2020 at 22:40

1 Answer 1


Both your log-likelihood equation and your R code have some errors:

$$ \begin{split} \eta_i & = \mathbf{X}_i \boldsymbol{\beta} \quad \textrm{(so far so good)} \\ p_i & = \exp(\eta_i)/(1+\exp(\eta_i)) = 1/(1+\exp(-\eta_i)) \\ L_i & = y_i \log p_i + (1-y) \log(1-p_i) \\ L & = \sum L_i \end{split} $$

In R you can use plogis() for the logistic ($1/(1+\exp(-\eta_i))$):

p <- plogis(X %*% beta)
sum(y*log(p) + (1-y)*log(1-p))

(or dbinom(y, prob=p, size=1, log=TRUE))

If your coefficients include an intercept, $\mathbf{X}$ should include a column of 1s at the corresponding position (usually the first column). You can check various intermediate results with predict(fitted_model, type="link") ($\eta$), predict(fitted_model, type="response") ($p$).

  • $\begingroup$ Thanks for your response, I had a couple of errors in my log-likelihood function, I have corrected these, my output and your output both give the same answer, however, this answer is incorrect. I suspect that the way I have multiplied x and beta may be wrong as beta includes the intercept. $\endgroup$ Commented Nov 6, 2020 at 23:35
  • $\begingroup$ Thanks again, I'm slightly confused over how to use predict(), does it give me as output, my x variables multiplied by my betas? Thanks $\endgroup$ Commented Nov 8, 2020 at 18:12
  • $\begingroup$ yes, if you specify type="link" (the default) $\endgroup$
    – Ben Bolker
    Commented Nov 8, 2020 at 18:21

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