# Cox Proportional Hazard Partial Likelihood rigorous derivation

I'm trying to find a rigorous derivation of the partial likelihood for Cox-PH. I find the hand-wavy explanation of Cox in "Partial Likelihood 1975" unsatisfactory:

There are 2 points which are difficult for me:

1. How to arrive to $$P(\text{subj } S_i \text{ had an event at } T_i |\text{ someone from the risk set } R_i \text{ had an event at }T_i)$$ (which I'll use in 2)? Specifically: how do I partition the information or the random variables in a way that I arrive to these conditional probabilities? What assumptions are being made along the way to allow me to only focus on these probabilities and not the others? (e.g., the $$\beta$$'s don't affect the times of the events?)
2. How to arrive from $$P(\text{subj } S_i \text{ had an event at } T_i |\text{ someone from the risk set } R_i \text{ had an event at }T_i)$$ to $$\frac{exp(\beta^Tz_i)}{\sum_{k\in R_i}exp(\beta^T z_k)}$$?

For 2 I'm guessing it's something along the lines of:

$$P(\text{subj } S_i \text{ had an event at } T_i |\text{ someone from the risk set } R_i \text{ had an event at }T_i) = \\ \frac{P(\text{subj } S_i \text{ had an event at } T_i \cap \text{ someone from the risk set } R_i \text{ had an event at }T_i)}{P(\text{ someone from the risk set } R_i \text{ had an event at }T_i)} = \\ \frac{P(\text{subj } S_i \text{ had an event at } T_i )}{P(\text{ someone from the risk set } R_i \text{ had an event at }T_i)} = \frac{h(T_i|z_i)}{\sum_{k \in R_i}h(T_i|z_k)}$$

But am not sure about the last equality.

So, this is my attempt (I've also made a video about this on my YouTube channel, you can find it here). If you find a better way, or if you spot an error, let me know.

It will be divided into two parts - 1) getting from the likelihood to the partial likelihood, and 2) developing the individual terms in the partial likelihood.

Notations:

• $$T_i$$ - a random variable signifying the time of an event or a censor (technically $$\min(X_i,C_i)$$)
• $$\delta_i$$ - a binary random variable that signifies an event or a censor (technically $$\chi_{\{X_i)
• $$Z_i$$ - a set of covariates that are associated with $$T_i$$
• $$t_i$$ - the value that the random variable $$T_i$$ got; we will assume that the variables are ordered by their time, i.e., that $$t_1\le t_2 \le...\le t_n$$
• $$\delta(t_i)=1$$ - someone from our dataset had an event at time $$t_i$$
• $$R(t_i)$$ - the risk set at $$t_i$$
• $$D$$ - the set of events

For brevity I will omit the individual deltas from the derivation, so $$T_i=t_i$$ is (when appropriate) the event $$T_i=t_i, \delta_i=\epsilon_i$$ (where $$\epsilon_i$$ denotes the binary value of the random variable).

Note that: $$\{\delta(t_i)=1\}=\{T_1=t_i \cup ... \cup T_n=t_i\} \\ R(t_i) = \{T_1 (The 1st row omits the $$\delta$$'s, the 2nd doesn't have them)

# Part 1: from the likelihood to the partial likelihood

Suppose we have 3 subjects, the 1st and 3rd experienced events, and the 2nd a censor.

The full likelihood, conditioned on the Z's and on the fact that the times are ordered (I omit this from the notation), will be:

$$\mathcal L(\beta) = P(T_1=t_1, T_2=t_2, T_3=t_3)$$

Note that this is equal to

$$= P(T_1=t_1, \delta(t_1)=1, R(t_1), T_2=t_2, T_3=t_3, \delta(t_3)=1, R(t_3))$$

Because the events added, are a superset of the events already present, and if $$A\subset B \Rightarrow A\cap B = A$$:

• $$\{T_1=1, \delta_1=1\} \subseteq \{\delta(T_1)=1\}$$
• Since we ordered the variables, $$\{T_1=t_1\} = \{T_1=t_1, T_2\ge t_1, ... , T_n \ge t_1\} \subseteq \{R(t_1)\}$$

