Solution: Unbiased variance estimator in the two sample tests

Note that under the null hypothesis, the counting processes \(N_1(t)\) and \(N_2(t)\) have intensity processes \(\lambda_1(t) = \alpha(t)Y_1(t)\) and \(\lambda_2(t) = \alpha(t)Y_2(t)\), respectively, so that the aggregated process \(N(t) = N_1(t) + N_2(t)\) has intensity process \(\lambda(t) =\alpha(t)Y(t) = \alpha(t)(Y_1(t) + Y_2(t))\), and decomposition \(N(t) = \int_0^t\alpha(s)Y(s)ds+M(t)\). We can now work with the expression for \(V_{11}(t_0)\) from \((3.55)\): \begin{align*} V_{11}(t_0) &= \int_0^{t_0} \frac{L(t)^2}{Y_1(t)Y_2(t)}dN(t) \\ &= \int_0^{t_0} \frac{L(t)^2}{Y_1(t)Y_2(t)}\alpha(t)Y(t)dt + \int_0^{t_0} \frac{L(t)^2}{Y_1(t)Y_2(t)}dM(t) \\ &= \langle Z_1\rangle(t_0) + \int_0^{t_0} \frac{L(t)^2}{Y_1(t)Y_2(t)}dM(t),\end{align*} where we've used the expression for the predictable variation process of \(Z_1(t_0)\) in \((3.54)\), which is what we seek to estimate with \(V_{11}(t_0)\). The integrand in right term is a predictable process (by the properties of \(Y\) and the assumptions on \(L\)), and thus the stochastic integral is a mean zero martingale, meaning that our estimator is unbiased.