U = F (Y_{n})

, no overflow in the first

k

subsequent years has probability

U^{k}

. Averaging as in part (c) yields

P (τ > k) = n \int_{0}^{1} u^{n + k - 1} d u = \frac{n}{n + k}, k \geq 0.

(1)

The tail-sum formula for expectation therefore gives

E τ = \sum_{k = 0}^{\infty} P (τ > k) = n \sum_{k = 0}^{\infty} \frac{1}{n + k} = \infty.

(2)

The answer is $E τ = \infty$ . This does not contradict almost-sure finiteness: the tail probability tends to zero, but too slowly for its sum to converge.

 Ancestors (10)

10F
Paper 2
ia
2010
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
List of universities
 Home

 View article source

 Discussion (0) New discussion

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

Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 10F / d

Solution

 Ancestors (10)

 Discussion (0) New discussion

 Articles by others on the same topic (0)

Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 10F / d

Solution

 Ancestors (10)

 Discussion (0)  Subscribe (1) New discussion

 Articles by others on the same topic (0)

 Discussion (0) New discussion