Using the "chain rule" of probability, we will decompose the new likelihood as follows:

$$\mathcal L(\beta) = P(T_1=t_1, \delta(t_1)=1, R(t_1), T_2=t_2, T_3=t_3, \delta(t_3)=1, R(t_3)) \\ =P(T_1=t_1, \delta(t_1)=1, R(t_1))P(T_2=t_2 | T_1=t_1, \delta(t_1)=1, R(t_1)) \\ \cdot P(T_3=t_3, \delta(t_3)=1, R(t_3)|T_1=t_1, \delta(t_1)=1, R(t_1), T_2=t_2)$$

The assumption made by Cox, that $$h_0$$ is arbitrary, and can be 0 for all times except events make us unable to reason about $$\beta$$ from $$T_2$$ (a censor), so we will collapse this probability to $$P_2$$. We will decompose the 1st term as follows:

$$P(T_1=t_1, \delta(t_1)=1, R(t_1)) = P(T_1=t_1 | \delta(t_1)=1, R(t_1))P(\delta(t_1)=1, R(t_1))$$

We will ignore the 2nd part and collapse it into $$P_1$$ (note, that technically we could compute it, but this will not be true for the subsequent terms). We will do the same for the 3rd term:

$$P(T_3=t_3, \delta(t_3)=1, R(t_3)|T_1=t_1, \delta(t_1)=1, R(t_1), T_2=t_2) \\ =P(T_3=t_3|T_1=t_1, ..., \delta(t_3)=1, R(t_3))P(\delta(t_3)=1, R(t_3) | ...)$$

Note that here the 2nd part is not computable, as it does depend on $$h_0$$ (we know the subjects survived until after $$t_2$$ and now require to compute the probability that they all survived until $$t_3$$ and that then someone had an event at $$t_3$$). So we will collapse this into $$P_3$$. Also note, that the for the 1st part, given $$R(t_3)$$ we don't care anymore about the past events or risk sets.

So, overall we get:

$$\mathcal L (\beta) = P(T_1=t_1|R(t_1),\delta(t_1)=1)P(T_3=t_3|R(t_3),\delta(t_3)=1)P_1P_2P_3$$

We will call the first terms who are not collapsed the partial likelihoods, and extrapolate to a general case:

$$\mathcal {PL}(\beta) = \prod_{i\in D} P(T_i=t_i|R(t_i),\delta(t_i)=1)$$

# Part 2: deriving the individual term

$$P(T_i=t_i|R(t_i),\delta(t_i)=1) = \\ P(T_i=t_i|\{T_1=t_i \cup ... \cup T_n=t_i\} \cap \{T_1

Notice that for all the $$T$$'s before $$i$$ we will get the empty set $$T_1 = t_i \cap T_1 < t_1 = \emptyset$$, so for brevity we will omit them.

$$= \frac{P(T_i=t_i,\{T_i=t_i \cup ... \cup T_n=t_i\} | \{T_i\ge t_i,...,T_n\ge t_i\})}{P(\{T_i=t_i \cup ... \cup T_n=t_i\}|\{T_i\ge t_i,...,T_n\ge t_i\})}$$

In the numerator the 1st event is a subset of the 2nd, so the 2nd cancels. Now, assuming independence, the event $$T_i=t_i$$ is only dependent on $$T_i \ge t_i$$. Also, the probability of a union is equal to the sum of probabilities. So we get:

$$= \frac{P(T_i=t_i | T_i\ge t_i)}{\sum_{j\in R(t_i)} P(T_j=t_i |T_j\ge t_i)}$$

Assuming discrete times, each probability is equal to the hazard, according to the definition of the discrete hazard. So we get:

$$= \frac{h_{T_i}(t_i)}{\sum_{j\in R(t_i)} h_{T_j}(t_i)} = \frac{h_0(t_i)e^{\beta^T Z_i}}{\sum_{j\in R(t_i)} h_0(t_i)e^{\beta^T Z_j}} = \frac{e^{\beta^T Z_i}}{\sum_{j\in R(t_i)} e^{\beta^T Z_j}}$$

# Conclusion

Putting everything together we got that the partial likelihood is equal to:

$$\mathcal {PL}(\beta) = \prod_{i\in D} \frac{e^{\beta^T Z_i}}{\sum_{j\in R(t_i)} e^{\beta^T Z_j}}$$