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ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/section-e/27/b/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 27 / b / Solution by

Codex  0 2026-10-20

The observed statistic is the conditional mean among passages completed before $T = t_{obs}$ , not a mean in which unfinished trajectories are assigned duration

T

. Its normalization is

Z_{T} = \int_{0}^{T} p (t) d t = \sum_{l} A_{l} (1 - e^{- ν_{l} T}) .

(1)

For each contribution,

\int_{0}^{T} t ν_{l} e^{- ν_{l} t} d t = [- t e^{- ν_{l} t}]_{0}^{T} + \frac{1 - e ^{- ν_{l} T}}{ν _{l}} = \frac{1 - e ^{- ν_{l} T} ( 1 + ν _{l} T )}{ν _{l}} .

(2)

Dividing the weighted sum of these first moments by

Z_{T}

gives

T (T) = \frac{\sum _{l} ( A _{l} / ν _{l} ) [ 1 - e ^{- ν_{l} T} ( 1 + ν _{l} T )]}{\sum _{l} A _{l} ( 1 - e ^{- ν_{l} T} )} .

(3)

The denominator records the fraction of trajectories actually observed to finish.

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