= Solution
Use the allowed ratio-of-expectations approximation where $G(u)>0$. The unconditional numerator and denominator from part (b) give
$$
\mathbb E\widehat H_{NA}(t)\approx\int_0^t\frac{G(u)\sum_iF_i(u)\,dH_i(u)}{G(u)\sum_iF_i(u)}=\int_0^t\frac{\sum_i e^{-H_i(u)}\,dH_i(u)}{\sum_i e^{-H_i(u)}}.
$$
The common censoring <survivor function> cancels in this approximation. By differentiating the logarithm, its value is \b[the population integrated hazard]:
$$
\boxed{-\log\left(\frac1n\sum_{i=1}^n e^{-H_i(t)}\right)=\overline H(t).}
$$
This is not generally the arithmetic mean of the $H_i(t)$; it is the <cumulative hazard function> of the population mixture from part (a). The cancellation requires a common independent censoring law and adequate observation of the time range.
\b[The approximation establishes approximate unbiasedness, not exact finite-sample unbiasedness.] For a direct counterexample, take one uncensored individual with <exponential distribution> of rate $\lambda$. The <Nelson–Aalen estimator> is then $\mathbf1_{T\le t}$, whose <expectation> is $1-e^{-\lambda t}$, whereas $\overline H(t)=\lambda t$. Thus it is not an <unbiased estimator> in general.
More explicitly, for independent individuals with a common event hazard $h$, the exact expected estimator is
$$
\mathbb E\widehat H_{NA}(t)=\int_0^t\mathbb P(Y(u)>0)h(u)\,du=\int_0^t\left[1-\{1-F(u)G(u)\}^n\right]h(u)\,du.
$$
This exhibits <finite-sample bias of Nelson–Aalen estimation> and dependence on $G$ when empty <risk sets> are possible. In a heterogeneous population, replacing the random ratios by ratios of their means introduces an additional approximation. On adequately observed intervals with large <risk sets>, the population-mixture expression is the appropriate large-sample target.
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