Variance stabilizing transformation I ran into this statistics question:
In many cases, an estimator $T_n$ for $\theta$ is asymptotically normal distributed, but its asymptotic variance depends on $\theta$:
$\sqrt{n}(T_n - \theta) \to N(0; \sigma^2(\theta))$ in distribution.
It can be of interest to find a transformation $g$ that stabilizes the variance, i.e. that is such that the asymptotic variance of the transformed estimator $g(T_n)$ no longer depends on $\theta$. Discuss how one can find such a transformation.
Frankly, I have no clue... I hope you guys can help! Thanks!
 A: There does not seem to be a need to find another estimator.
By the formulation of the problem, we implicitly assume that $T_n$ is a consistent estimator.
"Its asymptotic variance depends on $\theta$" means that $\text{Avar}$ is a function of $\theta$, so let's write it this way: $\sigma^2(\theta) = f(\theta)$. I assume that the functional form of $f()$ is known or can be derived.
Since $T_n$ is a consistent estimator, we have $\text{plim}T_n = \theta $. Under the usual regularity conditions we have then that for some function $h()$ of $T_n$, $\text{plim}h(T_n) = h(\text{plim}T_n) = h(\theta)$.
Then take the asymptotic variance function, which has a known form, but insert as argument $T_n$ and not $\theta$: $f(T_n)$. We will have
$$ \text{plim}f(T_n) =  f(\text{plim}T_n) = f(\theta) = \sigma^2(\theta) $$
In other words $f(T_n)$ is a consistent estimator of the asymptotic variance. When we face unknown quantities, we replace them by estimates as best as we can. Since we have a consistent estimator of the unknown quantity we use the former instead of the latter, and we perform approximate inference (which was approximate to begin with) by using
$$\sqrt{n}(T_n - \theta) \sim_a N(0; \sigma^2(\theta)) \approx N(0; f(T_n))$$
