• 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.

CM2A September 2022 Q9

M

msm

Keen member
A pension fund has been offered two investment opportunities.
Asset A gives an annual return of 3B%, where B is a binomial random variable with
parameters n = 4 and p = 0.4.
Asset B gives an annual return of 4P%, where P is a Poisson random variable with
parameter μ = 2.
Calculate the following three measures of investment risk for each asset:
(a) Variance [1]
(b) Semi-variance [4]
(c) Shortfall probability versus a benchmark return of 4%.[2]

I can't understand the solution for this question? They seem to be using 81/625 and 216/625 to multiply but I can't imagine where those numbers came from? Could someone please explain how to solve parts b and c?
 
It's often the case that our ASET solutions provide the necessary explanation and detail that is sometimes missing from the Examiners' solutions. I know we're getting close to the exam now but I'd still say it's a worthwhile purchase with all the exam technique advice you'd also pick up just reading through it,

John
 
John Potter said:
It's often the case that our ASET solutions provide the necessary explanation and detail that is sometimes missing from the Examiners' solutions. I know we're getting close to the exam now but I'd still say it's a worthwhile purchase with all the exam technique advice you'd also pick up just reading through it,

John
Click to expand...
I'll definitely do that next paper, but for now could you please explain this question? Even if it doesn't explain those numbers - how would one solve this question, please?
 
msm said:
I'll definitely do that next paper, but for now could you please explain this question? Even if it doesn't explain those numbers - how would one solve this question, please?
Click to expand...
nevermind, i got it. It was just 0.6^4
 
Hi Tricia

B has a Bin(4, 0.4) distribution.

If you write the 0.4 as 2/5 and 0.6 as 3/5 then:

P(B = 0) = (2/5)^0 * (3/5)^4 = 81/625

P(B = 1) = 4 * (2/5)^1 * (3/5)^3 = 216/625

Hope this helps
Anna
 
You must log in or register to reply here.
Back
Top