13 replies to this topic

[bushindo]

Spoiler for try my luck

I picked p in my draw. Lets estimate n = p. What would have been the likelihood that I got the max number ? : 1/p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p

 

 

> 

 

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

Spoiler for i think

It must be p only because then the Probabability of picking out p = 1/p, for n>p, the probability decreases.

 

You are both correct. bona_gold_star.gif

 

nakulendu posted first, so I marked his answer as the solution.

 

 

 

The solutions above are making some assumptions about the properties of underlying distribution of N, which may or may not be warranted depending on your interpretation of the OP.

Spoiler for

Implicit within the explanations above is the assumption that P(N) = P(N+1). The explanation goes on to say that since P(p|N) decreases as N increases (without bound, presumably), then it makes sense that p is the most optimal choice.

From an earlier post, the full Bayesian expression for P(N|p) is

Note that if we assume that that N is uniformly distributed between p and infinity (which is an improper prior distribution), then the denominator of the term above becomes the harmonic series, which does not converge. If we assume that N is uniformly distributed between p and some large finite M, then the solution is warranted.

So, choosing N=p is the correct solution IF P(N) is uniformly distributed AND P(N) -> 0 as N approaches infinity.

[bonanova]

Bushindo, Thanks.
I learn so much from you.

Is there also a maximum likelihood estimate of the underlying assumptions

that would make the maximum likelihood estimate of n well defined?

[Debasis]

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

 

It seems like this problem is dependent upon what is a reasonable a priori distribution for N.

Spoiler for

Let P(N) be the prior distribution for the number N. Let P(p|N) be the conditional probability for p given N; we know that P(p|N) = 1/N. We wish to compute the conditional probability P(N|p), which can be expanded using Bayes theorem

bayesian.png

 

So, if we are given a well-defined probability function P(N), we can easily compute the number N that gives that maximum probability P(N|p).

 

The problem is that we are not given what P(N) is. One reasonable choice would be to assume that P(N) is the improper uniform distribution from p to infinity (every number has an equal chance of occuring). However, if we plug that improper distribution into the equation above, we would be able to cancel out P(N), but we'd end up with a harmonic series in the denominator, which does not converge.

 

OMG!you are one hell of a methematician...lol

[bushindo]

Bushindo, Thanks.
I learn so much from you.

Is there also a maximum likelihood estimate of the underlying assumptions

that would make the maximum likelihood estimate of n well defined?

 

Thanks for the kind words, bonanova, I also learned much from you over the years. The two wonderful examples (among many) that come quickest to mind are Hole in a Sphere and Maiden Escape, both of which taught me completely new ways of thinking.

 

Spoiler for

The comment I made earlier is that the solution to this puzzle will depend on what kind of prior information (or assumption) we have about the probability distribution P(N), since the Bayesian derivation forces us to explicitly use P(N) in the calculation of the answer.

My interpretation of what you are asking is that is there a maximum likelihood estimate for the distribution P(N) such that P(N|p) is well-defined (and my impression is that this is a way to bypass the requirement to know exact values for P(N)). It is possible to assume the distribution P(N) as belonging to some well-defined probability class (exponential, gaussian, negative exponential, etc.) and then use maximum likelihood to find the best-fit curve within the assumed class. Of course, there is no free lunch. We just moved the prior assumption backwards a bit so that we make assumptions about the class of probability curves P(N) belongs to.

You may be asking is there a maximum likelihood estimate for estimating what class of probability function P(N) most likely is in, and that is a much tougher problem. If we have several candidate probability classes (again, exponential, binomial, gaussian, etc.) then we can easily compare between them and find the most likely probability class for P(N). The problem is that there are an infinite number of possible probability class (and most don't have nice tractable parametric forms), and finding the best class over this domain is not possible. So, if we restrict the possible probability classes to a finite number (note that we are making some assumptions here again), we can then solve the problem.