3
$\begingroup$

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:

enter image description here

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\chi_{\{X_i<C_i\}}$)
  • $Z_i$ - 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: $$ \{\delta(t_i)=1\}=\{T_1=t_i \cup ... \cup T_n=t_i\} \\ R(t_i) = \{T_1<t_i,...,T_{i-1}<t_i, T_i\ge t_i,...,T_n\ge t_i\} $$ (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

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

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:

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

$\endgroup$

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.