how do you calculate moments of truncated distributions? E(s) = E(N) x E(x | x>100) so why do we have to multiply by P(X > 100) in addition to this formula? See solutions to revision booklet 3 page 171 for details. I see that the terms cancel but I don't understand where the extra term comes from in order to cancel the denominator within the integral....
You're introducing an excess of 100, so all claims below 100 will disappear. Hence you need to multiply by the probability that the claims are actually more than 100.
Thanks for the response Ian - but doesn't the E automatically consider this by calculating E[X-100 GIVEN X>100]?
Very short answer: No. But that's CT6, so don't worry about it too much. Short answer: Remember that Pr (A given B) = Pr (A intersection B) / Pr (B) Longer answer:
