이전 챕터에서 우리는 \(k\)개의 population의 Hazard가 모두 같은지에 대해 Hypothesis testing을 하는 일반적인 이론에 대해서 공부했다. 즉, 각 population들의 counting process \(N_1, \ldots N_k\)에 대해, pooled sample을 통해 구한 hazard \(\hat A\)와, 각 counting processes들에 대한 hazard \(\hat A_h\), \(h=1,\ldots k\)를 비교하는 방법을 관찰했다. 이 두 hazard의 차이에 대한 (weight sum)\(^2\)를 quadratic form으로 마치 ANOVA와 비슷하게 test statistic을 construct하여 비교한다고 볼 수 있다. 이번 챕터에서는 이 두개의 estimate들의 차이가 어떻게 martingale이 되는지를 보이고, 또한 어떻게 Martingale property를 이용할 수 있는지 살펴볼 것이다. 나아가, 도출한 test statistic을 이용하여 어떻게 hypothesis test를 할 수 있는지까지 다루어 볼 것이다.
Here, we want to compare \(\hat A_h\) and \(\hat A\) only on the set where \(Y_h(s)>0\). So, we define \[ \tilde{A}_h(t)= \int_0^t J_h(s)d\hat A(s)= \int_0^t \frac{J_h(s)J(s)}{Y_\cdot (s)}dN_\cdot (s)= \int_0^t \frac{J_h(s)}{Y_\cdot (s)}dN_\cdot (s), \] where \(J_h(s)=I(Y_h(s)>0)\) and \(J(s)=I(Y_\cdot(s)>0)\).
이 \(\tilde{A}_h\)는 \(\hat{A}\)에 매우 근접한 값이다(small perturbation). 우리는 앞으로 \(\hat A\) 대신에 \(\tilde{A}_h\)를 사용하여 \(\hat A_h\)와 비교할 것이다.
두 번째 equality는 \(d\hat A(s)= \frac{J_\cdot(s)}{Y_\cdot(s)}dN_\cdot(s)\)를 이용하여 구할 수 있다.
Suppose that the null hypothesis is true.
Consider \(\hat A_h(t)-\tilde{A}_h(t)\). Let \(A_h^*(t)= \int_0^t J_h(s)dA(s)\). We have \[\begin{align*} \hat A_h(t)-\tilde{A}_h(t)&=\left(\hat A_h(t)- A_h^*(t)\right)-\left(\tilde{A}_h(t)- A_h^*(t)\right)\\ &=\left(\hat A_h(t)- \int_0^t \frac{J_h(s)Y_h(s)}{Y_h (s)}dA (s)\right)-\left(\tilde{A}_h(t)- \int_0^t \frac{J_h(s)Y_\cdot(s)}{Y_\cdot (s)}dA (s)\right)\\ &=\int_0^t \frac{J_h(s)}{Y_h (s)}dM_h (s)- \int_0^t \frac{J_h(s)}{Y_\cdot (s)}dM_\cdot (s). \end{align*}\] Thus (if the null hypothesis is true), \(\hat A_h(t)-\tilde{A}_h(t)\) is the difference between two martingales, and therefore is also a martingale.
마지막 equation은 아래와 같이 유도할 수 있다: \[\begin{align*} \hat{A}_h(t)-A_h^*(t) &=\int_0^t\frac{J_h(s)}{Y_h(s)}dN_h(s)-\int_0^t\frac{J_h(s)Y_h(s)}{Y_h(s)}dA(s)\\ &=\int_0^t\frac{J_h(s)}{Y_h(s)}(dN_h(s)-Y_h(s)dA(s))=\int_0^t\frac{J_h(s)}{Y_h(s)}M_h(s),\\ \tilde{A}_h(t)-A_h^*(t) &=\int_0^t\frac{J_h(s)}{Y_\cdot(s)}dN_\cdot(s)-\int_0^t\frac{J_h(s)Y_\cdot(s)}{Y_\cdot(s)}dA(s)\\ &=\int_0^t\frac{J_h(s)}{Y_\cdot(s)}(dN_\cdot(s)-Y_\cdot(s)dA(s))=\int_0^t\frac{J_h(s)}{Y_\cdot(s)}M_\cdot(s). \end{align*}\]
즉 \(\hat{A}_h(t)-\tilde{A}_h(t)\)는 martingale이다.
