Xjyuk

Suppose we have a (measurable and suitably well-behaved) set $Ssubseteq Bsubsetmathbb R^n$, 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 $alphainmathbb R$, 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 $p_1,ldots,p_ksimtextrmUniform(B)$. Then we can use the Monte Carlo estimate $$lambda(S)approx hatlambda:= frac#p_iin Sklambda(B).$$
But, while $hatlambda$ is an unbiased estimator of $lambda(S)$, I don't think it's the case that $e^-alphahatlambda$ is an unbiased estimator of $e^-alphalambda(S)$. Is there some way to modify this algorithm?

Suppose that you have the following resources available to you:

1. You have access to an estimator $hatlambda$.


2. 
   $hatlambda$ is unbiased for $lambda ( S )$.


3. 
   $hatlambda$ is almost surely bounded above by $C$.


4. You know the constant $C$, and


5. You can form independent realisations of $hatlambda$ as many times as you'd like.

Now, note that for any $u > 0$, the following holds (by the Taylor expansion of $exp x$):

beginalign
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
endalign

Now, do the following:

1. Sample $K sim textPoisson ( u )$.


2. Form $hatlambda_1, cdots, hatlambda_K$ as iid unbiased estimators of $lambda(S)$.


3. Return the estimator

$$hatLambda = e^u -alpha C cdot left(frac alpha u right)^K cdot prod_i = 1^K left C - hatlambda_i right.$$

$hatLambda$ is then a non-negative, unbiased estimator of $lambda(S)$. This is because

beginalign
mathbfE left[ hatLambda | K right] &= e^u -alpha C cdot left(frac alpha u right)^K mathbfE left[ prod_i = 1^K left C - hatlambda_i right | K right] \
&= e^u -alpha C cdot left(frac alpha u right)^K prod_i = 1^K mathbfE left[ C - hatlambda_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
endalign

and thus

beginalign
mathbfE left[ hatLambda right] &= mathbfE_K left[ mathbfE left[ hatLambda | K right] right] \
&= mathbfE_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 mathbfP ( 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 )
endalign

by the earlier calculation.

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 $alpha^prime = alphalambda(B).$

For a sample size of $n,$ let $t_n$ be any (unrandomized) estimator of $exp(-alpha lambda(S)) = exp(-(alphalambda(B)) p) = exp(-alpha^prime p).$ The expectation is

$$E[t_n(X)] = sum_x=0^n binomnxp^x (1-p)^n-x, t_n(x),$$

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.