COURSE 4 EXAM & ANSWERS POSTED

[Anonymous]

Go to the SOA website!

[Sydney Bristow]

Here's the link:

http://www.soa.org/eande/course4_1101.pdf

[Hagbard Celine]

Well....WELL.

Are people going to post their scores? What will the pass mark be...

[Sydney Bristow]

I've got 27 if these "preliminary" answers stick...

[Higher Authority]

I'm screwed! 20 answer's correct. I guess I'll be visiting test 4 again.

<font size=-1>[ This Message was edited by: needo on 2001-11-07 14:53 ]</font>

[Hagbard Celine]

24...hoping they stick!

[New at pd]

30

[Anonymous]

28!!!

[Hagbard Celine]

So.

What's the pass mark gonna be?

It was 22 last time, right? They wouldn't dream of increasing it by 3 hopefully!

[LB]

I got 26. Will this be enough?

[Joe]

27, party people!! At last!!

Screw you SOA guys, I'm going home...

[gloid]

31

[Anonymous]

Only 19. I'm a goner.

[scout]

29 - hoping that I stored everything correctly...

[Anonymous]

27!! So what do you think is the pass mark?

[erickson]

29.

[NYHeel]

I got 23. Is that enough? Was the pass mark 22 last sitting?

[M.]

15. Man, this test sucks. I fell for the trick answer too many times. Oh well.

[Weatherman]

How did the popular answer key do?

[C8 Guy]

28.

[Mo' 5!]

***

[Retired Bookworm]

21 Here...(grumble) hey 4xC4L, can I inherit your Username when you're done with it?

[Bullseye]

I got a 21. ( I think ) Therefore, I am also hoping for a pass score of 21...but I doubt it.

[Anonymous]

On 2001-11-07 13:35, Casual Bookworm wrote:
21 Here...(grumble) hey 4xC4L, can I inherit your Username when you're done with it?

No way, I want it!

[Lara Croft]

Good job, Hagbard!

Looks like you will be joining me for Part 7 in the spring!

[Hagbard Celine]

I hope so Lara. I worry about counting my eggs right now though.

[Anonymous]

I'm unable to access the SOA website. Could someone please post the aswers here?

[Anonymous]

Course 4: November 2001 -41- STOP
Test Item Key
1 B
2 E
3 B
4 B
5 B
6 E
7 C
8 D
9 A
10 A
11 D
12 A
13 B
14 E
15 C
16 E
17 E
18 E
19 B
20 B
21 D
22 B
23 D
24 C
25 A
26 C
27 C
28 A
29 E
30 C
31 D
32 B
33 D
34 D
35 D
36 A
37 E
38 B
39 E
40 C

[Course 4 Escapee]

On 2001-11-07 13:35, Casual Bookworm wrote:
21 Here...(grumble) hey 4xC4L, can I inherit your Username when you're done with it?


Well, I got 16 right. So far, I've had a 4, 5, 3 and this time probably a 2 or 3. So I think I'll be holding on to my title of shame. :sad:

[VP]

Popular vs. actual didn't do to well.

"blue misses" - popular choice was between 40 and 60%
#2 - 22% chose correctly, 54% chose the most popular wrong answer (22-54)
#6 - 15-48
#14- 14-40
#17- 25-45
#27- 40-44

"red misses" - popular choice was more than 60%
#23- 22-62

"black misses" - popular choice was less than 40%
#24- 5-32 (5%!!!! of the 59 got it right!)
#35- 18-34

[Weatherman]

Was anyone protesting #24? Or do you know why the answers are wrong?

[Lloyd Dobler]

I would like to protest #24 because I didn't get it right. I would also like to protest #6 because E makes no sense to me. The first choice sounds much better. I say throw em both out & throw out #1 as well since I f****ed up that calculation somehow. Someone please explain #24 & #6 to me. Thanks.

[Anonymous]

How I thought about the reason that E is right and A is wrong for # 6 is that you have fitted F(x) > real F(x) on the right tail. That means (1 - fitted F(x)) < (1 - real F(x)), and fitted S(x) < real S(x). So the fitted distribution is too thin on the right.

