Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 6 / b / iv

Table of contents
- Solution iv

Solution

With administrative censoring at

c

, one patient is observed to experience

A

with probability

\int_{0}^{c} θ e^{- (θ + ϕ) t} d t = \frac{θ}{θ + ϕ} (1 - e^{- (θ + ϕ) c}),

so

E [v_{+}] = n \frac{θ}{θ + ϕ} (1 - e^{- (θ + ϕ) c}) .

The observed time is

min (T_{A}, T_{B}, c)

, whose mean follows from the tail-sum formula for expectation:

E [min (T_{A}, T_{B}, c)] = \int_{0}^{c} e^{- (θ + ϕ) t} d t = \frac{1 - e ^{- (θ + ϕ) c}}{θ + ϕ} .

Hence

E [x_{+}] = n \frac{1 - e ^{- (θ + ϕ) c}}{θ + ϕ}, \frac{E [ v _{+} ]}{E [ x _{+} ]} = θ .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version