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silicon2006@hotmail.com science forum beginner
Joined: 19 May 2006
Posts: 6

Posted: Fri May 19, 2006 3:04 am Post subject:
A decision making problem



''A' and 'B' are two sets.
'A'={x1, x2, x3}, where x1, x2 and x3 are Gaussian random variables
with the same standard deviation s but different mean a1, a2, a3. It
is known that a1>a2>a3. The specific magnitude of s is unknown.
Similarly, 'B'={y1, y2, y3}, with corresponding mean b1>b2>b3. The
standard deviation 's' is the same as in set 'A'.
It is known that a1>b1, a2>b2 and a3>b3.
Now, we have two sets of measurements from 'A' and 'B'. The results
are:
P={p1>p2>p3}, Q={q1>q2>q3}. We try to figure out which measurement is
from 'A' and which is from 'B'.
Remember that p1, p2, p3 and x1, x2, x3 (or y1, y2, y3) may not in the
same order. For example, p1 might be the random variable with mean b3
in set 'B'.
A simple way to tell whether measurement 'P' or 'Q' is from set 'A' is
to compare the average
pp=(p1+p2+p3)/3, qq=(q1+q2+a3)/3 and see which one is larger. Is it the
best judgement? or there is another approach which best utilizes the
information? 

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illywhacker science forum beginner
Joined: 25 Jul 2005
Posts: 9

Posted: Fri May 19, 2006 7:09 pm Post subject:
Re: A decision making problem



Hi.
silicon2006@hotmail.com wrote:
Quote:  ''A' and 'B' are two sets.
'A'={x1, x2, x3}, where x1, x2 and x3 are Gaussian random variables
with the same standard deviation s but different mean a1, a2, a3. It
is known that a1>a2>a3. The specific magnitude of s is unknown.
Similarly, 'B'={y1, y2, y3}, with corresponding mean b1>b2>b3. The
standard deviation 's' is the same as in set 'A'.
It is known that a1>b1, a2>b2 and a3>b3.
Now, we have two sets of measurements from 'A' and 'B'. The results
are:
P={p1>p2>p3}, Q={q1>q2>q3}. We try to figure out which measurement is
from 'A' and which is from 'B'.
Remember that p1, p2, p3 and x1, x2, x3 (or y1, y2, y3) may not in the
same order. For example, p1 might be the random variable with mean b3
in set 'B'.
A simple way to tell whether measurement 'P' or 'Q' is from set 'A' is
to compare the average
pp=(p1+p2+p3)/3, qq=(q1+q2+a3)/3 and see which one is larger. Is it the
best judgement? or there is another approach which best utilizes the
information?

Yes. If understand correctly, the set P is from A and Q from B or
viceversa, but both are not from the same set. So there are two
outcomes, so it suffices to calculate the probability of one: let's say
P from A and Q from B. Call this O. You are also given the means. Call
the set of all the means m.
Then you want to calculate
p(O  P, Q, m) = p(P, Q  O, m) p(O  m) / p(P, Q  m) .
The prior probability of O given m can presumably be set to 1/2, unless
you have any reason to think that one is more likely than the other.
Since
p(P, Q  m) = p(P, Q  O, m) p(O  m) + p(P, Q  ~O, m) p(~O  m) ,
where ~O means 'not O', i.e. P > B and Q > A, we have that
p(O  P, Q, m) = p(P, Q  O, m) / ( p(P, Q  O, m) + p(P, Q  ~O, m)) .
(*)
To calculate p(P, Q  O, m), you have to introduce some more variables:
the assignments a of measurements from P to distributions in A, the
assignments b of measurements from Q to distributions in B, and s:
p(P, Q  O, m) = \int ds \sum_{a, b} p(P, Q  O, m, a, b, s) p(a, b, s
 O, m)
= \int ds \sum_{a} p(P, Q  O, m, a, b, s) p(a) p(b) p(s)
= \int ds p(s) \sum_{a} p(P  O, m, a, s) \sum_{b} p(Q  O, m, b, s) ,
where various reasonable independence assumptions have been made.
Assuming every assignment is equally likely a priori (there are 27 for
each of P and Q, assuming more than one measurement can come from each
distribution), and taking p(s) = 1/s (Jeffreys' prior), gives
p(P, Q  O, m) = (1/27) \int (ds/s) \sum_{a} p(P  O, m, a, s) \sum_{b}
p(Q  O, m, b, s) .
Now for a given assignment and knowing the means and variance, you can
calculate the probability of the data: it is just the appropriate
Gaussian distribution. The integrand will this be the product of two
sums of 27 Gaussian distributions: this gives 27 x 27 terms. The
biggest term in each of the sums will be that in which the exponent in
the Gaussian minimizes the Euclidean distance between the means and the
data. The other terms will probably be small, and it may be sufficient
to keep this one term in each sum.
In any case, the product is also a sum of Gaussians. The integral over
s takes the same form in each, and can be performed. The final sum
gives p(P, Q  O, m).
Repeat this for the opposite choice, ~O, and then calculate equation
(*). The most likely option, O or ~O, is the one with higher
probability, but the value of this probability also gives you an idea
of how certain this conclusion is.
illywhacker. 