[Joe]

We could always complain about questions. For instance, the simulation question where they asked the number of simulations. Personally, I derived both answers (800 and 1000) and decided to go with 800, wrongly. I'm pissed off. This is what I despise about the exams. They test not only the knowledge, but they set traps in which we could fall even if we understand the subject.

I even went to a Mahler seminar and asked him when to use n or n-1 in such simulation or credibility questions. His reply was: "it shouldn't make a difference in the answer choice. A good setup for range of answers would include both answers in the same range."

So there you have it. I'm OK because I got more than enough to pass, but if I had 21 or 22 and dodging such a glitch would have made me successful, I'd be livid. I fully understand your frustrations.

[Anonymous]

Man I hate this course. It amazes me that I can study and think I know the material way better than I did before, only to be stumped so many times. There were so many problems where I was close but couldn't light the cigar. Scored 21.
Although there seems to be some debate on difficulty between this sitting and past ones (I thought this one was at least as hard as May01), it will be obvious to the SOA statisticians that do the grading. I'll bet there are many q's that the distribution by selected answer is even (guesses). I'm hoping that this will lead to a lower pass mark and a 6 for me. Just look at how poorly Roger's popular key did (thanks anyway though RL...it's a great webpage). Good luck to those that study their tails off.
DD (C4DD)

[Anonymous]

You're right, I'm just frustrated with this test.


<font size=-1>[ This Message was edited by: Pita on 2001-11-08 10:46 ]</font>

[New at pd]

I'm sorry, but I don't see any issue with number 17. It is plainly described in Actex, ASM, HTP, and Cox that the sample variance is to be used when determining the number of data values to generate.

[Hagbard Celine]

It doesn't matter that it is clearly described in those 4 manuals...they are not on the syllabus.

The question is whether it is clearly described in the Simulation text.

I didn't read it carefully enough one way or the other to know, but it really doesn't matter how well those other guys spell it out.

[AK]

It is clearly stated on p.115 of Ross that the sample variance should be used.

<font size=-1>[ This Message was edited by: AK on 2001-11-08 10:42 ]</font>

[gloid]

Just like they say in the study manuals, they are supposed to be used as a "supplement" to the actual material. I don't think you should protest a problem because you didn't want to read the textbooks.

[Anonymous]

--------------------------------------------------------------------------------
FYI:This is what Abe Weishauss had to say. In case you don't know him, he is the ASM manual guy.

Sorry for the delay, I didn't get a chance to go through all the questions last
night. I think the exam is at least as difficult as the previous ones, so the
pass mark will probably be no higher.

Regarding question 24, the book says that the residuals "resemble" white noise,
not "constitute", making C false (!) The question appears to be defective,
since in (E), the word "autocorrelations" after "residuals" is missing.
Certainly the fact the residuals (epsilon-hats) is more than 2/sqrt(T) doesn't
indicate anything about the correctness of the specification.

[Gore Tree Lover]

The Soa is a beautiful organization - they give me such pleasure. I like how one word makes the difference (constitute vs. resemble). They resemble not consitute the "society of a**es".

<font size=-1>[ This Message was edited by: oolong on 2001-11-08 12:30 ]</font>

<font size=-1>[ This Message was edited by: oolong on 2001-11-08 12:30 ]</font>

[FoxtrotFool]

Has anyone figured out Course 4 #23?

I thought:
EPV = (1.06 + 1.12)/2 = 1.09
VHM = ((1.8-2.05)^2 + (2.3-2.05)^2)/2=.0625
k=EPV/VHM = 17.44
Z = 1/(1+k) ~ 0.054

What did I do wrong?

[Actuary321]

On 2001-11-08 12:31, FoxtrotFool wrote:
Has anyone figured out Course 4 #23?

I thought:
EPV = (1.06 + 1.12)/2 = 1.09
VHM = ((1.8-2.05)^2 + (2.3-2.05)^2)/2=.0625
k=EPV/VHM = 17.44
Z = 1/(1+k) ~ 0.054

What did I do wrong?


