= Solution
Let $N(t)=\sum_iN_i(t)$ and $Y(t)=\sum_iY_i(t)$ be the total observed <counting process> and size of the <risk set>. When the individual <cumulative hazard functions> are all $H_0$, summing the conditional intensity equations gives
$$
\mathbb E(dN(t)\mid\mathcal H_{t-})=Y(t)\,dH_0(t).
$$
Replace the expected increment by its observed value and divide by the predictable <risk set> size. Thus \b[the Nelson–Aalen estimator in integral form] is
$$
\boxed{\widehat H_{NA}(t)=\int_0^t\frac{\mathbf1_{\{Y(u)>0\}}}{Y(u)}\,dN(u),}
$$
with the integrand defined as zero when $Y(u)=0$. This is an integral against the step <counting process>, so each event time contributes its event count divided by the <risk set> just before that time. The <Nelson–Aalen estimator> estimates integrated hazard over times with observable risk; it does not supply information after observation has ended.
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