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Probability Old Exam P Forum

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  #131  
Old 05-11-2008, 10:04 AM
victorandjenny victorandjenny is offline
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Dear Sam,
I'd like to thank you first for your effort of training us. I benefited a lot although I started solving your problems today.
I found that my answer of this question is different from your solution. The process is the same, just the final step of calculation is different.
Best wishes.
Victor

(0.5-1/e)/root of (2/e-eEXP-2)=0.17
N=square of (2.33/0.17)=187.85.
So my answer is N should be at least 188.


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Originally Posted by sam_broverman View Post
The question can be found at
http://www.sambroverman.com/pqw/apr21-08p.pdf




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Last edited by victorandjenny; 05-11-2008 at 04:28 PM..
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  #132  
Old 05-11-2008, 04:27 PM
victorandjenny victorandjenny is offline
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Dear Sam,
Since I am not a native speaker, I was confused by this question. Did they each put one dollar in the pot, or each put $0.5 only. Surely I got the right answer, because there's no answer for the latter understanding.
Regards.
Victor

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Originally Posted by sam_broverman View Post
You can find the question at:
http://www.sambroverman.com/pqw/mar19-07p.pdf


Sam Broverman
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  #133  
Old 05-11-2008, 07:54 PM
victorandjenny victorandjenny is offline
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Default A question for Professor Broverman

Dear Professor Broverman,
I have a question for this problem.
When we are trying to find out the value of c, we integrate f(x) over x to get 1. However, should we consider f(0)=c, which is not included in the two parts of the integration (-indefinite, 0) and (0, +indefinite)? If yes, the c=1/3 but not 1/2.
Thank you in advance.
Victor



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  #134  
Old 05-11-2008, 09:19 PM
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Gandalf Gandalf is offline
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Quote:
Originally Posted by victorandjenny View Post
Dear Professor Broverman,
I have a question for this problem.
When we are trying to find out the value of c, we integrate f(x) over x to get 1. However, should we consider f(0)=c, which is not included in the two parts of the integration (-indefinite, 0) and (0, +indefinite)? If yes, the c=1/3 but not 1/2.
Thank you in advance.
Victor
For a continuous distribution, like this one, there is density but no probability at a point.

The integration could be considered as (-infinity, 0] and (0,+infinity) or (-infinity,0) and [0,+infinity]. The integrals give the same result either way.
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  #135  
Old 05-11-2008, 09:49 PM
victorandjenny victorandjenny is offline
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Dear Gandalf,
Thank you very much for your clarification. I understand it now.
Best wishes.
Victor


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Originally Posted by Gandalf View Post
For a continuous distribution, like this one, there is density but no probability at a point.

The integration could be considered as (-infinity, 0] and (0,+infinity) or (-infinity,0) and [0,+infinity]. The integrals give the same result either way.
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  #136  
Old 05-12-2008, 09:20 AM
sam_broverman sam_broverman is offline
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Gandalf,

Thank you for answering the question.

victorandjenny,

For the Mar 19/07 question, they both start out
each putting $1 in the pot. Then more is added
by the person who tosses the dice first.

For the Apr 21/08 question, it appears that I
made an error. I will post a correction.

SB
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  #137  
Old 05-12-2008, 09:20 AM
sam_broverman sam_broverman is offline
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Default May 12, 2008 Exam P Question

The question can be found at
http://www.sambroverman.com/pqw/may12-08p.pdf




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  #138  
Old 05-12-2008, 04:20 PM
victorandjenny victorandjenny is offline
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Dear Professor Broverman,
Thank you very much for your clarification.
I am suing your manual for SOA P (Actex) and I really like your book.
Best wishes.
Victor



Quote:
Originally Posted by sam_broverman View Post
Gandalf,

Thank you for answering the question.

victorandjenny,

For the Mar 19/07 question, they both start out
each putting $1 in the pot. Then more is added
by the person who tosses the dice first.

For the Apr 21/08 question, it appears that I
made an error. I will post a correction.

SB
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  #139  
Old 07-11-2008, 01:07 PM
gametime50 gametime50 is offline
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Default may 12/08 question

Dear Professor Broverman,
I was trying to solve this problem and could not find how or why it turned out the way it did. Here is the part that i did not understand.

var(E(X|Y))={2=E(X|Y=0) prob 1/3 =(2-4)^2 (1/3)(2/3) = 8/9
{4=E(X|Y=1) prob 2/3

I do not understand the solution to that part or how it was arrived at. Thank you for any help.
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  #140  
Old 07-11-2008, 03:28 PM
Actuarialsuck Actuarialsuck is offline
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I believe that's the Bernoulli shortcut?
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