# How to generate unbiased estimate of 1/E[x] using one random draw?

Posted by Yuling Yao on Apr 19, 2022.       Tag: computing

Quiz: you are given ONE random draw $x$ that was drawn from a density $p(x)$. Could you produce an unbiased estimate of $1/E_p[X]$?

Apart from mathematical fun, this type of problem comes out in stochastic approximation, in which we needs an unbiased estimate using a very small number of Monte Carlo draws. The unbiasedness here means that this sampling step will be repeated many times, but each time you are only shown one sample point $x$, and we wish the estimate to be unbiased under repeated sampling.

Here, the obvious wrong answer is to use $1/x$. You can try

n=10000
x=rbeta(n,2,2)
mean(1/x)


It is clear that E $[1/x]= 3$ while our desired quantity 1/E $[x]= 2$. Indeed it is also clear that E $[1/x] >$ 1/E $[x]$ for positive $x$.

How about some Taylor series expansion? something like $1/E[p(X)] = 1- (E(x)-1) + O(E(x)-1)^2$? It is legitimate but then you get some crude approximation $2-x$, provided that I believe that $E(x) \approx 1$, which we typically do not know in the first place.

I find one solution from rejection sampling. The idea is that self-normalized importance sampling is only unbiased asymptotically, while rejection sampling is always unbiased even if you have MC size 1.

Here is the method. To make it work, I need to know the upper bound of $x$, it has to be a bounded variable. Say the upper bound is $c$. Each time I saw a realization $x$, then independent I generate a random number from uniform (0,1). If $u$ is smaller than $x/c$, accept, and report $1/x$. If $u$ is larger than $x/c$, do not report any estimate.

Then whenever I report the accepted $1/x$, it is an unbiased estimate of 1/E$[x]$.

Here is a demo code for $p(x)=$ Beta$(2,2)$:

n=1000000
x=rbeta(n,2,2)
u=runif(n,0,1)
Unbiased_estiamte=rep(NA, n)
Unbiased_estiamte[u<x]= (1/x)[u<x]