Suppose we conduct Buffon's needle experiment, randomly dropping a needle of length $L\in(0,1]$ between horizontal lines which are a distance of $1$ apart. It is well known that the probability that the needle intersects one of these lines equals $p=2L/\pi$, so that we can estimate $\pi$ by $2L/\hat p=2Ln/m$, where $n$ is the size of the Monte Carlo experiment, and where $m$ is the number of intersections.
I am looking for the optimal length $L$ of the needle, making the estimate for $\pi$ as accurate as possible. It seems reasonable that when an intersection is as likely as no intersection, so when $L=\pi/4$, the estimation is most efficient (although I am not sure it this is correct, and if so, why).
More specifically, I'm looking for the variance of $2L/\hat p$. This seems to come down to estimate the variance of the reciprocal of a binomial r.v., for which no nice expression exist. Probably there is some other approach I'm not seeing. Any help is much appreciated.
N.B.: this is just for theoretical purposes; I don't mind to use $\pi$ in the process of estimating $\pi$ itself.
Best Answer
You can also estimate $1/\pi$ by $\hat p/2L,$ whose variance is easier to calculate.
For large enough $R,$ the distribution of $\hat p$ around $p$ will be concentrated near $p$ and very nearly symmetric near $p.$ When you suppose that a $95\%$ confidence interval is $\hat p\pm1.96\frac{2L}\pi\left(1-\frac{2L}\pi\right)/\sqrt R,$ you are assuming that the distribution is symmetric within that interval around $p$.
The symmetric CI around $\hat p$ will map to an asymmetric CI around $1/\hat p$ when you use $1/\hat p$ as an estimator for $\pi,$ but this just means that the symmetric CI that you want around $1/\hat p$ comes from an asymmetric CI around $\hat p$. If your estimate of $\pi$ has a decent amount of precision, the asymmetry of the desired CI around $\hat p$ is not even very much. It seems reasonable to suppose that the value of $L$ that minimizes the asymmetric CI is the same as that which minimizes the symmetric CI. This may not be perfectly accurate, but neither is the use of a normal distribution in place of the actual binomial distribution of $\hat p$.