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I have just started learning on the Cox proportional hazards model. I understand that the hazard function is the multiplicative of the baseline hazard rate $h_0(t)$ and the hazard rates dependence on the covariates $\exp(\beta x)$ where $\beta$ are the coefficients and $x$ are the covariates. This gives rise to the hazard function $h(t|x) = h_0(t)\exp(\beta x)$ in which a change in covariates gives rise to a multiplicative scaling on the hazard which stays constant over time.

When finding maximum likelihood estimates of the regression parameters $\beta$, the MLE estimate maximises the probability of observing the given set of survival times.

The probability that a subject $i$ fails in time $t_j$ is should be given by $$\mathbb P(\text{subject i fail in $t_j$} + \Delta t|\text{survive up till $t_j$}) = h_0(t_j)\exp(\beta x_i)\Delta t$$ where $\Delta t$ is a fine interval. The likelihood is the same equation without the $\Delta t$. I think this should be correct but I might be wrong.

However, the probability that subject $i$ fails in time $t_j$ is usually normalised over the other subjects that have probability of failing at time $t_j$ as well. Why do we have to normalise by the summing over the probability of all subjects at risk of failing at time $t_j$?

The form is usually written as $$\frac{h(t_j|x_i)}{\sum_k h(t_j|x_k)}$$ where $k$ represents the number of people at risk in time $t_j$.

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Rewrite your last expression in terms of both the baseline hazard $h_0(t)$ and the covariate-associated hazard ratios:

$$\frac{h_0(t_j)\exp(\beta x_j)}{\sum_k h_0(t_j)\exp(\beta x_k)}= \frac{\exp(\beta x_j)}{\sum_k \exp(\beta x_k)}$$

where $k$ represents the people at risk in time $t_j$. That's the value of the proportional-hazards assumption: the baseline hazard function just factors out of the further calculations that estimate the coefficient values.

You shouldn't call the result a "maximum likelihood estimate"; as AdamO notes in another answer, it's based on a "partial likelihood" as the procedure doesn't take into account the baseline hazard.

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  • $\begingroup$ Oh I see I think I get it. We cannot maximise the numerator $h_0(t_j)exp(\beta x)$ because we don’t know $h_0(t)$ So we divide by the denominator to get rid of it? $\endgroup$
    – calveeen
    May 18, 2021 at 16:04
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    $\begingroup$ @calveeen that was Cox's insight in devising proportional-hazards regression, as I understand it. $\endgroup$
    – EdM
    May 18, 2021 at 16:19
  • $\begingroup$ I think my confusion comes from what the $\frac{exp(\beta x_j)}{\sum_k exp(\beta x_k)}$ means because $h(t_j | x_i)$ is the probability of subject $i$ failing at time $t_j$. What does dividing by the sum of the probability of all other subjects that have not failed at time $t_j$ mean? other than getting rid of the baseline hazard $h_0(t)$ $\endgroup$
    – calveeen
    May 19, 2021 at 2:06
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    $\begingroup$ @calveeen I think of the ratio at the end of your question as the fraction of the "total hazard among those at risk" (the denominator) associated with the individual who had the event at time $t_j$. (If all individuals have the same hazard, this ratio represents the fractional drop in a Kaplan-Meier curve at time $t_j$.) Optimizing to find the $\beta$ values can be thought of as trying to align higher-hazard cases with earlier event times. With proportional hazards, you don't need further to take the baseline hazard into account. $\endgroup$
    – EdM
    May 19, 2021 at 11:36
  • $\begingroup$ Thank you I think it clears things up $\endgroup$
    – calveeen
    May 19, 2021 at 12:57
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The $h()$ is not a probability, it is a hazard, although they are monotonically related. The Cox model is not a full likelihood procedure, it maximizes a partial likelihood. Even though we don't directly estimate the hazard function as a nuisance parameter (which would be a conditional likelihood approach), we pretend we know what order people enter and leave the cohort, and who fails or is censored. This grouping, called the risk sets, is the key to the "normalizing" factor here. If we had a logistic regression, we would simply use the covariates to predict the probability of being a "case" in an analysis. However, since the sums of hazards don't normalize to any constant or have any bounds or constraints aside from being positive, we need to consider how many others are in a risk set to sort of rank the risk of a particular subject being a case in a particular risk set.

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  • $\begingroup$ Yes the hazard is a likelihood. In my post I multiplied the likelihood by a small constant $\Delta$ To get a probability. Why can’t we find parameters $\beta$ that maximise this probability. We can drop the small $\Delta$ and maximise the likelihood. However we maximise a term that is normalised over the probability of any subject failing at time $t_j$. $\endgroup$
    – calveeen
    May 18, 2021 at 16:00

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