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e_hobsbawm@hotmail.com
science forum beginner
Joined: 27 Jun 2006
Posts: 3
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 Posted: Tue Jun 27, 2006 10:25 am Post subject:
Easy Reasoning
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I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric |
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Stephen Montgomery-Smith
science forum Guru
Joined: 01 May 2005
Posts: 487
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| Posted: Tue Jun 27, 2006 4:28 pm Post subject:
Re: Easy Reasoning
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e_hobsbawm@hotmail.com wrote:
| Quote: | I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric
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Of course the real problem is to decide what the prior distribution is.
I have also thought hard about this kind of problem, tried many
different plausible prior distributions, and basically come up with
junky answers. My sense is that the Bayesian approach doesn't work well
with this kind of problem.
But if I am completely wrong, I hope someone will set me straight.
Stephen |
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Robert B. Israel
science forum Guru
Joined: 24 Mar 2005
Posts: 2151
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| Posted: Tue Jun 27, 2006 9:53 pm Post subject:
Re: Easy Reasoning
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In article <R0dog.1045394$xm3.470260@attbi_s21>,
Stephen Montgomery-Smith <stephen@math.missouri.edu> wrote:
| Quote: | e_hobsbawm@hotmail.com wrote:
I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric
Of course the real problem is to decide what the prior distribution is.
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Yes. The prior distribution for
a) how many balls are in the urn
b) how many of those balls are of each colour
c) what you are going to do with the ball you removed
d) whether there will be a "next ball"
....
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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e_hobsbawm@hotmail.com
science forum beginner
Joined: 27 Jun 2006
Posts: 3
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| Posted: Wed Jun 28, 2006 6:37 am Post subject:
Re: Easy Reasoning
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Robert Israel skrev:
| Quote: | In article <R0dog.1045394$xm3.470260@attbi_s21>,
Stephen Montgomery-Smith <stephen@math.missouri.edu> wrote:
e_hobsbawm@hotmail.com wrote:
I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric
Of course the real problem is to decide what the prior distribution is.
Yes. The prior distribution for
a) how many balls are in the urn
b) how many of those balls are of each colour
c) what you are going to do with the ball you removed
d) whether there will be a "next ball"
...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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When I start I have the assumption that all colors are equally
probable, i.e probability of picking a black ball is 10%. I put the
black ball back into the urn and repeat the experiment. Can I say
something about the probability of picking a black ball next time also?
My prior distribution (guess) could e.g. be that there are 50 balls in
the urn, 5 of each color.
//Fredrik
Best Regards
Eric |
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Stephen Montgomery-Smith
science forum Guru
Joined: 01 May 2005
Posts: 487
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| Posted: Wed Jun 28, 2006 3:15 pm Post subject:
Re: Easy Reasoning
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Robert Israel wrote:
| Quote: | In article <R0dog.1045394$xm3.470260@attbi_s21>,
Stephen Montgomery-Smith <stephen@math.missouri.edu> wrote:
e_hobsbawm@hotmail.com wrote:
I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric
Of course the real problem is to decide what the prior distribution is.
Yes. The prior distribution for
a) how many balls are in the urn
b) how many of those balls are of each colour
c) what you are going to do with the ball you removed
d) whether there will be a "next ball"
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I realise that I was thinking "Bayesian statistics" not "Bayesian
probability." |
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Robert B. Israel
science forum Guru
Joined: 24 Mar 2005
Posts: 2151
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| Posted: Wed Jun 28, 2006 5:32 pm Post subject:
Re: Easy Reasoning
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e_hobsbawm@hotmail.com wrote:
| Quote: | Robert Israel skrev:
In article <R0dog.1045394$xm3.470260@attbi_s21>,
Stephen Montgomery-Smith <stephen@math.missouri.edu> wrote:
e_hobsbawm@hotmail.com wrote:
I want to do a very simple Baysian reasoning.
Suppose I have an urn with balls with 10 different colors. I take a
random ball out of the urn and discover that it is black. What is the
probability that next one also will be black?
Best Regards
Eric
Of course the real problem is to decide what the prior distribution is.
Yes. The prior distribution for
a) how many balls are in the urn
b) how many of those balls are of each colour
c) what you are going to do with the ball you removed
d) whether there will be a "next ball"
...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
When I start I have the assumption that all colors are equally
probable, i.e probability of picking a black ball is 10%. I put the
black ball back into the urn and repeat the experiment. Can I say
something about the probability of picking a black ball next time also?
My prior distribution (guess) could e.g. be that there are 50 balls in
the urn, 5 of each color.
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There's nothing Bayesian about it if you specify that there are 5 of
each colour.
As Stephen noted, there is a difference between "Bayesian statistics"
and
"Bayesian probability"
A "Bayesian probability" situation might be the following:
There is a large population of balls, of which 1/10 are black. A
random sample of 50 balls is taken from this population and placed
in the urn. Thus each ball has probability 1/10 of being black,
independent of the others. You take a ball from the urn, see that it
is black, and put it back. When you take another ball, its
probability of being black is now 1/50 * 1 + 49/50 * 1/10 = 59/500
(i.e. with probability 1/50 you pick the first ball again, while with
probability 49/50 you get a different ball which has probability
1/10 of being black).
A "Bayesian statistics" situation might be the following:
There is a large population of balls, with 10 possible colours of
which one is black.
Let p_i be the fraction of balls that are colour #i (with black = #1).
I have no idea what fraction of the balls are each colour, so my
prior distribution for (p_1,...,p_10) is uniform on the simplex
{(x_1,...,x_10): all x_i >= 0, sum_i x_i = 1}.
In particular, the density for p_1 is f(x) = 9 (1-x)^8.
Now a random sample of 50 balls is taken from this population
and placed in the urn. You take a ball, put it back, and draw
another ball. Let A be the event that the first ball is black,
B the event that the second ball is black. Then
P(A) = int_0^1 x f(x) dx = 1/10
P(A and B) = int_0^1 (x/50 + 49/50 x^2) f(x) dx = 109/5500
P(B|A) = P(A and B)/P(A) = 109/550.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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