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 Short-Term Actuarial Math Old Exam C Forum

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#1
06-15-2018, 04:31 PM
 RockOn Member SOA Join Date: Dec 2013 Studying for Exam C Favorite beer: Polygamy Porter Posts: 83
SOA #304

Hello!
Can someone please explain to me how we know that the sum of the alphas add to 1?

Thank you.

I'm having trouble posting an image of the problem, so sorry.
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#2
06-15-2018, 04:57 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,398

In the limit as y goes to infinity, all the cdfs go to one.
#3
06-15-2018, 05:55 PM
 RockOn Member SOA Join Date: Dec 2013 Studying for Exam C Favorite beer: Polygamy Porter Posts: 83

Quote:
 Originally Posted by Academic Actuary In the limit as y goes to infinity, all the cdfs go to one.
Thank you for the response!
Yes, I understand that the cdfs sum to 1.

What confuses me is that the density function associated with each alpha is different, ie, each exponential density has a different theta. I can see how the alphas would sum to 1 if the density function associated with each alpha used the same theta but they don't.??
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#4
06-15-2018, 07:31 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,398

Quote:
 Originally Posted by RockOn Thank you for the response! Yes, I understand that the cdfs sum to 1.
The cdfs don't sum to 1. The cdfs weighted by the alphas sum to 1. If the alphas didn't sum to 1 then the cdf of the mixed distribution would not be 1 in the limit.
#5
06-15-2018, 07:45 PM
 RockOn Member SOA Join Date: Dec 2013 Studying for Exam C Favorite beer: Polygamy Porter Posts: 83

Quote:
 Originally Posted by Academic Actuary The cdfs don't sum to 1. The cdfs weighted by the alphas sum to 1. If the alphas didn't sum to 1 then the cdf of the mixed distribution would not be 1 in the limit.

Right. Still confused.

Say that:

alpha 1 = .1
alpha 2 = .2
alpha 3 = .3
alpha 4 = .4

Each one of these alphas are being multiplied by a different density function (i.e. and exponential with different thetas). How does the fact that the alphas sum to one ensure that the cdf will be one?

I'm so confused.
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FAP: MOD 1| MOD 2| MOD 3| MOD 4| MOD 5| IA| MOD 6| MOD 7| MOD 8| FA |
#6
06-15-2018, 09:00 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,398

Quote:
 Originally Posted by RockOn Right. Still confused. Say that: alpha 1 = .1 alpha 2 = .2 alpha 3 = .3 alpha 4 = .4 Each one of these alphas are being multiplied by a different density function (i.e. and exponential with different thetas). How does the fact that the alphas sum to one ensure that the cdf will be one? I'm so confused.
Each unweighted density integrates to 1. Each weighted density integrates to the alpha weight. The sum of the weighted densities will integrate to 1.
#7
06-15-2018, 09:01 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 3,012

Y is a mixture of 4 different distributions, namely exponentials with means theta_1, theta_2, theta_3, and theta_4 respectively. The alpha_i are probabilities of selecting each of those 4 different components of the mixture, and as such sum to 1 because the sum of the probabilities is 1.
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#8
06-17-2018, 06:18 PM
 RockOn Member SOA Join Date: Dec 2013 Studying for Exam C Favorite beer: Polygamy Porter Posts: 83

Hello and thank you for all the responses. They've all been very helpful and I think I have a much better understanding of this problem. Your explanations all make sense.
I hope that I can quickly assimilate this type of problem should it come up on an exam.

Again, thank you all.
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