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Double T
December 31st 2005, 03:37 PM
I have a question about P/1 exam, i hope someone can give me a clue, thanks!!
That is one of the example in Actex P/1 exam study manual, example 10-3 p. 239.

The probability of a claim amount given a claim occurs is given, qi, and the mean and variance of the conditional distribution of claim amount given a claim occurs is given, E[ Bi ], Var [ Bi ]. When the loss random variable is given in this form, we have for policy i, E[ Xi ] = qi*E [Bi], and Var [Xi] = qi*(1-qi)*(E[Bi])^2 + qi*Var[Bi]

My question is how the equation Var[Xi] comes from?? Can someone can explain to me?? Thanks for all your help, your advise is very important to me!!

T

Sam Broverman
January 1st 2006, 11:39 AM
We can find Var[Bi] from the relationship
Var[Bi] = E[Xi^2] - ( E[Xi] )^2 .
We note that Xi is either equal to Bi (prob. qi) or 0 (prob. 1-qi).
This means that Xi is a "mixture" of Bi and 0, with a "mixing weight"
of qi for Bi and a mixing weight of 1-qi for 0.
Bi and 0 are the "component distributions" of the mixture.

The moments of a mixed distribution are found by applying the
mixing weights to the corresponding moments of the component distributions
This means that E[Xi] = qi*E[Bi] + (1-qi)*0 = qi*E[Bi] ,
and E[Xi^2] = qi*E[Bi^2] + (1-qi)*(0^2) = qi*E[Bi^2] .
Then Var[Xi] = qi*E[Bi^2] - (qi*E[Bi])^2 .

Finally, we note that Var[Bi] = E[Bi^2] - (E[Bi])^2 , so
that E[Bi^2] = Var[Bi] + (E[Bi])^2 . If we substitute this
expression for E[Bi^2] into the equation
Var[Xi] = qi*E[Bi^2] - (qi*E[Bi])^2 ,
we get the result
Var[Xi] = qi*(Var[Bi] + (E[Bi])^2) - (qi*E[Bi])^2
= qi*Var[Bi] + (qi - qi^2)*(E[Bi]^2)
= qi*Var[Bi] + qi*(1-qi)*(E[Bi]^2) .

I have a question about P/1 exam, i hope someone can give me a clue, thanks!!
That is one of the example in Actex P/1 exam study manual, example 10-3 p. 239.

The probability of a claim amount given a claim occurs is given, qi, and the mean and variance of the conditional distribution of claim amount given a claim occurs is given, E[ Bi ], Var [ Bi ]. When the loss random variable is given in this form, we have for policy i, E[ Xi ] = qi*E [Bi], and Var [Xi] = qi*(1-qi)*(E[Bi])^2 + qi*Var[Bi]

My question is how the equation Var[Xi] comes from?? Can someone can explain to me?? Thanks for all your help, your advise is very important to me!!

T

Double T
January 1st 2006, 08:31 PM
Thank you Dr. Broverman, your explaination is very clear and detail.

I have one more question, what do u think about the P1 Exam in Feb 06. This is the second time i take P exam, do u think it gonna be similar as last exam?? Can u give me any advise??

Thank you and happy new year

T

Sam Broverman
January 1st 2006, 10:33 PM
I didn't notice any change in the exam catalog description
for Exam P from the Fall 2005 to the Spring 2006 catalogs,
so there is no indication from the SOA that there will be any change in
the nature of the exam.

It is possible that there will be some administrative changes
for the actual conducting of the exam to deal with some
complaints that arose from last fall's computer based exam.

Happy New Year to you, T. I wish you the best of luck on
the February exam.

pluviosilla
September 15th 2008, 11:28 AM
It seems to me that the following problem qualifies as a mixted distribution, BUT IT HAS NO JUMPS. How does it fit in the weighting paradigm? Consider this:

Claim payouts for two employees follow the same uniform distribution:
X ~ Uni(1000, 5000) and Y ~ Uni(1000, 5000), but we must also consider the fact that the probability of an employee incurring a loss is only 0.4. This situation is represented by the following PDF (according to the ACTEX actuarial study guide):

f(x) = { 0.6, for X = 0 and 0.0001 for 1000 <= X <= 5000}. Likewise for f(y).

I interpret this to be a mixed distribution. On the one hand, you have the uniform distribution of the payout, which is continuous. On the other hand, you have a degenerate discrete distribution that is {0 for 0 <= X < 1000 and 1 for 1000 <= X <= 5000). The author (professor Broverman) clearly derived the 0.0001 constant coefficient by multiplying the 0.4 weighting factor by the PDF for the Uni(1000, 5000): 0.4 * 1 / 5000 - 1000 = 0.4 * 1 / 4000 = 0.0001.

