German Tank Problem - Bayesian Analysis

Bayesian Analysis

The Bayesian approach to the German Tank Problem is to consider the probability that the number of enemy tanks is equal to, when the number of observed tanks, is equal to the number, and the largest of the serial numbers is equal to .

For brevity is written

The rule for conditional probability gives

The expression is the conditional probability that the largest tank serial number observed is equal to, when the number of enemy tanks is known to be equal to, and enemy tanks have been observed. It is

$(m|n,k) = \begin{cases} \frac{\binom{m - 1}{k - 1}}{\binom{n}{k}} &\text{if } k \le m \le n\\ 0 &\text{otherwise} \end{cases}$

where the binomial coefficient is the number of -sized samples from an -sized population.

The expression is the probability that the maximum serial number is equal to m once k tanks have been observed but before the serial numbers have actually been observed. can be removed from the formula by the following trick.

$\begin{align} (m|k) &=(m|k)\cdot 1 \\ &=(m|k){\sum_{n=0}^\infty(n|m, k)} \\ &=(m|k){\sum_{n=0}^\infty(m|n, k)\frac {(n|k)}{(m|k)}} \\ &=\sum_{n=0}^\infty(m|n, k)(n|k) \end{align}$

The expression is the probability that the total number of tanks is equal to n when k tanks have been observed but before the serial numbers have actually been observed. Assume that it is some discrete uniform distribution

$(n|k) = \begin{cases} \frac 1{\Omega - k} &\text{if } k \le n < \Omega \\ 0 &\text{otherwise} \end{cases}$

The upper limit must be finite, because the function

$f(n)=\lim_{\Omega\rarr\infty} \begin{cases} \frac 1{\Omega - k} &\text{if } k \le n < \Omega \\ 0 &\text{otherwise} \end{cases}$

which is not a probability mass function.

Then

$(n|m,k) = \begin{cases} \frac{(m|n, k)}{\sum_{n=m}^{\Omega - 1} (m|n, k)} &\text{if } m \le n < \Omega \\ 0 &\text{otherwise} \end{cases}$

If, then the unwelcome variable disappears from the expression.

$(n|m,k) = \begin{cases} 0 &\text{if } n < m \\ \frac{(m|n, k)}{\sum_{n=m}^\infty(m|n, k)} &\text{if } n \ge m \end{cases}$

For k≥1 the mode of the distribution of the number of enemy tanks is m.

For k≥2, the probability that the number of enemy tanks is equal to, is

$\Pr(N=n|M=m\ge k,K=k\ge 2) = \begin{cases} 0 &\text{if } n < m \\ \frac k{k - 1}\frac {\binom{m - 1}{k - 1}}{\binom n k} &\text{if } n \ge m \end{cases}$

and the probability that the number of enemy tanks, is greater than, is

$\Pr(N>n|M=m\ge k,K=k\ge 2) = \begin{cases} 1 &\text{if } n < m \\ \frac {\binom{m - 1}{k - 1}}{\binom n {k - 1}} &\text{if } n \ge m \end{cases}$

For k≥3, has the finite mean value:

For k≥4, has the finite standard deviation:

These formulas are derived below.

Read more about this topic: German Tank Problem