= Solution
For a <counting process> $N(t)$ adapted to its observed history $\mathcal F_t$, a predictable <counting-process intensity> $\lambda(t)$ specifies
$$
E\{dN(t)\mid\mathcal F_{t-}\}=\lambda(t)\,dt,
$$
or more generally $N(t)-\int_0^t\lambda(u)\,du$ is a <local martingale>. In <survival analysis>, let $N_i(t)=\mathbf1\{x_i\le t,v_i=1\}$ record failures and $Y_i(t)=\mathbf1\{x_i\ge t\}$ record the <risk set>, including subjects at their own observation time. Under independent <right censoring> and a common <hazard function> $h$, the aggregate intensity is $Y(t)h(t)$, where $N=\sum_iN_i$ and $Y=\sum_iY_i$.
Writing $H(t)=\int_0^th(u)\,du$, the conditional increment equation is $E(dN\mid\mathcal F_{t-})=Y\,dH$. Replacing $dH$ by the observed increment $dN/Y$ gives the <Nelson–Aalen estimator>
$$
\boxed{\widehat H(t)=\int_0^t\frac{\mathbf1\{Y(u)>0\}}{Y(u)}\,dN(u)=\sum_{t_j\le t}\frac1{Y(t_j)},}
$$
where $t_j$ ranges over the distinct observed event times. Censored observations remove subjects from subsequent <risk sets> but do not produce hazard jumps.
For these data the <Nelson–Aalen estimator> gives
$$
\begin{array}{c|rrrrrr}
\text{observation time}&3&4&5&6&9&13\\\hline
\text{risk count just before}&6&5&4&3&2&1\\
\text{hazard increment}&1/6&0&0&1/3&1/2&1\\
\widehat H(\text{observation time})&1/6&1/6&1/6&1/2&1&2
\end{array}
$$
Thus the six fitted <cumulative hazards> sum to $3(1/6)+1/2+1+2=4$, the number of observed failures.
The <event-count identity for Nelson–Aalen cumulative hazards> follows by exchanging the finite sums. With everyone entering at time zero,
$$
\boxed{\sum_{i=1}^n\widehat H(x_i)=\sum_{j:\text{event}}\frac{\sum_i\mathbf1\{x_i\ge t_j\}}{Y(t_j)}=\sum_{j:\text{event}}1=d.}
$$
This includes each failure subject in its own <risk set>. A delayed-entry dataset requires a different risk indicator and is not covered by this particular identity.
Under a <constant hazard survival model>, $\widehat H(x_i)=\widehat\theta x_i$. Imposing the same identity gives the <events divided by exposure estimator>
$$
\boxed{\widehat\theta=\frac d{\sum_i x_i}=\frac4{40}=0.1\ \text{per time unit}.}
$$
Its fitted <cumulative hazards> at the six times are $0.3,0.4,0.5,0.6,0.9,1.3$, again summing to four. It is also the <maximum-likelihood estimate> under independent <right censoring>, since the parameter-dependent <likelihood> is $\theta^d e^{-\theta\sum_i x_i}$. This is a sensible estimate if the exponential survival model is appropriate, but four failures give little precision. The event-count identity by itself does not validate constant hazard; the large final jump in the <Nelson–Aalen estimator> also reflects a <risk set> of one, rather than by itself proving an increasing hazard.
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