TIA PE4 #25

[oofta]

You observe two loss for amounts 5 and 7 for one insured. Find the Bulmann credibility estimate for the third loss for that insured.


This is not the entire question. I was able to find a,v, mu, and xbar just fine. I am curious why we set n=2 in P_c? I was thinking that there was only one exposure since there is only one insured.

[truonda]

I just remember that the n in Buhlmann's k is always the same as the n used to calculate xbar. Hope that helps.

[killjacker]

One exposure is not equal to one insured, it's equal to one insured-year. You have 2 insured-years, so there are 2 exposure units.

[home_alone]

Originally Posted by truonda

I just remember that the n in Buhlmann's k is always the same as the n used to calculate xbar. Hope that helps.

Be careful. This is not always the case. There are some situations where it is not true.

[truonda]

Originally Posted by home_alone

Be careful. This is not always the case. There are some situations where it is not true.

Can you provide an example? (so I don't fall into my own trap)

EDIT: I know that if you need to calculate an estimate for a specific group then n becomes the n used to calculate xbar for that group. Could this be what you were talking about?

[home_alone]

Originally Posted by truonda

Can you provide an example? (so I don't fall into my own trap)

EDIT: I know that if you need to calculate an estimate for a specific group then n becomes the n used to calculate xbar for that group. Could this be what you were talking about?

Originally Posted by MoProblems_MoMoney

I always make a last minute formula sheet / reminder list to look over before an actuarial exam. I thought I'd share what's on mine and see if anyone else has something similar.

• Match Buhlmann n to the # of "units" where Pc is calculated per "unit"
• Look for conditional variance formula (any time a parameter is unknown with a known distribution)
• Read NA-KM problems for NA/KM and F/S/H, carefully
• Bootstrap E[(A-B)^2] where A is a stupid estimator and varies in each case, B is constant and a smart estimator
• "Like X^2 but Smirnoffed Darling is needy" -- Likelihood and chi square are robust and do not need adjustment but Smirnoff and Anderson Darling need a lot adjusting of critical values
• "Darling is a slut" - no adjustment as n->infinity and heavy on tail weight estimation.
• Watch S(x) for >= vs > bounds
• In any deductible problem, look for CONDITIONING!
• Be slow and thorough!

I don't have an example but here is what I got from previous discussion on this issue. See the first bullet point.

[truonda]

I still don't think there is anything wrong with my logic.

[home_alone]

If it does work for you then you keep it then. It is yours. But remember it does not work in semiparametric cases for example and some other cases too.