# Unbiased sample from a product

Let $X = (x_1,\ldots,x_n)$ be an i.i.d sample from distribution $F%$ and let $y = \prod_{i=1}^n x_i$

Can we derive a randomized, unbiased. estimator $\hat{y}$ of $y$ that on average considers only a subsample of $X$?

Weak conditions may be imposed on $F$, for instance we may assume it has finite mean and variance, we may also assume that $\forall i, x_i >0$, but otherwise nothing is known about $F$.

• What precisely do you mean? E.g., since $y$ is a random variable, it can't have an estimator (in standard statistical terminology), only a predictor. Should this be some function of the x_i or ("randomized") a Markov kernel? And what does "on average" mean here? – Lutz Mattner Sep 16 '14 at 14:28
• I mean that $\hat{y}$ is a random variable such that $E(\hat{y}) = E(y)$, that there exists a sampling procedure for $\hat{y}$ that examines a random subsample of $X$, and that this subsample has on expectation less than $n$ elements. – Arthur B Sep 16 '14 at 14:39
• And the answer to my problem seems to be the generalized poisson estimator. – Arthur B Sep 16 '14 at 14:40

The solution is the Poisson estimator. Let $a_i$ = $\log x_i$. We would like to approximate $\exp (\sum_{i=1}^n a_i) = \exp (n \hat{a})$
Draw $\kappa \sim \textrm{Poisson}(\lambda)$, then draw with replacement $a_{i_1}, \ldots, a_{i_\kappa}$ and compute $y = e^{\lambda}\prod_{j=1}^{\kappa} n a_{i_j}$
$$E(y) = \sum_{k=0}^{\infty} \frac{1}{k!} E\left(\prod_{j=1}^{k} n a_{i_j}\right) = e^{n\hat{a}}$$