Re: urn model
- From: hrubin@xxxxxxxxxxxxxxxxxxxx (Herman Rubin)
- Date: 10 Jun 2008 19:04:10 -0400
In article <8c120584-4771-419f-9b64-073d9377ce1e@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
randerson1184 <randerson1184@xxxxxxxxx> wrote:
On Jun 9, 4:26 pm, randerson1184 <randerson1...@xxxxxxxxx> wrote:
On Jun 9, 3:08 pm, isabelle...@xxxxxxxxxxx wrote:
On Jun 9, 3:54 pm, randerson1184 <randerson1...@xxxxxxxxx> wrote:
On Jun 9, 2:51 pm, randerson1184 <randerson1...@xxxxxxxxx> wrote:
On Jun 9, 2:46 pm, Paul Rubin <ru...@xxxxxxx> wrote:
isabelle...@xxxxxxxxxxx wrote:
On Jun 9, 2:36 pm, Paul Rubin <ru...@xxxxxxx> wrote:
isabelle...@xxxxxxxxxxx wrote:
Hi everybody,You need to rethink that last line a bit. If necessary, try assuming a
I am having a hard time writing the likelihood of the following urn
model. I would need to find the maximum likelihood estimator of p, so
I only need those terms of the likelihood that are functions of p. I
have not seen very many urn problems in my textbooks, if there is a
reference with urn problems, I will gladly read it.
We continue to draw balls with replacement from an urn containing both
black and white balls until two balls of the same color are drawn. Let
X be the number of draws (including the first draw) we take until we
stop. We are not told how many black and white balls that we see, but
only the value of X. Let p be the unknown proportion of white balls in
the urn.
I greatly appreciate your help.
Here is my work so far.
X can take only two values: 2 or 3.
The sample space is
BB, X=2
WW, X=2
BWB, X=3
BWW, X=3
WBB, X=3
WBW, X=3
When X=2, the likelihood can be proportional to p^2 or (1-p)^2.
When X=3, the likelihood can be proportional to p^1*(1-p)^2 or p^2*(1-
p)^1.
value such as 0.1 for p and see what you come up with for the
probabilities of X=2 and X=3.
/Paul
I am not sure how to do this. I do not have a method to write the
probabilities. Here is an attempt.
P[X=2]=P[BB or WW]=P[BB]+P[WW]=constant1*{p^2 + (1-p)^2}
P[X=3]=P[BWB]+P[BWW]+P[WBW]+...
=constant2*{2*p^1*(1-p)^2 + 2*p^2*(1- p)^1}
You are on the right track. Why do feel the need for constant1 and
constant2? You have six possible sequences listed above. Write the
probability of each in terms of p, and see where that takes you.
/Paul
I got that X~Ber(2p-2p^2-p^3)
I defined a success as X=3 for which the probability is the 2p-2p^2-
p^3
So from here, I just use the likelihood function to find the MLE of p
right?
I think I need to make the simple transformation Y=X-2 and then
declare it bernoulli...
I am finding
X-2~Bernoulli(2p-2p^2)
thanks a lot for your help
Yep, you're right, 2p - 2p^2. Let me know what you get for your MLE.
Alright, I re-read the thread, and I couldn't figure out why I thought
you were trying to find an MLE. Perhaps it's because you used the term
"likelihood." You're just trying to find an estimator for 'p' right?
I am a frequentist (go ahead, do your worst) so I would tend to go
with MLE or UMVUE (which I should have memorized for the bernoulli
case by now). If we were to find a bayes estimator, what prior would
we place on p? Beta?
If beta, how do you go about selecting values for your parameters for
your prior distribution? Hyperpriors?
If you have a stopping rule which depends only on the number
of black and white balls, the MLE is the proportion of balls
of each color found until that point. The same holds for all
Bayes estimates. The best unbiased estimate is the proportion
of paths ending at that point.
As to what prior (and loss) should be chosen, it is the user's
prior. The loss-prior combination should not depend on the
probability distribution, unless one can obtain a robustness
theorem showing that the assumed prior is good for what the
real problem involved.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558
.
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