1
$\begingroup$

I would like to estimate the function value of the sigmoid over an expectation, that is: \begin{equation} \sigma(\mathbb{E}_{p(x)}[f(x)]), \end{equation} where $\sigma(x) = \frac{1}{1 + e^{-x}}$, and $p(x)$ is the one we only access its samples but cannot evaluate its density.

To estimate $\sigma(\mathbb{E}_{p(x)}[f(x)])$, we could use $\sigma(\frac{1}{n}\sum_{i=1}^n f(x_i))$, but it is biased. My question is, how could we define an unbiased estimator.

$\endgroup$
3
  • $\begingroup$ You need to postulate a specific family of distributions governing the random samples $x_i.$ What would that be? $\endgroup$
    – whuber
    Commented Apr 28, 2022 at 12:31
  • $\begingroup$ @whube Let's discuss the case when p(x) is Bernoulli or categorical distribution $\endgroup$
    – jzin
    Commented Apr 29, 2022 at 3:51
  • $\begingroup$ The solution to that is well known and instructive. $\endgroup$
    – whuber
    Commented Apr 29, 2022 at 11:39

1 Answer 1

1
$\begingroup$

No unbiased estimator exists, when $p(x)$ is Categorical distribution:

Unbiased estimator of exponential of measure of a set?

For the binomial distribution, why does no unbiased estimator exist for $1/p$?

Thank @whuber for the given link.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.