Variance of MLE of a function of bernoulli parameter

Question

Given $m$ i.i.d. Bernoulli( $\theta$ ) r.v.s $X_{1}, X_{2}, \ldots, X_{m},$ I'm interested in finding the variance (Not asymptotic) of estimator of $(1-\theta)^{1 / k},$ when $k$ is a positive integer.

Optimizing the likelihood function implies $\hat{\theta}=\sum_{i=1}^{m} \frac{X_{i}}{m} .$ Thus, using the invariance property, the required MLE estimator is $\left(1-\sum_{i=1}^{m} \frac{X_{i}}{m}\right)^{1 / k}$.

I am wondering is there a way of finding the variance of this estimator?

I know that asymptotically MLE is efficient and achieves CRLB. Thus, asymptotic variance is easy to calculate using Fisher information.

I am interested in finding the variance in the non-asymptotic case? Is it even possible? Any tips are appreciated.

Answer 1

First note that $S = m\hat{\theta} \sim Binomial(m,\theta)$. We are seeking to find:

$$Var(\hat{\theta}) = E[ (1-\hat{\theta})^{2/k}] - E[ (1-\hat{\theta})^{1/k}]^2.$$

Using the fact that $\hat{\theta} = S/m$ and the pmf of $S$ is $f_S(s) = {m \choose s} \theta ^s (1-\theta)^{m-s},$ we can write:

$$Var(\hat{\theta}) = \sum_{j=0}^m(1-j/m)^{2/k}f_S(j)- [\sum_{j=0}^m(1-j/m)^{1/k}f_S(j)]^2$$ $$Var(\hat{\theta}) = \sum_{j=0}^m(1-j/m)^{2/k}{m \choose j} \theta ^j (1-\theta)^{m-j} - [\sum_{j=0}^m(1-j/m)^{1/k}{m \choose j} \theta ^j (1-\theta)^{m-j}]^2.$$