If \(K_h\) is a function, then \[ Z_h(t)= \int_0^t K_h(s) d(\hat{A}_h-\tilde{A}_h)(s) \] provides a reasonable measure of the difference between \(\hat{A}_h\) and \(\tilde{A}\). If \(K_h\) is predictable, then \(Z_h\) is a martingale.
Most of the tests that have been proposed are based on such stochastic integrals, where \(K_h\) has the particular form \[ K_h(s)=Y_h(s)K(s), \] where
\(K(s)\) is bounded,
\(K(s)\) depends only on \(\{(N_\cdot(u), Y_\cdot (u)), u\le s\}\)
여기서부터 위에서 언급했던 vector \((Z_1(t),\ldots,Z_h(t))'\)의 covariance 를 구하는 과정을 보일 것이다.
Observe that we have \[\begin{align*} Z_h(t) &= \int_0^t K_h(s) d(\hat{A}_h-\tilde{A}_h)(s)\\ &=\int_0^t K(s)Y_h(s) (d\hat{A}_h(s)-d\tilde{A}_h(s))\\ &=\int_0^t K(s)Y_h(s) \left(\frac{J_h(s)}{Y_h(s)}dN_h(s)-J_h(s)d\hat{A}(s) \right)\mbox{ }\mbox{ }\mbox{ }\left(\because \tilde{A}_h(t)=\int_0^t J_h(s)d\hat{A}(s)\right)\\ &=\int_0^t K(s)Y_h(s) \left(\frac{J_h(s)}{Y_h(s)}dN_h(s)-J_h(s)\frac{J_\cdot(s)}{Y_\cdot(s)}dN_\cdot(s)\right)\\ &=\int_0^t K(s)J_h(s)dN_h(s)-\int_0^t K(s)J_h(s)J_\cdot(s)\frac{Y_h(s)}{Y_\cdot(s)}dN_\cdot(s)\\ &=\int_0^t K(s)J_h(s)dN_h(s)-\int_0^t K(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dN_\cdot(s) \mbox{ }\mbox{ }\mbox{ }\left(\because J_h(s)J_\cdot(s)= J_h(s)\right)\\ &=\left(\int_0^t K(s)J_h(s)dN_h(s)-\int_0^tK(s)J_h(s)Y_h(s)dA(s) \right)-\left(\int_0^tK(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dN_\cdot(s)-\int_0^tK(s)J_h(s)Y_h(s)dA(s)\right)\\ &=\int_0^t K(s)J_h(s)(dN_h(s)- Y_h(s)dA(s))-\left(\int_0^tK(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dN_\cdot(s)-\int_0^tK(s)J_h(s)Y_h(s)\frac{Y_\cdot(s)}{Y_\cdot(s)}dA(s)\right)\\ &=\int_0^t K(s)J_h(s)(dN_h(s)- Y_h(s)dA(s))-\left(\int_0^tK(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}\left(dN_\cdot(s)- Y_\cdot(s)dA(s)\right)\right)\\ &=\int_0^t K(s)J_h(s)dM_h(s)-\int_0^tK(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dM_\cdot(s). \end{align*}\] Keep in mind that \(N_\cdot= \sum_{h=1}^k N_h\), \(M_\cdot= \sum_{h=1}^k M_h\), \(Y_\cdot= \sum_{h=1}^k Y_h\). Let \[\delta_{hl}=\begin{cases}1 & \mbox{ if }h=l, \\0 & \mbox{ otherwise. }\end{cases}\] Then, we have
\[\begin{align*} Z_h(t) &= \int_0^t K(s)J_h(s)dM_h(s)-\int_0^tK(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dM_\cdot(s)\\ &= \int_0^t K(s)J_h(s)dM_h(s)-\sum_{l=1}^k \int_0^t K(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)} dM_l(s)\\ &= \sum_{l=1}^k \int_0^t K(s)J_h(s)\delta_{hl} dM_l(s)-\sum_{l=1}^k \int_0^t K(s)J_h(s)\frac{Y_h(s)}{Y_\cdot(s)}dM_l(s)\\ &= \sum_{l=1}^k \int_0^t K(s)J_h(s)\left(\delta_{hl} -\frac{Y_h(s)}{Y_\cdot(s)} \right) dM_l(s). \end{align*}\]
