On the full likelihood of a transformed sample and the partial likelihood

Question

I am following the 1975 paper by Cox entitled Partial Likelihood.

Consider a vector $y$ of observations represented by a random variable $Y$ having density $f_Y(y;\theta)$ and suppose that $Y$ is transformed into the sequence $(X_1,S_1,X_2, S_2 \dots,X_m, S_m)$ then the full likelihood of this sequence is

$$\prod_{j=1}^m f_{X_j | X^{(j-1)} , S^{(j-1)}} (x_j| x^{(j-1),} s^{(j-1)}; \theta ) \prod_{j=1}^m f_{S_j | X^{(j)} , S^{(j-1)}} (s_j| x^{(j),} s^{(j-1)}; \theta )$$

• Are we transforming $Y$ into a sequence by, as an example, applying a function $f:R \rightarrow R^n$ to $Y$ or is it a different transformation?
• How is the full likelihood obtained? I tried repeated conditioning in the case $m=4$ but in Cox formula the marginal densities are conditioned only on the previous term.

Answer 1

Q1. Are we transforming $Y$ into a sequence? Not in the way you're implying by your function $f:R \rightarrow R^n$. Remember that $Y$ is a vector. Think of the transformation as simply being a way of decomposing, partitioning or rewriting $Y$.

Example 1 later in the paper makes this idea clearer. Here, Cox considers a discrete time discrete state stochastic process with a special state denoted $0$. In this case $Y$ would be a full description of the state occupied at each time point. Alternatively, we can rewrite $Y$ as $(X_1,S_1,X_2, S_2 \dots,X_m, S_m)$ where each $X_i$ describes the path of the process in between visits to state $0$, while $S_i$ counts the number of time steps spent at $0$ on the $i$'th visit. It's clear that the full description of the process can be recovered from the rewritten version.

Q2. How is the full likelihood obtained? Look very carefully at the notation below equation (2) in the paper and you'll see that e.g. $X^{(j)}:=(X_1, X_2,\dots,X_j)$ so the conditioning is not only on the previous term.

For example, in the case $m=4$, we get the full likelihood $f(x_4,s_4,x_3,s_3,x_2,s_2,x_1,s_1)$ as

$\eqalign{ &=f(x_4,s_4,x_3,s_3,x_2,s_2 | x_1,s_1) f(x_1,s_1) \\ &= f(x_4,s_4,x_3,s_3,x_2,s_2 | x_1,s_1) f(s_1 | x_1) f(x_1)\\ &= f(x_4,s_4,x_3,s_3 | x_2,s_2,x_1,s_1) f(x_2,s_2 | x_1, s_1)f(s_1 | x_1) f(x_1)\\ &=f(x_4,s_4,x_3,s_3 | x_2,s_2,x_1,s_1) f(s_2 | x_2,x_1, s_1)f(x_2 | x_1,s_1)f(s_1 | x_1) f(x_1)\\ &=f(x_4,s_4 | x_3,s_3, x_2,s_2,x_1,s_1) f(x_3,s_3 | x_2,s_2,x_1,s_1) \times f(s_2 | x_2,x_1, s_1)f(x_2 | x_1,s_1)f(s_1 | x_1) f(x_1) }$

You can see that, when you rewrite each of the first two terms above by conditioning (on $x_4$ and $x_3$ respectively), the resultant expression is equivalent to equation (2) in the paper in the case $m=4$.