# Does a non-degenerate transformation exist so that $\bar{X}$'s asymptotic variance is constant? [closed]

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?

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

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