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Suppose that I observe $k=4$ tanks with serial numbers $2,6,7,14$. What is the best estimate for the total number of tanks $n$?

I assume the observations are drawn from a discrete uniform distribution with the interval $[1,n]$. I know that for a $[0,1]$ interval the expected maximum draw $m$ for $k$ draws is $1 - (1/(1+k))$. So I estimate $\frac {k}{k+1}$$(n-1)≈$ $m$, rearranged so $n≈$ $\frac {k+1 }{k}$$m+1$.

But the frequentist estimate from Wikipedia is defined as:

$n ≈ m-1 + $$\frac {m}{k}$

I suspect there is some flaw in the way I have extrapolated from one interval to another, but I would welcome an explanation of why I have gone wrong!

rjt90
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  • You seem to have arrived near the frequentist estimate instead of the Bayesian one. –  Jul 30 '13 at 19:01
  • Just wondering, how do you arrive at $1-\frac{1}{1+k}$? – angryavian Jul 30 '13 at 19:42
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    the exact expected value of the maximum serial number given $k$ out of $n$ tanks is $\frac{k}{k+1}(n+1)$. Note when $k=n$ you expect $n$. The continuous cases is the limit of this case. – Thomas Andrews Jul 30 '13 at 19:54
  • I've just tagged that section of the Wikipedia article as questionable, and commented on the discussion page here. – Michael Hardy Jul 30 '13 at 21:24
  • I suspect that by "frequentist" estimate, whoever wrote that meant "best unbiased estimate". Not that an MLE would also be a frequentist estimate, but would be different from that. – Michael Hardy Jul 30 '13 at 21:33

2 Answers2

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Just seen what went wrong. I accidentally put in a plus sign instead of a minus sign. Ugh:

$n≈$ $\frac {k+1}{k}$$m+1$ should be $n≈$ $\frac {k+1}{k}$$m-1$.

This is the same as the frequentist formula.

rjt90
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What Thomas mentioned in the comment is correct.

The German Tank Problem is different from drawing random samples from a uniform distribution, in which case the expected sample maximum is $\frac{k}{k+1}(n-1)+1$. (maximum of k samples in $U(1, n)$)

Instead, we are sampling without replacement - we can never obtain two tank samples with the same serial number.

Lorry
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