Does Cox Regression have an underlying Poisson distribution?

Question

Our small team was having a discussion and got stuck. Does anyone know whether Cox regression has an underlying Poisson distribution. We had a debate that maybe Cox regression with constant time at risk will have similarities with Poisson regression with a robust variance. Any ideas?

Answer 1

Yes, there is a link between these two regression models. Here is an illustration:

Suppose the baseline hazard is constant over time: $h_{0}(t) = \lambda$. In that case, the survival function is

$S(t) = \exp\left(-\int_{0}^{t} \lambda du\right) = \exp(-\lambda t)$

and the density function is

$f(t) = h(t) S(t) = \lambda \exp(-\lambda t)$

This is the pdf of an exponential random variable with expectation $\lambda^{-1}$.

Such a configuration yields the following parametric Cox model (with obvious notations):

$h_{i}(t) = \lambda \exp(x'_{i} \beta)$

In the parametric setting the parameters are estimated using the classical likelihood method. The log-likelihood is given by

$l = \sum_{i} \left\{ d_{i}\log(h_{i}(t_{i})) - t_{i} h_{i}(t_{i}) \right\}$

where $d_{i}$ is the event indicator.

Up to an additive constant, this is nothing but the same expression as the log-likelihood of the $d_{i}$'s seen as realizations of a Poisson variable with mean $\mu_{i} = t_{i}h_{i}(t)$.

As a consequence, one can obtain estimates using the following Poisson model:

$\log(\mu_{i}) = \log(t_{i}) + \beta_0 + x_{i}'\beta$

where $\beta_0 = \log(\lambda)$.