Positive estimator

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 $(X_1, \ldots, X_N)$.

• I would like to seek some clarification on this problem. First of all, the use of unbiased estimator is a little unconventional here. Usually, one thinks of (and defines!) an unbiased estimator as a sequence of functions for each fixed $n$ 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. – cardinal Jun 24 '12 at 15:07
• So, could you please specify the (additional) constraints you seek to place on the problem? (Note also, it is interesting that the distribution of $N$ has both its mode and median at 1 and a mean value of $e$, so we don't normally use very many elements in the sequence.) – cardinal Jun 24 '12 at 15:08
• to put it another way: I know how to simulate random samples from the random variable $X$ that has $\mu$ as mean (it takes a finite amount of computer time for each draw) and I would like to generate a random sample from a random variable Y that has $e^\mu$ as a mean. The cost of generating the random variable $Y$ must take a finite amount of computer time (on average). – Alekk Jun 24 '12 at 22:37
• It has recently been proved that such an algorithm cannot be constructed. – Alekk Jul 30 '13 at 5:17
• @Alekk could you share with us the proof or reference to it? – Memming Jul 17 '14 at 17:26