N=Number of claims in experience ? 1.8+2.3 = 4.1 therefore Z=N/(N+K) = 4.1/(4.1+17.44) = 0.190343574

or was it N=Number of classes in experence = 4 and Z=4.(4+17.44) = 18.6567164

either way answer is D = .19

I got this correct but can't remember if I did it correctly or guessed.

PS I saw this question in another thread.

[FoxtrotFool]

thanks for pulling it oof the other thread. I didn't see it in the other thread...

Can someone explain it further?

To me, the number of "claims" doesn't really make sense. The question states that "There are an equal number of business and pleasure use drivers" Which I would believe that means that a driver either use the car for business or pleasure. And expected claim would be 2.05, not 4.1.

The total number of group doesn't make much sense either. Anyone knows a reference to it?

[Actuary321]

Actually I think I might have done it this way:

EPV = (1.06 + 1.12)/2 = 1.09
VHM = ((1.8-2.05)^2 + (2.3-2.05)^2)/(2-1)=0.125
k=EPV/VHM = 8.72
N=(1.8+2.3)/2=2.05
Z = 2.05/(2.05+8.72) = 0.190343547

but I really don't remember. I know I calculated EPV as above and VHM as above and as in the earlier post with 2 in the denominator. But what I got for K and how I picked D, I don't remember.

[Anonymous]

i did it the slow way .. and split the two up. then there is no confusion what N is. (has to be 1).

the split between rural and urban is 20%:80%.

To find v-hat :

V(Xj|theta) =
Rural Business 0.5 p=10% (20% * 0.5)
Urban Business 1.0 p=40%
Rural Pleasure 0.8 p=10%
Urbal Pleasure 1.0 p=40%

So E(V(Xj|theta)) = 0.5*.1+...+1.0*0.4 = 0.93

To find a-hat :

(Xj|theta) =
Rural Business 1.0 p=10%
Urban Business 2.0 p=40%
Rural Pleasure 1.5 p=10%
Urban Pleasure 2.5 p=40%

Then E(Xj^2|theta) = 4.425, E(Xj|theta) = 2.05 ==&gt; V(E(Xj|theta)) = 4.425 - 2.05^2 = 0.2225

So k=v/a=0.93/0.2225=4.1798

Z=1/(1+4.1798)=0.1930

This is definitely the slow way ... but seems to work.

[FoxtrotFool]

Hmm... Actuary321, your solution does seems more reasonable... however...

The mean and variance seemed to be theoritical, as supposed to estimated, because doesn't empirical estimation of VHM need to be:

((1.8-2.05)^2+(2.3-2.05)^2) - EPV/2

which leads to a negative VHM and k

[Actuary321]

Opps, sorry I was looking at the wrong list of answers, turns out I didn't get it right after all. I picked A as well.

This is where that other post's theory can come into play:

I didn't know how to do it (and still don't) so I made up a formula and got the correct answer.

[LB]

On 2001-11-08 13:37, FoxtrotFool wrote:
thanks for pulling it oof the other thread. I didn't see it in the other thread...

Can someone explain it further?

To me, the number of "claims" doesn't really make sense. The question states that "There are an equal number of business and pleasure use drivers" Which I would believe that means that a driver either use the car for business or pleasure. And expected claim would be 2.05, not 4.1.

The total number of group doesn't make much sense either. Anyone knows a reference to it?


N is the total number of claims. If you pick someone out of a this group his expected is 2.05. But the total exposure(claims) for both groups in one year is the sum of the two.

[Jen]

Seriously, how many correct answers can score a pass??

I guess mine is beteen 26-29, will that make me pass at all? I am so nervous......

[Actuary321]

Go ahead and start studying for your next exam Jen, you passed.

[Anonymous]

Hey, Jen. I also think that I got between 27 - 29. Problem is, I didn't take my answers out of the exam so I don't know for sure. What if I messed up on around 5 that I thought I got right? Especially the theory questions, seems like there were more of those this time and I usually can't remember what I put for those. Anyway, 22 or 23 should pass, so here is to beginning studying for 6.

