Assume the <infectivity profile> is separable:
$$
\beta_{t,\tau}=R_tg_\tau,
\qquad g_\tau\geq0,
\qquad \sum_{\tau\geq1}g_\tau=1.
$$
Then $g_\tau$ is the discretized <generation-interval distribution>, $R_t$ is the <instantaneous reproduction number>, and the <infectious disease renewal equation> becomes
$$
\Delta_t=R_t\Lambda_t,
\qquad
\Lambda_t=\sum_{\tau=1}^{t}g_\tau\Delta_{t-\tau}.
$$
Hence $R_t=\Delta_t/\Lambda_t$ whenever the total infectiousness $\Lambda_t$ is positive.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 1 / b / ii / Solution

The <instantaneous reproduction number> at time $t$ is
$$
R_t^{\mathrm{inst}}=\sum_{\tau\geq1}\beta_{t,\tau}.
$$
It is the expected number of secondary infections that one infected individual would produce if the transmission conditions at time $t$ applied throughout that individual's infectious life. The <case reproduction number>, also called the effective reproduction number in this question, is
$$
R_t^{\mathrm{case}}=\sum_{\tau\geq1}\beta_{t+\tau,\tau},
$$
the expected number actually produced by a person infected at $t$ as calendar-time conditions subsequently change. The instantaneous quantity is easier to estimate in real time because it depends on current and past <incidence>; the case quantity depends on future conditions.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 1 / b / i / Solution

{title2=$R_t^{\mathrm{inst}}$}

The instantaneous reproduction number freezes the transmission conditions at calendar time $t$ and counts the expected secondary infections produced over a complete infectious lifetime under those conditions.


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