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Let $X_i$ be an i.i.d sequence of random variable with finite mean $\mu$ and variance $\sigma^2$.

Define $\bar{X} = n^{-1}\sum_{i=1}^n X_i$.

I'm asked to find a transformation $h(x)$ such that $h(\bar{X})$'s asymptotic variance is constant.

By the CLT we know $\bar{X} \to N(\mu,\sigma^2/n)$, which is equivelant to

$$\sqrt{n}(\bar{X} - \mu) \to N(0,\sigma^2)$$

My question is whether a non-degenerate transformation (that doesn't involve $n$ since $h(x)$ is stated a fixed function) exists for this to be the case?

Clearly if I simply take $h(x) = c$ for some constant $c$, $h(\bar{X})$ has asymptotic variance $0=$ constant. But what about a non-degenerate case?

I don't see how I can construct a non-degenerate function with asymptoticly constant variance since $\bar{X}$ has decreasing variance in $n$ and there so would (?) $h(\bar{X})$, unless I take a sequence of functions indexed by $n$.

Basically my question is, does a non-degenerate case exist such that for a fixed function $h(x)$, $h(\bar{X})$ has asymptotically constant variance?

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closed as unclear what you're asking by whuber Oct 3 '18 at 15:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Since an asymptotic variance, if it exists at all, is a number, please explain to us what you mean by saying it is "constant." $\endgroup$ – whuber Oct 3 '18 at 15:00
  • $\begingroup$ You were probably asking for variance stabilising transformations where the asymptotic variance is independent of parameter. $\endgroup$ – StubbornAtom May 11 at 19:19