Estimating sample size given its smallest member I have a friend who has a list of n numbers from [0, x). They are integers (though x is so large that doesn't really matter), and are uniformly randomly distributed. My friend figures out the smallest member from this set, and calls it y.
I know x and y, but not the original list of numbers or n. How can I estimate n? How accurate will my estimate be?
 A: The bias of any estimator will go to $\infty$ as $n\to\infty$. This is because as $n\to\infty$, the probability that $y=0$ goes to $1$. You have to assign some arbitrary estimate $n_0$ to $y=0$, and then no matter what estimates you assign to the other values of $y$, as $n\to\infty$ the expected value of your estimate will go to $n_0$ as $n\to\infty$, whereas it should be going to $\infty$ along with $n$.
However, you can estimate $1/(n+1)$. This is messy and approximate for a discrete distribution, so I'll derive it for a uniform continuous distribution on $[0,1]$ and you can use it as an approximation for your discrete case.
The cumulative distribution function for $y$ in this case is $1-(1-y)^n$, the density is $n(1-y)^{n-1}$, and the expected value of $y$ is
$$
\int_0^1n(1-y)^{n-1}y\,\mathrm dy=\frac1{n+1}\;.
$$
Thus $y$ is a unbiased estimator for $1/(n+1)$. Its variance is
$$
\begin{align}
\langle y^2\rangle-\langle y\rangle^2
&=
\int_0^1n(1-y)^{n-1}y^2\,\mathrm dy-\frac1{(n+1)^2}
\\
&=
\frac2{(n+1)(n+2)}-\frac1{(n+1)^2}
\\
&=
\frac n{(n+1)^2(n+2)}
\\
&\sim\frac1{(n+1)^2}\;.
\end{align}
$$
Thus for large $n$ the standard deviation is also approximately $1/(n+1)$.
In the continuous case, you could use $n=y^{-1}-1$ to estimate $n$. The expected value of this estimator is infinite, but at least you'd have the right expected value for $1/(n+1)$. In the discrete case, you can't do this, since typically $y=0$ and then $y^{-1}$ isn't defined.
