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

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- Solution c

Solution

Only reaction 1 changes

X_{1}

. Starting from five molecules, its successive states are

5 \to 3 \to 1

, with rates

5^{2} = 25

and

3^{2} = 9

; state one is then absorbing for this channel. Thus

P (X_{1} (t) = 5) = e^{- 25 t} .

The probability of still being at three is the convolution of the first waiting time with an exponential distribution and survival of the second:

P (X_{1} (t) = 3) = \int_{0}^{t} 25 e^{- 25 s} e^{- 9 (t - s)} d s = \frac{25}{16} (e^{- 9 t} - e^{- 25 t}) .

Applying the law of total probability to the three possible states gives

g (t) = 1 - \frac{25}{16} e^{- 9 t} + \frac{9}{16} e^{- 25 t} .

It has

g (0) = 0

and tends to one as both reaction waiting times elapse.

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