 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-207/6/a/solution

For

g (u) = e^{- u}

\overline{S} (t) = \int_{0}^{\infty} e^{- u {1 + H_{0} (t)}} d u = \frac{1}{1 + H _{0} ( t )} .

\overline{h} (t) = - \frac{d}{d t} lo g \overline{S} (t) = \frac{h _{0} ( t )}{1 + H _{0} ( t )} .

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