By orthogonality of the martingales \(M_1,\ldots,M_k\) (it means \(\left<M_h,M_j\right>(t)=0\) implying \(M_h(t) M_j(t)\) is a martingale, i.e., \(E(M_h(t) M_j(t)|t-)=M_h(0)M_j(0)=0\) with \(M(0)=0\) for all \(t)\), we have \[\begin{align*} \left<Z_h, Z_j\right>(t)&= \sum_{l_1=1}^k \sum_{l_2=1}^k\left<\int_0^t K(s)J_{l_1}(s)\left(\delta_{hl_1} -\frac{Y_h(s)}{Y_\cdot(s)} \right)dM_{l_1}(s), \int_0^t K(s)J_{l_2}(s)\left(\delta_{hl_2} -\frac{Y_h(s)}{Y_\cdot(s)} \right)dM_{l_2}(s)\right>\\ \\ &= \sum_{l=1}^k \int_0^t K^2(s) \left(\delta_{hl} -\frac{Y_h(s)}{Y_\cdot(s)} \right)\left(\delta_{jl} -\frac{Y_j(s)}{Y_\cdot(s)} \right) J_h(s)J_j(s) d\left<M_l, M_l\right>(s)\\ &= \sum_{l=1}^k \int_0^t K^2(s) \left(\delta_{hl} -\frac{Y_h(s)}{Y_\cdot(s)} \right)\left(\delta_{jl} -\frac{Y_j(s)}{Y_\cdot(s)} \right) J_h(s)J_j(s) Y_l(s)dA(s) \mbox{ }\mbox{ }\mbox{ by Theorem 3}\\ &= \int_0^t [K^2(s)J_h(s)J_j(s)]\sum_{l=1}^k \left(\delta_{hj}Y_l(s) -\delta_{hl}Y_l(s)\frac{Y_j(s)}{Y_\cdot(s)}-\delta_{jl}Y_l(s)\frac{Y_h(s)}{Y_\cdot(s)} +\frac{Y_j(s)Y_h(s)Y_l(s)}{Y_\cdot(s)^2} \right) dA(s) \\ &= \int_0^t [K^2(s)J_h(s)J_j(s)] \left(\delta_{hj}Y_h(s) -Y_h(s)\frac{Y_j(s)}{Y_\cdot(s)}-Y_j(s)\frac{Y_h(s)}{Y_\cdot(s)} +\frac{Y_j(s)Y_h(s)}{Y_\cdot(s)} \right) dA(s) \\ &= \int_0^t [K^2(s)J_h(s)J_j(s)] \left(\delta_{hj}Y_h(s) -\frac{Y_h(s)Y_j(s)}{Y_\cdot(s)}\right) dA(s) \\ &= \int_0^t [K^2(s)J_h(s)J_j(s)] \frac{Y_h(s)}{Y_\cdot(s)}\left(\delta_{hj}Y_h(s) -\frac{Y_j(s)}{Y_\cdot(s)}\right) Y_\cdot(s) dA(s) \\ \end{align*}\] The covariance of \(Z_h\) and \(Z_j\) is the expectation of \(\left<Z_h,Z_j\right>(t)\), and may be unbiasedly estimated by \[ \hat\sigma_{hj}(t)=\int_0^t [K^2(s)J_h(s)J_j(s)] \frac{Y_h(s)}{Y_\cdot(s)}\left(\delta_{hj}Y_h(s) -\frac{Y_j(s)}{Y_\cdot(s)}\right) dN_\cdot(s) \]
바로 위의 식에서 이전의 결과를 빼면 \(\int_0^t \mbox{(predictable function)}(dN_\cdot(s)-Y_\cdot(s)dA(s))=:U(t)\)이기 때문에 Theorem 5에 의해 martingale이다. 즉 \(E(U(t))=0\)이라서 \(\int_0^t \mbox{(predictable function)}Y_\cdot(s)dA(s)\)와 \(\int_0^t \mbox{(predictable function)}dN_\cdot(s)\)의 expectation은 같다. 때문에 \(\int_0^t \mbox{(predictable function)}(N_\cdot(s)\)도 covariance의 unbiased estimator이다.
