# Variance of MLE of a function of bernoulli parameter

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.

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