# German tank problem: comparing two estimators

The following estimators can be used for the german-tank-like problems.

If we collect a sample of size $k$ with sample mean $\bar{x}$ and highest number $m$, then we can estimate the population size with either $\hat{x}_{1}$ or $\hat{x}_{2}$:

$$\hat{x}_{1} = m + \frac{m}{k} - 1$$ $$\hat{x}_{2} = 2 \bar{x}$$

After computationally testing both on random datasets, with various sample sizes and various number of samples, it appears that:

• both are unbiased (both hit the true value on average)
• $\hat{x}_{1}$ is much more accurate than $\hat{x}_{2}$, especially when $k$ is relatively small ($k < 30$)

The reasoning behind $\hat{x}_{2}$ is obvious, while $\hat{x}_{1}$ is a little more obscure.

What I don't get is, how comes $\hat{x}_{1}$ is the more accurate of the two?

• Have you tried the wikipedia article? en.wikipedia.org/wiki/German_tank_problem#Derivation Dec 28, 2016 at 11:27
• Shouldn't the second estimator be 2bar{x}-1? Does that estimator behave a little better? Jan 29, 2018 at 21:57