 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-40/1/b/ii/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 1 / b / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

For individual

i

, let

X_{i} = min (T_{i}, c_{i})

and

Δ_{i} = 1_{{T_{i} \leq c_{i}}}

. The preceding expectation calculation applies separately at each fixed censoring time:

E [H (X_{i})] = θE [X_{i}] = 1 - e^{- θ c_{i}} = E [Δ_{i}] .

Summing and using linearity of expectation, with

D = \sum_{i} Δ_{i}

the failure count, gives

i = 1 \sum n E [H (X_{i})] = θ i = 1 \sum n E [X_{i}] = E [D] .

This identity needs the stated marginal exponential distributions, but not independence between individuals; independence will be needed for the product likelihood.

 Read the full article

 New to topics? Read the docs here!