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Rusty CT3 knowledge!

V

vix

Member
I am trying to prepare for CA2 but am having trouble with my dull and distant memory of CT3!

I need to find an upper limit, B, such that the expected value of an exponential distribution (lambda = L) is equal to a constant, Z. Unfortunately, the equation required to do this is not provided in the notes for the assignment so I am trying to resolve this with my rather rusty old stats knowledge.:eek:

So far, I have set the integral between 0 and B of xL.exp(-Lx) equal to Z and integrated by parts to obtain an equation in terms of the unknown, B.

The result I obtain is:

–B.exp(-LB) – (1/L).exp(-LB) + (1/L) = Z

But solving this for B does not give me a result of the order that I expected for B.

Can anyone tell me where I am going wrong?
 
Last edited by a moderator:
Solution Found

Having spent a good few hours on this last night, I finally found the problem.

Firstly, I should have been trying to find E(S) where E(S)=E(N).E(X).

Secondly, E(X) = [–B.exp(-LB) – (1/L).exp(-LB) + (1/L)] / P(X<B)

Using E(N) = Total number of payouts x P(X<B) then gives:

E(S) = Total number of payouts x [–B.exp(-LB) – (1/L).exp(-LB) + (1/L)]

It took some time but I'm pleased to say that my integration was at least on track.

My only question now is, why would the formula for this not be provided for a CA2 assignment if it is your Excel modelling and audit trail documentation that is being tested rather than your maths?:confused:
 
Thank you

Thanks Vix,

I finally get it!

I did my CA2 course last week, but the pre-course assignment we did had a v similar (identical?) calculation in it. I know we're not allowed talk about the course because they re-use assignments, so I was a bit stuck. Went to the course and found a couple of other people had had problems with the same calc, but couldn't find anyone to explain it to me. Indeed, I spent ages discussing it with the guy sitting next to me - thanks Mike!

Turns out I didn't need that calculation for the actual assessment, but still I guess in the long run it's nice to know where that comes from.

Thanks again
Louise
 
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