Suppose that one knows how to generate (independent) random samples $X_1, X_2, \ldots$ distributed as the random varable $X$ with $\mathbb{E}[X]=\mu \in \mathbb{R}$. It is then easy to construct an unbiased estimator of the quantity $e^{\mu}$. Indeed, it suffices to generate an integer random variables $N$ such that $\mathbb{P}[N \geq k] = \frac{1}{k!}$. The random variable $$Y=1 + X_1 + X_1X_2 + \ldots + \prod_{k \leq N-1} X_k + \prod_{k \leq N} X_k$$ satisfies $\mathbb{E}[Y]=e^{\mu}$.

**Question:** is it possible to construct an unbiased estimator of $e^{\mu}$ that is positive with probability $1$. If the random variable $X$ is lower bounded by a constant $C$ one can Taylor expand $\exp(C) \cdot \exp(x-C)$ and use the same idea as above. What about the case where $X$ is not lower bounded? One could try to Taylor expand $\exp(m) \cdot \exp(x-m)$ where $m = \min(X_1, \ldots, X_N)$ but in this case the estimator does not satisfy $\mathbb{E}[Y]=e^{\mu}$ anymore due to the dependence between the random variable $m$ and $(X_1, \ldots, X_N)$.

unbiased estimatoris a little unconventional here. Usually, one thinks of (and defines!) an unbiased estimator as asequenceof functions for eachfixed$n$ such that $\mathbb E Y_n = e^\mu$. Here, instead we consider only a single random $N$ for the entire sequence. It is then not much of a leap to ask if we can consider using the entire sequence to construct the estimator, but then your problem (quite trivially) has an affirmative answer. $\endgroup$ – cardinal Jun 24 '12 at 15:07