But NOTICE THAT THE CONTINUOUS DISTRIBUTION IS 0 AT X = 1000, which means that THERE IS NO JUMP.

This does not seem to fit the paradigm of mixed distributions as I have studied them in both the ASM and ACTEX manuals. Am I incorrect to call this a mixed distribution?

Sam Broverman
September 22nd 2008, 09:05 PM
Pluviosilla,

Can you give the question number for that question.

It looks like a joint distribution. A joint dist can also
be mixed in the sense of having some discrete probability
points and some continuous regions. The graph of the
cdf of a joint dist is in 3 dimensions since there is x, y
and F(x,y). There could be some disontinuities or jumps
in the 3-d surface of such a distribution.

Sam Broverman

pluviosilla
September 22nd 2008, 11:30 PM
Yes, professor Boverman, happy to provide the reference! Thanks for asking. The problem I refer to is problem #35, from Problem Set 8, on p. 190 of the 2005 edition of the ACTEX study manual.

After making the previous post, I noticed that you later (in Section 9) develop the idea of a "special case" of mixed distribution of X1 and X2 where X1 is the constant 0.

This special case often arises when dealing with the probability of a loss, because in addition to the distribution of the loss, you also have a Bernoulli-style probability that the loss occurs or does not occur. I assume this is the degenerate discrete portion of the mixed distribution. But, ...

(1.) I don't see how this fits with the paradigm of weighting components of mixed distributions according to the size of the jumps, because there is no jump where the discrete distribution is 1, and if there are no jumps
(2.) How to you justify the system of assigning weights to different components of the mixed distribution? This seems to assume that the various components can be added together in a linear fashion. It is almost like a vector product. Do we get this linear property whenever the components of the mixed distribution are mutually exclusive?

I know I'm probably not asking very good questions. I'm waving my hands in the dark. But I want to see a little more theoretical justification for this technique of weighting.

Thanks.

Sam Broverman
September 29th 2008, 10:46 PM
pluviosilla,

When a random variable has discrete points, there will always
be a "jump" in the cdf at each discrete point.

A random variable with some discrete points
and some continuous regions can be described
as a mixture of the discrete parts and the continuous parts.

The formal definition of a mixture of random variables X1 with
pdf or pf f1(x) and mixing weight c and X2 with pdf or pf f2(x) and
mixing weight 1-c is to define the new pdf/pf as
f(x) = c f1(x) + (1-x) f2(x) . If X1 is discrete and X2 is continuous
then we must be more careful about this definition. For each
point x of the discrete random variable X1, we define the
probability in the mixed distribution f(x) = c f1(x) (and
we ignore the density of the continuous X2). For each point x
of the continuous X2 that is not a point of probability for X1,
we define the density of the mixed distribution as (1-c) f2(x) .

For instance. Suppose that X1 is the discrete random variable
with values .25 and 1 with prob .5 at each point. And suppose the
X2 is the continuous random variable with pdf f2(x) = 2x for 0<=x<=1 .
And suppose we create the mixture dist with mxing weights
c = .4 for X1 and 1-c = .6 for X2 .
Then the mixed dist will have points of prob at x= .25 and x=1 with prob.
p(.25) = c f1(.25) = (.4)(.5) = .2 and p(1) = .2
and there will be density of (1-c) f2(x) = .6(2x) = 1.2x
for 0<=x<1 . At x=1 there will be prob. of .2 (not density).
If we draw the cdf of the mixed distribution, there will
be a jump at x=.25 and a jump at x=1.

SB

pluviosilla
October 3rd 2008, 07:13 AM
Thank you very much for this nice long reply, professor Broverman, and for the clear explanation.

Actually, I just had something of a breakthrough reading the solution to one of the problems in the ACTEX manual. From your solution, it became clear that
MIXED DISTRIBUTIONS ARE REALLY JUST A PROBABILITY SPACE THAT HAS BEEN PARTITIONED. Each distribution can be viewed as a conditional probability with WEIGHTS ARE JUST THE PROBABILITY OF THE PARTITION, thus:

f(x) = w1 * f1(x) + w2 * f2(x)

is really just:

f(x) = p(a) * p(x|a) + p(b) * p(x|b)

I've been reading a lot of exam prep material (not just ACTEX) and it wasn't until I read the solution to problem 23 of Section 9 of your manual that this simple fact became clear to me. Thank you!

So the "jumps" are not the essence of what makes a distribution "mixed". I was confused about this, because I didn't grasp the basic idea of a mixed distribution.

I suppose that if you want to understand why it is o.k. to treat the 1st & 2nd moments as weighted averages but not the variance, you can go back to the mathematics of conditional probability (and the Law of Total Probability) and derive the "rules" for handling mixed distributions. Correct?