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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  • $\begingroup$ 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