= Solution
Let
$$
a=e^{\beta z_{n-1}},\qquad b=e^{\beta z_n},\qquad c=e^{\beta z_{n-2}},
$$
and let $C$ be the common <partial likelihood> contribution from the first $n-3$ observations. Since $t_{n-2}>x_{n-2}$, the three possible complete-data tail orderings and their partial likelihoods are
$$
\begin{array}{c|c}
t_{n-2}<x_{n-1}<x_n&C\dfrac{c}{a+b+c}\dfrac{a}{a+b}\\[6pt]
x_{n-1}<t_{n-2}<x_n&C\dfrac{a}{a+b+c}\dfrac{c}{b+c}\\[6pt]
x_{n-1}<x_n<t_{n-2}&C\dfrac{a}{a+b+c}\dfrac{b}{b+c}.
\end{array}
$$
Their sum is
$$
C\left[
\frac{c}{a+b+c}\frac{a}{a+b}
+\frac{a}{a+b+c}\frac{c+b}{b+c}
\right]
=C\frac{a}{a+b}.
$$
When individual $n-2$ is <right-censored> at $x_{n-2}$, that individual leaves the <risk set> before the event at $x_{n-1}$, so the directly calculated <Cox partial likelihood> is also $C,a/(a+b)$. Summing over the unobserved compatible event orderings therefore reproduces the censored-data partial likelihood.
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