• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Q&A Bank 2 Q24

A

Annie79

Member
Please can someone help with the final stage of this question?

The question is about a Poisson claims process with given claim size density function which it asks you to derive the MGF for. I am happy with the solution up to the last step where it seems to put limits into the integrated expression.
What happens to the two infinity limits and why do the denominators change from (t-3) and (t-7) to (3-t) and (7-t)? Is there a quick trick to this?

The solution jumps from:
3/2[e^(t-3)x / (t-3)] upper limit infinity and lower limit of zero
+ 7/2[e^(t-7)x / (t-7)] with same limits and integrated wrt x

to the MGF of:
3/2(3-t) + 7/2(7-t)

Thanks!!
 
Dealing just with the first integral (same logic applies to second integral):

The integral converges only if t is less than 3 (otherwise you'll get e to the infinity). If t is less than 3 then the numerator of the integrand is e to the minus something. At x equals infinity this will tend to zero. So the answer becomes 3/2(t-3) times (e to the minus infinity minus 1), which is 3/2(t-3) times -1 which is 3/2(3-t).
 
Back
Top