Unbiased estimator of exponential of measure of a set? Suppose we have a (measurable and suitably well-behaved) set $S\subseteq B\subset\mathbb R^n$, where $B$ is compact.  Moreover, suppose we can draw samples from the uniform distribution over $B$ wrt the Lebesgue measure $\lambda(\cdot)$ and that we know the measure $\lambda(B)$.  For example, perhaps $B$ is a box $[-c,c]^n$ containing $S$.
For fixed $\alpha\in\mathbb R$, is there a simple unbiased way to estimate $e^{-\alpha \lambda(S)}$ by uniformly sampling points in $B$ and checking if they are inside or outside of $S$?
As an example of something that doesn't quite work, suppose we sample $k$ points $p_1,\ldots,p_k\sim\textrm{Uniform}(B)$.  Then we can use the Monte Carlo estimate $$\lambda(S)\approx \hat\lambda:= \frac{\#\{p_i\in S\}}{k}\lambda(B).$$
But, while $\hat\lambda$ is an unbiased estimator of $\lambda(S)$, I don't think it's the case that $e^{-\alpha\hat\lambda}$ is an unbiased estimator of $e^{-\alpha\lambda(S)}$.  Is there some way to modify this algorithm?
 A: Suppose that you have the following resources available to you:


*

*You have access to an estimator $\hat{\lambda}$.

*$\hat{\lambda}$ is unbiased for $\lambda ( S )$.

*$\hat{\lambda}$ is almost surely bounded above by $C$.

*You know the constant $C$, and

*You can form independent realisations of $\hat{\lambda}$ as many times as you'd like.


Now, note that for any $u > 0$, the following holds (by the Taylor expansion of $\exp x$):
\begin{align}
e^{-\alpha \lambda ( S ) } &= e^{-\alpha C} \cdot e^{\alpha \left( C - \lambda ( S ) \right)} \\
&= e^{- \alpha C} \cdot \sum_{k \geqslant 0} \frac{ \left( \alpha \left[ C - \lambda ( S ) \right] \right)^k}{ k! } \\
&= e^{- \alpha C} \cdot e^u \cdot \sum_{k \geqslant 0} \frac{ e^{-u} \cdot \left( \alpha \left[ C - \lambda ( S ) \right] \right)^k}{ k! } \\
&= e^{u -\alpha C} \cdot \sum_{k \geqslant 0} \frac{ u^k e^{-u} }{ k! } \left(\frac{ \alpha \left[ C - \lambda ( S ) \right]}{u} \right)^k
\end{align}
Now, do the following:


*

*Sample $K \sim \text{Poisson} ( u )$.

*Form $\hat{\lambda}_1, \cdots, \hat{\lambda}_K$ as iid unbiased estimators of $\lambda(S)$.

*Return the estimator


$$\hat{\Lambda} = e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \cdot \prod_{i = 1}^K \left\{ C - \hat{\lambda}_i \right\}.$$
$\hat{\Lambda}$ is then a non-negative, unbiased estimator of $\lambda(S)$. This is because
\begin{align}
\mathbf{E} \left[ \hat{\Lambda} | K \right] &= e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \mathbf{E} \left[ \prod_{i = 1}^K \left\{ C - \hat{\lambda}_i \right\} | K \right] \\
&= e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \prod_{i = 1}^K  \mathbf{E} \left[ C - \hat{\lambda}_i \right] \\
&= e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \prod_{i = 1}^K   \left[ C - \lambda ( S ) \right] \\
&= e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \left[ C - \lambda ( S ) \right]^K
\end{align}
and thus
\begin{align}
\mathbf{E} \left[ \hat{\Lambda} \right] &= \mathbf{E}_K \left[ \mathbf{E} \left[ \hat{\Lambda} | K \right] \right] \\
&= \mathbf{E}_K \left[ e^{u -\alpha C} \cdot \left(\frac{ \alpha }{u} \right)^K \left[ C - \lambda ( S ) \right]^K \right] \\
&= e^{u -\alpha C} \cdot \sum_{k \geqslant 0} \mathbf{P} ( K = k ) \left(\frac{ \alpha }{u} \right)^K \left[ C - \lambda ( S ) \right]^K \\
&= e^{u -\alpha C} \cdot \sum_{k \geqslant 0} \frac{ u^k e^{-u} }{ k! } \left(\frac{ \alpha \left[ C - \lambda ( S ) \right]}{u} \right)^k \\
&= e^{-\alpha \lambda ( S ) }
\end{align}
by the earlier calculation.
A: The answer is in the negative.
A sufficient statistic for a uniform sample is the count $X$ of points observed to lie in $S.$  This count has a Binomial$(n,\lambda(S)/\lambda(B))$ distribution.  Write $p=\lambda(S)/\lambda(B)$ and $\alpha^\prime = \alpha\lambda(B).$ 
For a sample size of $n,$ let $t_n$ be any (unrandomized) estimator of $\exp(-\alpha \lambda(S)) = \exp(-(\alpha\lambda(B)) p) = \exp(-\alpha^\prime p).$  The expectation is
$$E[t_n(X)] = \sum_{x=0}^n \binom{n}{x}p^x (1-p)^{n-x}\, t_n(x),$$
which equals a polynomial of degree at most $n$ in $p.$  But if $\alpha^\prime p \ne 0,$ the exponential $\exp(-\alpha^\prime p)$ cannot be expressed as a polynomial in $p.$  (One proof: take $n+1$ derivatives.  The result for the expectation will be zero but the derivative of the exponential, which itself is an exponential in $p,$ cannot be zero.)
The demonstration for randomized estimators is nearly the same: upon taking expectations, we again obtain a polynomial in $p.$
Consequently, no unbiased estimator exists.
