Solution to German Tank Problem Bb H D5ZzKMm VHGg JQq

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Is there a formal mathematical proof that the solution to the German Tank Problem is a function of only the parameters k (number of observed samples) and m (maximum value among observed samples)? In other words, can one prove that the solution is independent of the other sample values besides the maximum value?

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    $\\begingroup$ What you're asking is how to show that the sample maximum is sufficient for the parameter $\\theta$ specifying the upper bound of a discrete uniform distribution from 1 to $\\theta$. $\\endgroup$ – Scortchi 10 hours ago
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    $\\begingroup$ Fisher Neyman factorization theorem The likelihood function, probability of the $k$ observed samples (summarized by the maximum $m$) given the parameters $n$ (the number of tanks) can be completely written in terms of the $k$ and $m$ $$\\Pr(M=m | n,k) = \\begin{cases} 0 &\\text{if } m > n \\\\ \\frac{\\binom{m - 1}{k - 1}}{\\binom n k} &\\text{if } m \\leq n, \\end{cases}$$ Would that be an answer? $\\endgroup$ – Martijn Weterings 10 hours ago
  • $\\begingroup$ @Scortchi that is correct, thank you for rephrasing it in a clearer way for me. $\\endgroup$ – Bogdan Alexandru 9 hours ago
  • $\\begingroup$ @MartijnWeterings no; essentially I am asking (quoting Scortchi's comment above) for a proof that the sample maximum is sufficient for the solution without actually computing the solution. $\\endgroup$ – Bogdan Alexandru 9 hours ago
  • $\\begingroup$ So you are not looking for the Fisher Neyman factorization theorem as the proof? $\\endgroup$ – Martijn Weterings 9 hours ago

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Likelihood

Common problems in probability theory refer to the probability of observations $x_1, x_2, ... , x_n$ given a certain model and given the parameters (let's call them $\\theta$) involved. For instance the probabilities for specific situations in card games or dice games are often very straightforward.

However, in many practical situations we are dealing with an inverse situation (inferential statistics). That is: the observation $x_1, x_2, ... , x_k$ is given and now the model is unknown, or at least we do not know certain parameters $\\theta$.

In these type of problems we often refer to a term called the likelihood of the parameters, $\\mathcal{L(\\theta)}$, which is a rate of believe in a specific parameter $\\theta$ given observations $x_1, x_2, .. x_k$. This term is expressed as being proportional to the probability for the observations $x_1, x_2, .. x_k$ assuming that a model parameter $\\theta$ would be hypothetically true. $$\\mathcal{L}(\\theta,x_1, x_2, .. x_k) \\propto \\text{probability observations $x_1, x_2, .. x_k$ given $\\theta$ }$$

For a given parameter value $\\theta$ the more probable a certain observation $x_1, x_2, .. x_n$ is (relative to the probability with other parameter values), the more the observation supports this particular parameter (or theory/hypothesis that assumes this parameter). A (relative) high likelihood will reinforce our believes about that parameter value (there's a lot more philosophical to say about this).


Likelihood in the German tank problem

Now for the German tank problem the likelihood function for a set of samples $x_1, x_2, .. x_k$ is:

$$\\mathcal{L}(\\theta,x_1, x_2, .. x_k ) = \\Pr(x_1, x_2, .. x_k, \\theta) = \\begin{cases} 0 &\\text{if } \\max(x_1, x_2, .. x_k) > \\theta \\\\ {{\\theta}\\choose{k}}^{-1} &\\text{if } \\max(x_1, x_2, .. x_k) \\leq \\theta, \\end{cases}$$

Whether you observe samples {1, 2, 10} or samples {8, 9, 10} should not matter when the samples are considered from a uniform distribution with parameter $\\theta$. Both samples are equally likely with probability ${{\\theta}\\choose{3}}^{-1}$ and using the idea of likelihood the one sample does not tell more about the parameter $\\theta$ than the other sample.

The high values {8, 9, 10} might make you think/believe that $\\theta$ should be higher. But, it is only the value {10} That truly gives you relevant information about the likelihood of $\\theta$ (the value 10 tells you that $\\theta$ will be ten or higher, the other values 8 and 9 do not contribute anything to this information).


Fisher Neyman factorization theorem

This theorem tells you that a certain statistic $T(x_1, x_2, … , x_k)$ (ie some function of the observations, like the mean, median, or as in the German tank problem the maximum) is sufficient (contains all information) when you can factor out, in the likelihood function, the terms that are dependent on the other observations $x_1, x_2, … , x_k$, such that this factor does not depend on both the parameter $\\theta$ and $x_1, x_2, … , x_k$ (and the part of the likelihood function that relates the data with the hypothetical parameter values is only dependent on the statistic but not the whole of the data/observations).

The case of the German tank problem is simple. You can see above that the entire expression for the Likelihood above is already only dependent on the statistic $\\max(x_1, x_2, .. x_k)$ and the rest of the values $x_1, x_2, .. x_k$ does not matter.


Little game as example

Let's say we play the following game repeatedly: $\\theta$ is itself a random variable and drawn with equal probability either 100 or 110. Then we draw a sample $x_1,x_2,...,x_k$.

We want to choose a strategy for guessing $\\theta$, based on the observed $x_1,x_2,...,x_k$ that maximizes our probability to have the right guess of $\\theta$.

The proper strategy will be to choose 100 unless one of the numbers in the sample is >100.

We could be tempted to choose the parameter value 110 already when many of the $x_1,x_2,...,x_k$ tend to be all high values close to hundred (but none exactly over hundred), but that would be wrong. The probability for such an observation will be larger when the true parameter value is 100 than when it is 110. So if we guess, in such situation, 100 as the parameter value, then we will be less likely to make a mistake (because the situation with these high values close to hundred, yet still below it, occurs more often in the case that the true value is 100 rather than the case that the true value is 110).

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