Cox's 1972 publication Regression Models and Life Tables links logistic regression to an extension of the discrete time proportional hazard model. I do not understand how Equation (21) in the publication is derived.

Let $T$ be a discrete random variable that takes the values $t_1 < t_2 < ... $ with probabilities

$$f(t_j) = Pr\{T = t_j\}$$

In the discrete case the hazard is a probability and not a rate,

$$\lambda(t_j) = Pr\{T = t_j | T \geq t_j \} = \frac{Pr\{T = t_j , T \geq t_j \}}{Pr\{T \geq t_j \}} = \frac{f(t_j)}{S(t_j)}$$

I.e., in discrete time the hazard is just the probability of the event occurring at time $t_j$ given that the event has not occured up to, but not including, $t_j$.

As per the proportional hazard model we model the hazard for a particular vector of covariates/features $\mathbf{x}$ at time $t_j$ with

$$\lambda(t_j, \mathbf{x}) = \lambda_0(t_j)exp\{\sum_{i=1}^n\beta_ix_i\}$$

Since in discrete time the hazard is a probability we may be interested in looking at the odds:

$$\frac{\lambda(t_j, \mathbf{x})}{1 - \lambda(t_j, \mathbf{x})} = \frac{\lambda_0(t_j)}{1 - \lambda_0(t_j)}exp\{\sum_{i=1}^n\beta_ix_i\} $$

I don't understand how the right hand side of the above equation is derived.

Many thanks for any help!

  • $\begingroup$ Eq 21 doesn't derive from 9; it's proposed for the discrete case instead of 9. In the continuous case it gives equation 9. $\endgroup$
    – Glen_b
    Nov 28, 2014 at 0:01
  • $\begingroup$ I'm new to survival analysis and don't see how Eq 21 degenerates into 9. So given $$\frac{\lambda(t_j, \mathbf{x})}{1 - \lambda(t_j, \mathbf{x})} = \frac{\lambda_0(t_j)exp\{\sum_{i=1}^n\beta_ix_i\}}{1 - \lambda_0(t_j)} $$ You can see that the numerators follow the definition of proportional hazard (Eq 9). So you're saying that in continuous time the denominators equal one another? I.e.: $1 - \lambda(t_j, \mathbf{x}) = 1 - \lambda_0(t_j)$ which leads to $1 - \lambda_0(t_j)exp\{\sum_{i=1}^n\beta_ix_i\} = 1 - \lambda_0(t_j)$. I don't see how this can be equal in continuous time? $\endgroup$
    – Helix
    Nov 28, 2014 at 3:59

1 Answer 1


The answer lies in the $dt$ terms in Cox's equation 21, which your question omitted.

Cox says that in the continuous case equation 21, $$\frac{\lambda(t; z)\,dt}{1 - \lambda(t; z)\,dt} = e^{-z\beta} \frac{\lambda_0(t)\,dt}{1 - \lambda_0(t)\,dt},$$ reduces to equation 9, $$\lambda(t; z) = e^{z\beta} \lambda_0(t).$$

Cox notes that "in discrete time $\lambda_0(t; z) dt$ is a non-zero probability"--in particular, it's the probability of an event occurring between $t$ and $t + dt$, given survival up to time $t$. The continuous-time case corresponds to the limit in which you consider infinitesimally small discrete time slices, that is, $dt \to 0$. In that case, the $\lambda(t; z)\,dt$ and $\lambda_0(t)\,dt$ in the denominators go to zero, so the denominators becomes 1, and then canceling the $dt$ in the numerator yields equation 9.

  • $\begingroup$ Does anyone know of any reliable packages implementing this discrete-time Cox model? $\endgroup$
    – Lei Huang
    Mar 7, 2019 at 1:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.