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March 29, 2024, 06:06:32 am

Author Topic: My thread of questions  (Read 25756 times)  Share 

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khalil

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Re: My thread of questions
« Reply #30 on: August 28, 2009, 09:04:59 pm »
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,
« Last Edit: August 28, 2009, 09:10:59 pm by khalil »

TrueTears

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Re: My thread of questions
« Reply #31 on: August 28, 2009, 09:12:30 pm »
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that is true.
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

Flaming_Arrow

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Re: My thread of questions
« Reply #32 on: August 28, 2009, 09:12:40 pm »
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yes

EDIT: beaten
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khalil

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Re: My thread of questions
« Reply #33 on: August 28, 2009, 09:30:11 pm »
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Are we supposed to know the law of total probability?
Pr(B)= Pr(B|A)\times Pr(A)+ Pr(B|{A}')\times Pr({A}')

Flaming_Arrow

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Re: My thread of questions
« Reply #34 on: August 28, 2009, 09:30:48 pm »
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Are we supposed to know the law of total probability?
Pr(B)= Pr(B|A)\times Pr(A)+ Pr(B|{A}')\times Pr({A}')

yep
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khalil

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Re: My thread of questions
« Reply #35 on: August 28, 2009, 09:31:58 pm »
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Even memorise it?

TrueTears

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Re: My thread of questions
« Reply #36 on: August 28, 2009, 09:32:13 pm »
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Isn't that kinda intuitive that formula?
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

khalil

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Re: My thread of questions
« Reply #37 on: August 28, 2009, 09:39:29 pm »
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lol. well its not in my intuition. Can you please explain how it is spontaneous to your liking?

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Re: My thread of questions
« Reply #38 on: August 28, 2009, 09:41:13 pm »
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You can use a tree-diagram to visualise that equation. The pathways ending in B are given by the terms on the RHS.

Flaming_Arrow

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Re: My thread of questions
« Reply #39 on: August 28, 2009, 09:43:22 pm »
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refer to diagram
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khalil

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Re: My thread of questions
« Reply #40 on: August 28, 2009, 09:49:25 pm »
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Ok thanks. Wow, Flaming_Arrow, did you prepare those diagrams right now...looks like you live up to ur username

Flaming_Arrow

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Re: My thread of questions
« Reply #41 on: August 28, 2009, 09:49:44 pm »
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Ok thanks. Wow, Flaming_Arrow, did you prepare those diagrams right now...looks like you live up to ur username

ye i did
2010: Commerce @ UoM

khalil

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Re: My thread of questions
« Reply #42 on: August 28, 2009, 09:52:09 pm »
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So that means that Pr(B|A')Pr(A')=Pr(A n B')?

Flaming_Arrow

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Re: My thread of questions
« Reply #43 on: August 28, 2009, 09:54:22 pm »
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So that means that Pr(B|A')Pr(A')=Pr(A n B')?

na

Pr(B|A')Pr(A')=Pr(B n A') \ Pr(A') * Pr(A') = Pr(B n A')
2010: Commerce @ UoM

khalil

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Re: My thread of questions
« Reply #44 on: August 29, 2009, 12:31:17 pm »
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Pr(A) = Pr(A|B) Pr(B) + Pr(A|B') Pr(B')

Does Pr(B')=Pr(A)?