Can someone help me to show that $$ \hat{A}(t, \beta_0) = \sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} dN_i(s) $$ is a martingale. The setup is the Cox proportional hazard model in a semiparametric manner.
This is what I got so far:
$$ \sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} dN_i(s) = \sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} (dM_i(s|X) + d\Lambda_i(s|X) ) $$ $$ = \sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} (dM_i(s|X) + d \int_0^s Y_i(u) \alpha_0 e^{X_i^T \beta_0} du ) $$ $$ =\sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} dM_i(s|X) + \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} \sum_{i=1}^{n} Y_i(s) e^{X_i^T \beta} \alpha_0(s)ds $$ $$ =\sum_{i=1}^{n} \int_0^t\frac{1}{\sum_{j}^n Y_j(s) e^{X_i^T \beta_0}} dM_i(s|X) + \int_0^t \alpha_0(s)ds $$