Re: A simple but confusing question
From: Ian Jermyn (Ian.Jermyn_at_sophia.inria.fr)
Date: 09/27/04
- Next message: Osher Doctorow: "Chaos , Catastrophes, and Fractals Simplified by PI"
- Previous message: Shanyu Zhao: "A simple but confusing question"
- In reply to: Shanyu Zhao: "A simple but confusing question"
- Messages sorted by: [ date ] [ thread ]
Date: Mon, 27 Sep 2004 13:58:29 +0200
Let w_{n} be the proposition 'the nth ball is white'. Let W_{n} be the
proposition w_{n} & w_{n - 1} & ... & w_{2} &w_{1}.
What you want to know is:
Pr(w_{n + 1} | W_{n}) .
Introduce the proportion of white balls in the box before a draw takes
place. Call it a. I do not use the notation p because it is not a
probability. Assume that the number of balls in the box is so large that we
can treat a as a real number between 0 and 1, and that the number of balls
we are drawing is not sufficient to change the number of balls in the box
swignificantly, so that we are effectively doing 'sampling with
replacement'. (You can analyse the full problem, but it is harder.)
Then
Pr(w_{n + 1} | W_{n}) = \int da Pr(w_{n + 1}, a | W_{n})
[by marginalisation]
= \int da Pr(w_{n + 1} | W_{n}, a) Pr(a | W_{n})
[by conditional probability]
= \int da Pr(w_{n + 1} | W_{n}, a) Pr(W_{n} | a) Pr(a) / Pr(W_{n})
[by Bayes'theorem] .
The first probability is just a, since knowing a, the probability that the
next ball is white is not altered by the fact that previous balls were white
(remember we are doing sampling with replacement). The second probability is
a^{n}. Now we come to the tricky part. What is the probability for the
proportion a before we have drawn any balls? The simplest answer is just to
say it is constant, and that is what I will do here. If you want a long
discussion of this point, see Ed Jaynes book, 'Probabiilty Theory: the Logic
of Science', which anyway is important reading. (In a real problem, you
would likely have some information about a, and things would be easier.)
So if we assume that Pr(a) = 1, we find that
Pr(W_{n}) = \int da Pr(W_{n} | a) Pr(a) = \int da a^{n} = 1 / (n + 1) ,
and that
Pr(w_{n + 1} | W_{n}) = \int da a. a^{n} (n + 1) = (n + 1) / (n + 2).
So for n = 100, this gives 101/102.
There is no question of bounding this probability. For any given prior
knowledge of a, it is fixed. For example, if one knows the exact value of a
a priori, then it is equal to a, and so can vary between 0 and 1.
Ian.
-- -------------------------------------------------- Ian Jermyn INRIA (Ariana) 2004 route des Lucioles B. P. 93 06902 Sophia Antipolis Cedex France T: +33 (0)4 9238 7683 F: +33 (0)4 9238 7643 E: Ian.Jermyn@sophia.inria.fr W: http://www-sop.inria.fr/ariana/personnel/Ian.Jermyn "Shanyu Zhao" <szhao@darkwing.uoregon.edu> a écrit dans le message de news:96a39245.0409270026.63fd76c@posting.google.com... > Hi, > > Here is the naive question: > > There are a large number of balls in a bucket, the white color balls > occupy p and the black balls (1-p). If p is unknown, when you pick 100 > balls from the bucket, find that all of them are white. Then you pick > the 101st ball, what is the probability that the ball still a white > one? > > If we can't calculate the probability, can we reach the conclusion > that the probability is greater than (100/101)? Or more precisely, > what is the bottom line of this probability. Intuitively it should be > above some value. > > Is this problem a parameter estimation or hypothesis testing? If we > use parameter estimation, clearly p=1, which means the probability is > 100%. But this is not true. > > I've always been confused by this problem. Please help me out! > Thanks alot! > > Shanyu
- Next message: Osher Doctorow: "Chaos , Catastrophes, and Fractals Simplified by PI"
- Previous message: Shanyu Zhao: "A simple but confusing question"
- In reply to: Shanyu Zhao: "A simple but confusing question"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
|
