Actuarial Outpost ASM Exam C, 14th edition, Problem 33.14
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#1
09-15-2012, 01:55 AM
 drsingh Member CAS SOA Join Date: Jul 2012 Posts: 454
ASM Exam C, 14th edition, Problem 33.14

Question 33.14

You are given the following claim settlement activity for a book of automobile claims as of the end of 2008:

1. Reported in 2006: 4 settled in 2006, 2 settled in 2007 and UNKNOWN number settled in 2008.

2. Reported in 2007: 5 settled in 2007 and 3 settled in 2008.

3. Reported in 2008: 6 settled in 2008

L=(year settled - year reported) is a random variable describing the time lag in settling a claim. L follows a Poisson distribution. Determine the maximum likelihood estimate of mean lag in settling a claim, E[L].
--------------

My approach:
Step 1: From the above, derive the below table
L Number of times seen in data
0 4+5+6=15
1 2+3=5
>=2 Unknown (data is truncated)

Step 2: There are only two categories of observed data, so use Bernouli shortcut for MLE (MLE will maintain relativity of the observations in the two categories of the data.)

Hence, MLE will give a lambda such that

P (L=1) : P(L=0) :: 5 : 15
Or, lambda = 5/15 = 1/3, given that it is Poisson, where p(1) = lambda * p(0).
("a:b :: c:d" means a/b is same as c/d)

The solution given in manual follows similar approach, but IGNORES the 6 claims reported in 2008 and settled in 2008 from the calculation of the total for the zero lag category of observed data. I don't understand the reasoning given in the manual.

Can someone elaborate more?

Last edited by drsingh; 09-15-2012 at 02:00 AM..
#2
09-15-2012, 08:15 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,881

Consider this problem, where the principle is similar.

Tom, Dick and Harry find a coin, which may not be a fair coin. They decide to experiment by flipping the coin, each choosing the number of times he would flip it.

Tom reports: I got 6 heads and 4 tails.
Dick reports: I got 9 heads and 9 tails.
Harry reports: Oh, was I supposed to count heads and tails? I didn't. I got 15 heads but didn't keep a record of the tails.

Would you calculate the MLE of p(heads) as (6+9+15)/(6+4+9+9+15)?

Same thing here: you have no reason to believe that the number reported in 2008, settled in 2009 will be 0, so you shouldn't calculate the MLE the same as if you knew it was 0.
#3
09-15-2012, 12:12 PM
 drsingh Member CAS SOA Join Date: Jul 2012 Posts: 454

Quote:
 Originally Posted by drsingh Question 33.14 You are given the following claim settlement activity for a book of automobile claims as of the end of 2008: 1. Reported in 2006: 4 settled in 2006, 2 settled in 2007 and UNKNOWN number settled in 2008. 2. Reported in 2007: 5 settled in 2007 and 3 settled in 2008. 3. Reported in 2008: 6 settled in 2008 L=(year settled - year reported) is a random variable describing the time lag in settling a claim. L follows a Poisson distribution. Determine the maximum likelihood estimate of mean lag in settling a claim, E[L]. -------------- My approach: Step 1: From the above, derive the below table L Number of times seen in data 0 4+5+6=15 1 2+3=5 >=2 Unknown (data is truncated) Step 2: There are only two categories of observed data, so use Bernouli shortcut for MLE (MLE will maintain relativity of the observations in the two categories of the data.) Hence, MLE will give a lambda such that P (L=1) : P(L=0) :: 5 : 15 Or, lambda = 5/15 = 1/3, given that it is Poisson, where p(1) = lambda * p(0). ("a:b :: c:d" means a/b is same as c/d) The solution given in manual follows similar approach, but IGNORES the 6 claims reported in 2008 and settled in 2008 from the calculation of the total for the zero lag category of observed data. I don't understand the reasoning given in the manual. Can someone elaborate more?
OK, so after posting, I decided to do this problem the long way. I first calculated the Likelihood function for each of the years and multiplied them all. In writing the expression for each year, after the usual numerator, I divided the denominator by the "total probability of the untruncated possibilities raised to the power of total number of observations" in each year.

Essentially, if we only have data for p(l=j through k) then

the numerator is: p(j)^n_j * ... * p(k)^n_k

and the denominator is: [ p(j) +....+p(k) ] ^(n_j + ... + n_k)

If we have data for all possibilities, denominator is 1; if only left truncation, denominator reduces to S(d)^total N, etc.
#4
09-15-2012, 12:14 PM
 drsingh Member CAS SOA Join Date: Jul 2012 Posts: 454

Quote:
 Originally Posted by Gandalf Consider this problem, where the principle is similar. Tom, Dick and Harry find a coin, which may not be a fair coin. They decide to experiment by flipping the coin, each choosing the number of times he would flip it. Tom reports: I got 6 heads and 4 tails. Dick reports: I got 9 heads and 9 tails. Harry reports: Oh, was I supposed to count heads and tails? I didn't. I got 15 heads but didn't keep a record of the tails. Would you calculate the MLE of p(heads) as (6+9+15)/(6+4+9+9+15)? Same thing here: you have no reason to believe that the number reported in 2008, settled in 2009 will be 0, so you shouldn't calculate the MLE the same as if you knew it was 0.
Thanks for this example. After the post last night, I was able to solve all such problems the "long way". Now with our example, and additional insight, I can do it the short way as well.
#5
12-09-2017, 05:32 PM
 gauchodelpaso Member SOA Join Date: Feb 2012 College: Eastern Michigan U Posts: 79

This problem is in recent Manuals listed in Chapter 35, Fitting discrete distributions, Exercise 35.14. I believe it is important for solving it with first principles to take in account that all the info after 2008 is unknown, so it is truncated from above and that you need to divide each Poisson probability by the cumulative F at the point of truncation. That will give immediately for 2008 that the probability of a Lag of 0 as 1, so the 6 in the sample doesn't matter. On top for the 2006 year the truncation goes all the way to 2008, so you don't know anything after 2007 for the claims reported in 2006. And then just remembering the fact of g(x) = f(x)/S(L) where L is the limit, and doing the loglikelihood of the remaining points it's a simple exercise., benefited by p0 = e^-(lambda) being a common factor for numerators and denominators of all g(x).

After grouping similar factors we get:
-14 ln(1+lambda) +5 ln(lambda) = ln(L),

getting its derivative = 0, you arrive to lambda = 5/9.

The author explains that it may be solved differently using some aspects of this particular problem.

1) the cases of lag 0 reported in 2008 don't count as their probability is p[2008(0)]/p[2008(0)] = (e^-lambda/e^-lambda) =1 and then log[g2008(0)] = 0.

2) for the remaing years 2006 and 2007 we have n0=9 cases of lag of 0 and n1=5 cases of lag of 1. Therefore there are no other probabilities to account for, like p2, as we truncated the only one as it's unknown. So the use of Bernouilli gives p1/p0= n1/n0

3) That coupled with the relationship between p1 an p0 for Poisson being the parameter, lambda=p1/p0=n1/n0=5/9.

I tried to follow the other explanation, and it seems to be similar, but got lost somewhere in the middle of the explanation. I assume that similarly some people may get lost in mine too, but between the two and the author's solution perhaps it may be understood. Good luck! I found that Mahler's treatment of the subject may help too. But the ASM table 32.1 is important to keep in mind for censored and truncated distributions.
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