= Solution
Put $q_t=\Pr(dN_+(t)=1\mid\mathcal H_{t-})$. A binary event increment has conditional <variance> $q_t(1-q_t)$. Since $q_t=Y_+(t)h(t)dt+o(dt)$, its squared mean is of second order, giving
$$
\boxed{\operatorname{Var}(dN_+(t)\mid\mathcal H_{t-})
=Y_+(t)dH(t)+o(dt).}
$$
Treating the predictable <risk set> size as known given the history, the <Nelson–Aalen estimator> increment therefore has first-order variance
$$
\boxed{\operatorname{Var}(d\widehat H(t)\mid\mathcal H_{t-})
=\frac{dH(t)}{Y_+(t)}+o(dt),\qquad Y_+(t)>0.}
$$
Substitute $d\widehat H=dN_+/Y_+$ for $dH$ to obtain the estimated increment variance $dN_+/Y_+^2$. Summing these estimated predictable variances gives \b[the usual <Nelson–Aalen variance estimator>]:
$$
\boxed{\widehat{\operatorname{Var}}(\widehat H(t))
=\int_0^t\frac{\mathbf1_{\{Y_+(u)>0\}}}{Y_+(u)^2}\,dN_+(u)
=\sum_{a_j\leq t}\frac1{Y_+(a_j)^2}.}
$$
The martingale $dN_+-Y_+dH$ has orthogonal increments, which justifies accumulating the predictable variances of the estimation error. This is an estimated sampling variance on the observed at-risk range, not a claim of exact finite-sample unbiasedness after the <risk set> becomes empty. For multiple events at an event time the usual extension replaces the numerator 1 by the event count.
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