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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!

Leo
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  • You need to normalize (in this case multiply by a scalar) $n^{1/2}(T_n-\theta)$ such that the variance converges to $1$. Have you got any ideas? – Dima McGreen Aug 15 '13 at 20:49
  • The word "asymptotic" should probably make it easier: You don't need the dependence on $\theta$ to disappear completely, except in the limit as $n\to\infty$. – Michael Hardy Aug 15 '13 at 22:36
  • It was probably intended that which function $g$ is, does not depend on $\theta$, since $\theta$ is unobservable. – Michael Hardy Aug 15 '13 at 22:37

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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

$$\sqrt{n}(T_n - \theta) \sim_a N(0; \sigma^2(\theta)) \approx N(0; f(T_n))$$

Alecos Papadopoulos
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