Orthogonality of martingales가 무슨 의미냐면, Orthogonality of \(M_h, M_j\)에 의해서 즉 \(\text{Cov}(dM_h(t), dM_j(t))=d\left<M_h,M_j\right>(t)=0\)이 된다. 그 의미는 즉 \(\left<Z_h,Z_j\right>(t)\)는 Cov의 estimator이고 이것의 cross product term들은 결국 expectation을 취하면 0이 될 것이기 때문에, 고려하지 않는다는 의미이다.
Let \(\hat\Sigma(t)\) be the matrix formed from the \(\hat\sigma_{hj}(t)\)’s. THe matrix \(\hat\Sigma(t)\) is singular: it is fairly clear that the \(Z\)’s satisfy one constraint (namely that they sum to 0), and so we expect that \(\hat\Sigma(t)\) will have rank \(k-1\).
Recall that if \(A\) is a matrix, then a generalized inverse of \(A\) is any matrix \(A^-\) which satisfies \(AA^-A=A\). A particular choice of generalized inverse is the Moore-Penrose inverse, which satisfies in addition the conditions, \(A^-AA^-=A^-\), and \(A^-A\) and \(AA^-\) are symmetric. It is usually denoted \(A^\dagger\).
A reasonable test statistic is \[ X^2(t)= \mathbf{Z}(t)'\hat\Sigma(t)^-\mathbf{Z}(t), \] where \(\hat\Sigma(t)^-\) is some generalized inverse of \(\hat\Sigma(t)\).
이전의 도출한 \(Z_h(t) = \sum_{l=1}^k \int_0^t K(s)J_h(s)\left(\delta_{hl} -\frac{Y_h(s)}{Y_\cdot(s)} \right) dM_l(s)\)에서 \(h\)에 대해 sum을 하게되면 \(\left(\delta_{hl} -\frac{Y_h(s)}{Y_\cdot(s)} \right)\) 부분이 0가 된다. 때문에 constraint 한개를 갖고 있다고 볼 수 있다. 이는 ANOVA에서의 \(X_{i\cdot}-X_{\cdot\cdot}\)과 같은 상황이다. 즉 이 constraint때문에 \(\hat\Sigma(t)\)가 full rank를 가질 수 없어서 invertible하지 않다. 이에 chi-square test statistic을 구하기 위해 generalized inverse를 사용한다.
하지만 이는 \(\mathbf{Z}(t)\)가 asymptotic normality의 특성을 가질 때 이용할 수 있는 property이다. 때문에 엄밀히 하기 위해서는 \(\mathbf{Z}(t)\)의 asymptotic distribution을 derive해보아야 한다. 이는 \(n\) times the integral in \(\left<Z_h,Z_j\right>(t)= \int_0^t [K^2(s)J_h(s)J_j(s)] \frac{Y_h(s)}{Y_\cdot(s)}\left(\delta_{hj}Y_h(s) -\frac{Y_j(s)}{Y_\cdot(s)}\right) Y_\cdot(s) dA(s)\)이 converges to “an integral in which the random quantities \(K, Y_h, Y_j, Y_\cdot\) have been replaced by deterministic functions”라고 가정할 때, martingale CLT를 이용하여 보일 수 있다. 그 다음 Cramer-Wold device를 이용하여 asymptotic normality를 최종적으로 보일 수 있다 (복잡해서 생략).
The general conclusion is that if \(H_0\) is true, then \[
n^{1/2}\mathbf{Z}(t)\stackrel{d}\rightarrow \mathcal{N}(0,\Sigma(t)),
\] where \(\Sigma(t)=(\sigma_{hj})\) where \(\sigma_{hj}\)’s are the quantities to which
the \(\hat\sigma_{hj}\)’s converge in
probability.