[FoxtrotFool]

On 2001-11-08 16:07, LB wrote:

On 2001-11-08 13:37, FoxtrotFool wrote:
thanks for pulling it oof the other thread. I didn't see it in the other thread...

Can someone explain it further?

To me, the number of "claims" doesn't really make sense. The question states that "There are an equal number of business and pleasure use drivers" Which I would believe that means that a driver either use the car for business or pleasure. And expected claim would be 2.05, not 4.1.

The total number of group doesn't make much sense either. Anyone knows a reference to it?


N is the total number of claims. If you pick someone out of a this group his expected is 2.05. But the total exposure(claims) for both groups in one year is the sum of the two.


LB, The question states:
Determine the Buhlmann credibility factor for a single driver.

A single driver is expected to have 2 claims. But I still think since we are talking about the claim frequency, a single driver is one exposure. (unless s/he was observed for 4 years) Therefore N=1......

[Jen]

Thanks, you guys. I guess it doesn't help much to keep worrying. I am going to talk to my boss about buying materials for next sitting. Good luck to everyone including myself :smile:

[LB]

On 2001-11-08 17:12, FoxtrotFool wrote:

On 2001-11-08 16:07, LB wrote:

On 2001-11-08 13:37, FoxtrotFool wrote:
thanks for pulling it oof the other thread. I didn't see it in the other thread...

Can someone explain it further?

To me, the number of "claims" doesn't really make sense. The question states that "There are an equal number of business and pleasure use drivers" Which I would believe that means that a driver either use the car for business or pleasure. And expected claim would be 2.05, not 4.1.

The total number of group doesn't make much sense either. Anyone knows a reference to it?


N is the total number of claims. If you pick someone out of a this group his expected is 2.05. But the total exposure(claims) for both groups in one year is the sum of the two.


LB, The question states:
Determine the Buhlmann credibility factor for a single driver.

A single driver is expected to have 2 claims. But I still think since we are talking about the claim frequency, a single driver is one exposure. (unless s/he was observed for 4 years) Therefore N=1......


See credibility study note page 8-54 last paragraph. They mention that N is the number of observations (claims) of the risk process.

[New at pd]

On 2001-11-08 12:05, Mary wrote:
--------------------------------------------------------------------------------
FYI:This is what Abe Weishauss had to say. In case you don't know him, he is the ASM manual guy.

Sorry for the delay, I didn't get a chance to go through all the questions last
night. I think the exam is at least as difficult as the previous ones, so the
pass mark will probably be no higher.

Regarding question 24, the book says that the residuals "resemble" white noise,
not "constitute", making C false (!) The question appears to be defective,
since in (E), the word "autocorrelations" after "residuals" is missing.
Certainly the fact the residuals (epsilon-hats) is more than 2/sqrt(T) doesn't
indicate anything about the correctness of the specification.


---------------------------------


Has anyone brought this up with the SoA?

[mikep65]

In response to the credibility question. Rogerlee posted the solution with respect to the way I have always seen it done. Your other thoughts on changing the N to something besides 1, does give the correct answer, but it isn't what they asked for.

N = 1

You have 4 cases. Equal weights for Business and Pleasure - given. You can then get the weights for the Rural and Urban from the data. Which gives you 80% and 20%.
That is the trick. Once you have four categories with their weights, the problem is easy and works just like all credibility problems. No tricks on what N is. Just setting up and getting the weights.

[Anonymous]

You are correct wrt the Business or Pleasure/Rural or Urban Buhlmann problem. This is analogous to dependent die/spinner problems we have seen.
What bothers me and makes me think the question is defective is that they gave you the overall mean and variance of Business and Pleasure for what looked to be across Rural and Urban. Unfortunately, I assumed that the variance was ok after the weighted means checked out. I know they give you accurate information that you may not need to solve the problem, but I do not think it is right that they give you misleading information. Any seconds?

[mikep65]

It isn't a defective question. Tricky maybe, but not defective.