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silicon2006@hotmail.com science forum beginner
Joined: 19 May 2006
Posts: 6

Posted: Mon May 29, 2006 4:13 pm Post subject:
Re: A decision making problem



Thanks for the argument!
What do you mean by 'the set of all the means m'? After you saying
'm'={a1, a2, a3, b1, b2, b3}?
But the only things we know about a1, a2, a3, b1, b2, b3 are that:
(1) a1>a2>a3
(2) b1>b2>b3
(3) a1>b1, a2>b2, a3>b3.
The specific magnitudes of a1, a2, a3, b1, b2, b3 are NOT known.
In other words, 'm' is NOT given.
illywhacker wrote:
Quote:  Hi.
silicon2006@hotmail.com wrote:
''A' and 'B' are two sets.
'A'={x1, x2, x3}, where x1, x2 and x3 are Gaussian random variables
with the same standard deviation s but different mean a1, a2, a3. It
is known that a1>a2>a3. The specific magnitude of s is unknown.
Similarly, 'B'={y1, y2, y3}, with corresponding mean b1>b2>b3. The
standard deviation 's' is the same as in set 'A'.
It is known that a1>b1, a2>b2 and a3>b3.
Now, we have two sets of measurements from 'A' and 'B'. The results
are:
P={p1>p2>p3}, Q={q1>q2>q3}. We try to figure out which measurement is
from 'A' and which is from 'B'.
Remember that p1, p2, p3 and x1, x2, x3 (or y1, y2, y3) may not in the
same order. For example, p1 might be the random variable with mean b3
in set 'B'.
A simple way to tell whether measurement 'P' or 'Q' is from set 'A' is
to compare the average
pp=(p1+p2+p3)/3, qq=(q1+q2+a3)/3 and see which one is larger. Is it the
best judgement? or there is another approach which best utilizes the
information?
Yes. If understand correctly, the set P is from A and Q from B or
viceversa, but both are not from the same set. So there are two
outcomes, so it suffices to calculate the probability of one: let's say
P from A and Q from B. Call this O. You are also given the means. Call
the set of all the means m.
Then you want to calculate
p(O  P, Q, m) = p(P, Q  O, m) p(O  m) / p(P, Q  m) .
The prior probability of O given m can presumably be set to 1/2, unless
you have any reason to think that one is more likely than the other.
Since
p(P, Q  m) = p(P, Q  O, m) p(O  m) + p(P, Q  ~O, m) p(~O  m) ,
where ~O means 'not O', i.e. P > B and Q > A, we have that
p(O  P, Q, m) = p(P, Q  O, m) / ( p(P, Q  O, m) + p(P, Q  ~O, m)) .
(*)
To calculate p(P, Q  O, m), you have to introduce some more variables:
the assignments a of measurements from P to distributions in A, the
assignments b of measurements from Q to distributions in B, and s:
p(P, Q  O, m) = \int ds \sum_{a, b} p(P, Q  O, m, a, b, s) p(a, b, s
 O, m)
= \int ds \sum_{a} p(P, Q  O, m, a, b, s) p(a) p(b) p(s)
= \int ds p(s) \sum_{a} p(P  O, m, a, s) \sum_{b} p(Q  O, m, b, s) ,
where various reasonable independence assumptions have been made.
Assuming every assignment is equally likely a priori (there are 27 for
each of P and Q, assuming more than one measurement can come from each
distribution), and taking p(s) = 1/s (Jeffreys' prior), gives
p(P, Q  O, m) = (1/27) \int (ds/s) \sum_{a} p(P  O, m, a, s) \sum_{b}
p(Q  O, m, b, s) .
Now for a given assignment and knowing the means and variance, you can
calculate the probability of the data: it is just the appropriate
Gaussian distribution. The integrand will this be the product of two
sums of 27 Gaussian distributions: this gives 27 x 27 terms. The
biggest term in each of the sums will be that in which the exponent in
the Gaussian minimizes the Euclidean distance between the means and the
data. The other terms will probably be small, and it may be sufficient
to keep this one term in each sum.
In any case, the product is also a sum of Gaussians. The integral over
s takes the same form in each, and can be performed. The final sum
gives p(P, Q  O, m).
Repeat this for the opposite choice, ~O, and then calculate equation
(*). The most likely option, O or ~O, is the one with higher
probability, but the value of this probability also gives you an idea
of how certain this conclusion is.
illywhacker. 


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