Re: probability question
Il s'agit d'un message multivolet au format MIME. --982FBF2E2FA5C1B960626D56 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Hi, This assertion is true Franck Matt Dobrin a écrit : Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in advance. -Matt -- _/_/_/_/_/ _/_/_/_/_/ _/ _/ Franck Corset _/ _/ Projet IS2 _/ _/ Inria Rhone-Alpes _/_/_/_/_/ _/ ZIRST, 655, avenue de l'Europe _/ _/ Montbonnot _/ _/ 38334 Saint Ismier cedex _/ _/ FRANCE _/ _/_/_/_/_/ http://www.inrialpes.fr/is2 --982FBF2E2FA5C1B960626D56 Content-Type: text/x-vcard; charset=us-ascii; name=Franck.Corset.vcf Content-Transfer-Encoding: 7bit Content-Description: Carte pour Franck Corset Content-Disposition: attachment; filename=Franck.Corset.vcf begin:vcard n:Corset;Franck tel;cell:0610487239 tel;home:0476700659 tel;work:0476615355 x-mozilla-html:FALSE org:Inria;Isère adr:;;18, rue Nicolas Chorier;Grenoble;;38000;France version:2.1 email;internet:[EMAIL PROTECTED] title:Doctorant fn:Franck Corset end:vcard --982FBF2E2FA5C1B960626D56-- = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: probability question
Il s'agit d'un message multivolet au format MIME. --F4F503AE2A9C358CB6A37D62 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Hi, This assertion is true Franck Matt Dobrin a écrit : Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in advance. -Matt -- _/_/_/_/_/ _/_/_/_/_/ _/ _/ Franck Corset _/ _/ Projet IS2 _/ _/ Inria Rhone-Alpes _/_/_/_/_/ _/ ZIRST, 655, avenue de l'Europe _/ _/ Montbonnot _/ _/ 38334 Saint Ismier cedex _/ _/ FRANCE _/ _/_/_/_/_/ http://www.inrialpes.fr/is2 --F4F503AE2A9C358CB6A37D62 Content-Type: text/x-vcard; charset=us-ascii; name=Franck.Corset.vcf Content-Transfer-Encoding: 7bit Content-Description: Carte pour Franck Corset Content-Disposition: attachment; filename=Franck.Corset.vcf begin:vcard n:Corset;Franck tel;cell:0610487239 tel;home:0476700659 tel;work:0476615355 x-mozilla-html:FALSE org:Inria;Isère adr:;;18, rue Nicolas Chorier;Grenoble;;38000;France version:2.1 email;internet:[EMAIL PROTECTED] title:Doctorant fn:Franck Corset end:vcard --F4F503AE2A9C358CB6A37D62-- = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: probability question
It's true. If you are concerned with proof, following the this belove according to conditional probability p(a|b)=p(a,b)/p(b) (1) P(A,B|C)=P(A,B,C)/P(C) (2) P(A,B,C)=P(A,C)*P(B|A,C) (3) P(A,C)=P(C)*P(A|C) WITH (2) AND (3) WE GET (4) P(A,B,C)=P(C)*P(A|C)*P(B|A,C) TAKING (1) AND (4) WE GET P(A*B|C)=P(A|C)*P(B|A*C)? Hope this help Nathaniel U¿ytkownik Matt Dobrin [EMAIL PROTECTED] napisa³ w wiadomo¶ci 9uh8ge$5hv$[EMAIL PROTECTED]">news:9uh8ge$5hv$[EMAIL PROTECTED]... Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in advance. -Matt = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: probability question
Michael Cohen ([EMAIL PROTECTED]) wrote:
: : On Wed, 1 Dec 1999, Yonah Russ wrote:
: :
: : how do you solve a problem like this one?
: : thanks in advance
: : ---
: : X is a chance variable such that X(omega)={1,2,3...,n}
: : and for every i in {1,2,3...n}, 4P(X=i+2)=5P(X=i+1)-P(X=i)
: :
: : find the breakdown of X.
: :
:
: Let's take n finite first. Let ^ denote "raised to power". Using difference
: equations, the solutions are constant and constant times (1/4)^i or linear
: combinations thereof: P(X=i) = a (1/4)^i + b where a and b are constants. The
: requirement that the P(X=i) sum to 1 forces a=3(1-bn)4^n/(4^n-1) and b is at most
: 1/n.
:
: If n is infinite, then b must be 0 (otherwise the sum of the P(X=i) is infinite).
: Then a=3. So P(X=i)=3 (1/4)^i.
:
One revision: At the end of the second to last paragraph, I should not have
required b be at most 1/n (a can be negative). We just need to be sure P(X=1)
and P(X=n) are non-negative --- the other values are in-between.
--
Michael P. Cohen home phone 202-232-4651
1615 Q Street NW #T-1 office phone 202-219-1917
Washington, DC 20009-6310 office fax 202-219-1736
[EMAIL PROTECTED]
Re: probability question
Donald F. Burrill ([EMAIL PROTECTED]) wrote:
: Haven't seen a public response to this question that I find credible,
: and am curious. Is the problem as described solvable?
: If n is finite, what meaning attaches to "P(X=n+2)" and "P(X=n+1)"?
: If n is infinite, shouldn't the description read, without reference
: to an apparently finite n, "X(omega)={1,2,3,...}"?
: Or should the description read "for every i in {1,2,3,...,n-2}"?
:
: On Wed, 1 Dec 1999, Yonah Russ wrote:
:
: how do you solve a problem like this one?
: thanks in advance
: ---
: X is a chance variable such that X(omega)={1,2,3...,n}
: and for every i in {1,2,3...n}, 4P(X=i+2)=5P(X=i+1)-P(X=i)
:
: find the breakdown of X.
:
Let's take n finite first. Let ^ denote "raised to power". Using difference
equations, the solutions are constant and constant times (1/4)^i or linear
combinations thereof: P(X=i) = a (1/4)^i + b where a and b are constants. The
requirement that the P(X=i) sum to 1 forces a=3(1-bn)4^n/(4^n-1) and b is at most
1/n.
If n is infinite, then b must be 0 (otherwise the sum of the P(X=i) is infinite).
Then a=3. So P(X=i)=3 (1/4)^i.
--
Michael P. Cohen home phone 202-232-4651
1615 Q Street NW #T-1 office phone 202-219-1917
Washington, DC 20009-6310 office fax 202-219-1736
[EMAIL PROTECTED]
