8
$\begingroup$

I take the bus to work every day. Every bus has a serial number, but unlike in the German Tank Problem, I don't know if they are numbered uniformly $1...n$.

Suppose the first $k$ buses are all different, but on day $k+1$ I take one I've been on before. What is the best estimate for the total number of buses?

$\endgroup$

1 Answer 1

5
$\begingroup$

Maximum likelihood estimate is the smallest $n$ for which $$\left( 1+\frac{1}{n} \right)^k \leq \frac{n}{n-k+1},$$ that gives a value of $n$ asymptotically equal to $\frac{k^2}{2}$, consistently with the Birthday Paradox. Not sure whether an unbiased estimate would be better for any practical purpose; maybe you do have an a priori distribution for which a Bayesian estimate makes sense?

$\endgroup$
2
  • $\begingroup$ It seems that the estimate $\frac{k(k+1)}{2}$ is unbiased (the only unbiased estimate, actually). $\endgroup$
    – Thorny
    Commented Feb 11, 2010 at 14:50
  • $\begingroup$ Maximum likelihood estimates often perform somewhat better than unbiased estimates. $\endgroup$ Commented Jun 21, 2010 at 23:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .