# Cox discrete time regression model question

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!

• 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. – Glen_b Nov 28 '14 at 0:01
• 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? – Helix Nov 28 '14 at 3:59

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.