Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 9F / iv

Table of contents
- Solution iv

Solution

Conditioning on

N

and using the moment-generating function of a unit exponential variable,

E (e^{θ Y}) = n = 1 \sum \infty p (1 - p)^{n - 1} (\frac{1}{1 - θ})^{n} = \frac{p}{1 - θ} m = 0 \sum \infty (\frac{1 - p}{1 - θ})^{m} = \frac{p}{p - θ},

where the geometric series converges precisely when

θ < p

. This is the moment-generating function of an exponential variable of rate

p

, so the geometric sum of exponential variables satisfies

Y \sim Exp (p) .

 Ancestors (10)

 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