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Jonas science forum addict
Joined: 08 Sep 2005
Posts: 77

Posted: Fri May 26, 2006 12:52 am Post subject:
Simple Proof



I am doing a bit of independent study for the first actuarial exam.
The textbook that I have asks that I prove that the complement of A and
B are independent if A and B are independent. I started out with the
following:
I want to show that the probability of the complement of A times the
complement of B is equal to the probability of the intercept of the
complement of A and B. Pr(\A)*Pr(\B)=Pr(\(AB))
[1Pr(A)]*[1Pr(B)]
1Pr(A)Pr(B)+Pr(A)Pr(B)
1Pr(A)Pr(B)+Pr(AB)
I am stuck at this point. If I were to have 1Pr(AB), in the last
step, I would be done. I can't figuere out where I am making my
mistake.
I also tried working with the notation for conditional probability to
prove that the complements are independent but I had no success.
Any good hints or suggestions that you can give would be much
appreciated. 

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Biba science forum beginner
Joined: 14 Feb 2006
Posts: 4

Posted: Fri May 26, 2006 7:46 am Post subject:
Re: Simple Proof



It goes like this:
A^c  complement of A
B^c  complement of B
A+B  union of A and B (not necesserly disjoint)
P(A+B)=P(A)+P(B)P(AB)
Then you have:
P(A^cB^c)=1P(A+B)
=1P(A)P(B)+P(A)P(B)
=1(1P(A^c))(1P(B^c))+(1P(A^c))(1P(B^c))
=11+P(A^c)1+P(B^c)+1P(B^c)P(A^c)+P(A^c)P(B^c)
=P(A^c)P(B^c)
which proves the claim.
I hope it helps 

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Jonas science forum addict
Joined: 08 Sep 2005
Posts: 77

Posted: Fri May 26, 2006 1:11 pm Post subject:
Re: Simple Proof



This helps very much so. Thanks for your help. I see where I made my
mistake. I think that the proof you give below can be shortened by
saying that the step after the second could be 1[P(A+B)] then it can
be said that P[(AB)^c].
Biba wrote:
Quote:  It goes like this:
A^c  complement of A
B^c  complement of B
A+B  union of A and B (not necesserly disjoint)
P(A+B)=P(A)+P(B)P(AB)
Then you have:
P(A^cB^c)=1P(A+B)
=1P(A)P(B)+P(A)P(B)
=1(1P(A^c))(1P(B^c))+(1P(A^c))(1P(B^c))
=11+P(A^c)1+P(B^c)+1P(B^c)P(A^c)+P(A^c)P(B^c)
=P(A^c)P(B^c)
which proves the claim.
I hope it helps 


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Jonas science forum addict
Joined: 08 Sep 2005
Posts: 77

Posted: Fri May 26, 2006 1:46 pm Post subject:
Re: Simple Proof



What I previously wrote was wrong. Please disregard it. 

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VijaKhara@gmail.com science forum beginner
Joined: 30 Sep 2005
Posts: 26

Posted: Sat May 27, 2006 10:20 pm Post subject:
Re: Simple Proof



Well, I know I am wrong but I cannot find out what is my wrong point.
To me , in Biba's proof, he doesn't use anything relating to the
assumtion that A and B are independent.
In BIBA's proof:
P(A+B)=P(A)+P(B)P(AB) this identity holds for every events.
P(A^cB^c)=1P(A+B) This holds because of De Morgan's law:
A^cB^c=(A+B)^c for all events
Plz Figure out where Biba used the assumtion that A and B are
independent to come to the result!
Thanks 

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Mike Terry science forum Guru Wannabe
Joined: 02 May 2005
Posts: 137

Posted: Sun May 28, 2006 12:13 am Post subject:
Re: Simple Proof



<VijaKhara@gmail.com> wrote in message
news:1148768450.895960.84020@y43g2000cwc.googlegroups.com...
Quote:  Well, I know I am wrong but I cannot find out what is my wrong point.
To me , in Biba's proof, he doesn't use anything relating to the
assumtion that A and B are independent.
In BIBA's proof:
P(A+B)=P(A)+P(B)P(AB) this identity holds for every events.
P(A^cB^c)=1P(A+B) This holds because of De Morgan's law:
A^cB^c=(A+B)^c for all events
Plz Figure out where Biba used the assumtion that A and B are
independent to come to the result!
Thanks

Please quote Biba's proof and we might be able to help you!
Mike. 

