Only reaction 1 changes $X_1$. Starting from five molecules, its successive states are $5\to3\to1$, with rates $5^2=25$ and $3^2=9$; state one is then absorbing for this channel. Thus
$$
\mathbb P(X_1(t)=5)=e^{-25t}.
$$
The probability of still being at three is the <convolution> of the first waiting time with an <exponential distribution> and survival of the second:
$$
\mathbb P(X_1(t)=3)
=\int_0^t25e^{-25s}e^{-9(t-s)}\,ds
=\frac{25}{16}\left(e^{-9t}-e^{-25t}\right).
$$
Applying the <law of total probability> to the three possible states gives
$$
\boxed{g(t)=1-\frac{25}{16}e^{-9t}
+\frac9{16}e^{-25t}.}
$$
It has $g(0)=0$ and tends to one as both reaction waiting times elapse.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-356/1/c/solution