Buhlmann cred for r.v.'s X1, X2, X3

[NewTubaBoy]

I pray for no thrown out questions.... because it just ups the stupid pass score.

[Abraham Weishaus]

Originally Posted by Tom Craven

The problem didn't state that X1, X2, X3 have the same mean, and, they were random variables, not observations of a single random variable. I think they weren't playing the role of past losses from policyholders of unknown classes, which are labeled X1, ... , Xn in the statement of the Buhlmann model on p. 557 of Loss Models. That would have been a list of past losses like X1=2, X2=6, X3=8, where X1, ... , Xn are i.i.d conditional on the class. As far as I could tell, in the test problem X1, X2, X3 are just 3 random variables, unrelated except for happening to have the same variance.

For example, X1 could be the r.v. representing a loss from a policyholder of class 1 (similarly for X2, X3). If we had see 3 years of loss data for a policyholder of class 1, that would have been 3 observations, all from the distribution of X1. (We could then have applied the Buhlmann model, with mu = E(E(Xi|i)) and n = 3, to determine a credibility estimate for the next observation of X1.)

But when they state, "Given an observation of X1 of (some number), the Buhlmann cred estimate of X2 is 1.4," it seems to me that the observation of (some number) of X1 couldn't have been used in the usual Buhlmann formula with n=1 to produce the credibility estimate for the next obs of X2, since X1 and X2 are not (necessarily) identically distributed. My interpretation was that the estimate 1.4 was produced from the Buhlmann formula with n=0, for the 0 observations of X2 that we are given. (I mean that in that sentence we are not yet given an observation of X2; later in the problem an observation of X2 is given.). So with n=0, Z=0, and 1.4 is mu = E(E(Xi|i)).

Unfortunately I'll never see the exact wording of the problem, but I doubt this interpretation is correct. One usually wouldn't refer to a Buhlmann credibility estimate when there is by design no credibility. (It could happen "accidentally" if v is infinite or a is 0.)

From what I read here, it sounds like 2 was the total variance (not the EPV), and they were testing you on backing out v and Z.

[Antimony]

Originally Posted by Tom Craven

I interpret "Var(Xi) = 2 for all i" as meaning the same thing as "the process variance is 2 for each of the 3 classes."

That's how I read it too. I don't really see how 2 could be the TV, because the wording of the phrase made it clear that it was talking about 3 different variances that were all equal to 2.

[TwinsFan1987]

Do both interpretations lead to a given answer?

[teapeaexcueexplustea]

When I said above that I had done the same thing, I actually misread the post. I saw that Var(Xi) was the process variance for each X, therefore the EPV=Var(Xi)

[Livan33]

Originally Posted by TwinsFan1987

Do both interpretations lead to a given answer?

Were there actual answers for this question? Or was it just ranges? I can't seem to remember.

Never mind, I just saw that Tom said his answer fell into one of the ranges.

[pi_rsquare]

I used Var(Xi) as Total Variance and I got 1.44 as the answer.

If you did it with other method, what answer did you get ?

[Surfohio]

Here is something interesting I tried.

The estimate of x3, assuming it has some credibility from past observations is

x3 = 2/(2+k)*1.25 + k/(2+k) * mu
where 1.25 is the average of actual observations x1 and x2.

Then, since 1/(1+k)*1 + k/(1+k)*mu = 1.4 <-- estimate of x2

mu = (1.4 - 1/(k+1))/(k/(k+1)) = (1.4k + .4)/k

Therefore

x3= 2.5/(2+k) + (1.4k+.4)/2+k = (2.9 + 1.4k)/(2+k) which is always between 1.4 and 1.45 regardless of k (since k is always positive).

Anyone remember the choices for this one?

[Tom Craven]

Originally Posted by Surfohio

Anyone remember the choices for this one?

I THINK that A was: lower than 1.5. Then the remaining choices were increments of .1, up to E being everything above, of course.