= Exponential survival risk over an interval
{title2=$\mathbb P(a<T\leq b\mid T>a)=1-e^{-\lambda(b-a)}$}
Let the <survival time> $T$ have an <exponential distribution> with rate $\lambda=0.2$ per year. Its <hazard function> is $h(t)=0.2$ per year, its <cumulative hazard> is $H(t)=0.2t$ for $t$ in years, and its <survival function> is $S(t)=e^{-0.2t}$. For a subject still event-free after five years, the <conditional probability> of an occurrence within the next two years is
$$
\begin{aligned}
\mathbb P(5<T\leq7\mid T>5)
&=1-\frac{S(7)}{S(5)}=1-e^{-0.4}\\
&\approx0.329680.
\end{aligned}
$$
The original population's <probability> of an occurrence between those same endpoints is instead $S(5)-S(7)=e^{-1}-e^{-1.4}\approx0.121282$. Dividing by $S(5)\approx0.367879$ recovers the conditional answer. The <memoryless property> makes the two-year conditional answer independent of the preceding five years.
For just $\Delta=0.01$ year, the exact conditional event <probability> is $1-e^{-0.002}\approx0.001998$, close to the instantaneous approximation $h(5)\Delta=0.002$. For two years that approximation gives $0.4$, whereas the exact answer is about $0.329680$. At the single exact time five, the event <probability> is zero. These are three different quantities: an instantaneous <hazard function> with inverse-time units, a short-<interval> approximation, and an exact finite-<interval> <probability>.
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