 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-32/4/b/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 4 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

Within month two, each individual contributes only the time spent at risk between times one and two. There are

104 - 6 - 1 = 97

complete one-month contributions. The six event contributions are

0.08, 0.16, 0.22, 0.50, 0.63, 0.72

, and the censored contribution is

0.69

. Consequently

T_{2} = 97 + (0.08 + 0.16 + 0.22 + 0.50 + 0.63 + 0.72) + 0.69 = 100 person-months .

The month-specific likelihood factor in a piecewise-exponential survival model is

θ_{2}^{6} e^{- θ_{2} T_{2}}

θ_{2} = \frac{6}{100} = 0.06 per month .

The eight individuals no longer at risk at time one contribute no month-two exposure. Using either all 112 individuals or all 104 as full-month observations would give the wrong denominator.

 Read the full article

 New to topics? Read the docs here!