They give you the overall mean and variance in order to calculate the weights.

You do not use the overall mean and variance in calculating the answer though. First you get the weights, then you use each mean and variance for the 4 possibilities: Business-rural, business-urban, pleasure-rural, pleasure-urban.

[FoxtrotFool]

On 2001-11-09 08:30, claude the mule wrote:
In response to the credibility question. Rogerlee posted the solution with respect to the way I have always seen it done. Your other thoughts on changing the N to something besides 1, does give the correct answer, but it isn't what they asked for.

N = 1

You have 4 cases. Equal weights for Business and Pleasure - given. You can then get the weights for the Rural and Urban from the data. Which gives you 80% and 20%.
That is the trick. Once you have four categories with their weights, the problem is easy and works just like all credibility problems. No tricks on what N is. Just setting up and getting the weights.


Cloude, your method is one that I didn't think of before, and I suspect that it is what SOA has in mind.

With that your solution in mind, I would argue that the difference is a matter of semantics. The statement of questions separate the population into 2 groups in statement 2. There is no clear indications that a driver is to only drive in Rural environment or Urban environment. So, I can argue that a driver is expected to drive 20% Rural, and 80% urban.

If the question has a third statement that says "a driver drives only in Rural condition or Urban condition", then I don't think there can be a dispute that .19 is the right answer.

[mikep65]

I do not agree. I'm not sure that it matters whether or not individual drivers could fall into different categories. All we care about is the probabilities of each category, and they give you enough information for that.

Business and Pleasure is 50/50 split - given. The use of the weighted means gives you the split for Rural and Urban -
p(1.0) + (1-p)(2.0) = 1.8

It doesn't matter what mixture makes this true. Only that it is true, and it is since it is given.

[FoxtrotFool]

I know SOA often gives more infomation than necessary to solve a problem.

If the population is only divide into two groups - Business and Pleasure. We still need to know that Rural and Urban separation is independent. Otherwise you can't really use the more finely defined groups.

Had they not give the Rural/Urban infomation, there is a clear way to solve the question. And I would think without knowing the independence of Rural vs. Urban, the same solution applies.


<font size=-1>[ This Message was edited by: FoxtrotFool on 2001-11-09 11:13 ]</font>

[Anonymous]

From Abe Weishauss FYI:
I looked at 23. They should have made it more clear that urban and rural
drivers are 2 separate categories. You're expected to treat this as a 4
category example, and to ignore the expected variances (which confuse you);
you
must calculate the expectation of the process variance based on the 4
processes.

[FoxtrotFool]

Thanks Mary.

Thanks Cloude again for the correct way of solving Q23.

I guess we have some authority to state that it is not a clear question, so I'll stop. :smile:

[gloid]

There was a similar problem to this urban/rural problem on a previous exam. It dealt with driving a car into a wall or into a lake with a few crash test dummies in them. This was pretty much the same idea (I think it was Bayesian analysis, not BC), you really have 4 groups. They tell you that half are business and half are pleasure, you only need to find the proportion of rural/urban to get the rest of the relevant information. The weighted averages give you this info. Why else would they give all the information about rural/urban?

[FoxtrotFool]

The reason that I would have prefered to have a little more infomation is because without stating Rural and Urban is independent, you can't do Buhlmann in 4 groups. (You can specify by either stating the conditional probability of Rural and Urban is independent, or by saying that Rural and Urban drivers are disjoint)

The crash dummy's question stated pretty clearly that it's a four group problem.

Oh, I'll answer a question with a question...(you asked why would they provide rural/urban split if the infomation is not necessary...) Why would they provide the total claim variance if the infomation is not necessary? SOA exams often gives useless infomation.

[Anonymous]

Go FtF Go. I'm with you...like I said (tried to say) in an earlier post, if they're going to try and give you irrelevant info wrt solving the question, they better be sure to include all of the caveats. I'm submitting more than just a survey for this one.

By the way, with the independence stated I completely agree this is a credibility problem with 4 distinct classes (rows).