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VijaKhara@gmail.com science forum beginner
Joined: 30 Sep 2005
Posts: 26

Posted: Sun May 28, 2006 2:59 am Post subject:
Re: Simple Proof



Hi there, the following is his proof:
It goes like this:
A^c  complement of A
B^c  complement of B
A+B  union of A and B (not necesserly disjoint)
P(A+B)=P(A)+P(B)P(AB)
Then you have:
P(A^cB^c)=1P(A+B)
=1P(A)P(B)+P(A)P(B)
=1(1P(A^c))(1P(B^c))+(1P(A^c))(1P(B^c))
=11+P(A^c)1+P(B^c)+1P(B^c)P(A^c)+P(A^c)P(B^c)
=P(A^c)P(B^c)
which proves the claim.
Thanks
Mike Terry wrote:
Quote:  VijaKhara@gmail.com> wrote in message
news:1148768450.895960.84020@y43g2000cwc.googlegroups.com...
Well, I know I am wrong but I cannot find out what is my wrong point.
To me , in Biba's proof, he doesn't use anything relating to the
assumtion that A and B are independent.
In BIBA's proof:
P(A+B)=P(A)+P(B)P(AB) this identity holds for every events.
P(A^cB^c)=1P(A+B) This holds because of De Morgan's law:
A^cB^c=(A+B)^c for all events
Plz Figure out where Biba used the assumtion that A and B are
independent to come to the result!
Thanks
Please quote Biba's proof and we might be able to help you!
Mike. 


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Mike Terry science forum Guru Wannabe
Joined: 02 May 2005
Posts: 137

Posted: Sun May 28, 2006 2:07 pm Post subject:
Re: Simple Proof



<VijaKhara@gmail.com> wrote in message
news:1148785192.053586.65810@g10g2000cwb.googlegroups.com...
Quote:  Hi there, the following is his proof:
It goes like this:
A^c  complement of A
B^c  complement of B
A+B  union of A and B (not necesserly disjoint)
P(A+B)=P(A)+P(B)P(AB)
Then you have:
P(A^cB^c)=1P(A+B)

=1P(A)P(B)+P(AB) # as you note below
=1P(A)P(B)+P(A)P(B) # INDEPENDENT EVENTS
Quote:  =1P(A)P(B)+P(A)P(B)
=1(1P(A^c))(1P(B^c))+(1P(A^c))(1P(B^c))
=11+P(A^c)1+P(B^c)+1P(B^c)P(A^c)+P(A^c)P(B^c)
=P(A^c)P(B^c)

The above is correct but maybe over complex? How about just:
=1P(A)P(B)+P(A)P(B) # INDEPENDENT EVENTS
=(1P(A))(1P(B))
=P(A^c)P(B^c)
Mike.
Quote: 
which proves the claim.
Thanks
Mike Terry wrote:
VijaKhara@gmail.com> wrote in message
news:1148768450.895960.84020@y43g2000cwc.googlegroups.com...
Well, I know I am wrong but I cannot find out what is my wrong point.
To me , in Biba's proof, he doesn't use anything relating to the
assumtion that A and B are independent.
In BIBA's proof:
P(A+B)=P(A)+P(B)P(AB) this identity holds for every events.
P(A^cB^c)=1P(A+B) This holds because of De Morgan's law:
A^cB^c=(A+B)^c for all events
Plz Figure out where Biba used the assumtion that A and B are
independent to come to the result!
Thanks
Please quote Biba's proof and we might be able to help you!
Mike.



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Biba science forum beginner
Joined: 14 Feb 2006
Posts: 4

Posted: Mon May 29, 2006 9:09 am Post subject:
Re: Simple Proof



Mike Terry wrote:
How about just:
Quote: 
=1P(A)P(B)+P(A)P(B) # INDEPENDENT EVENTS
=(1P(A))(1P(B))
=P(A^c)P(B^c)

Sure, I just wanted to write down every step of the proof, in details. 

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