An application of the Delta method

Question

I have the sample $Y_{i}\overset{i.i.d}{\sim} N(0,\sigma_{y}^{2})$ of a population with normal distribution with mean $0$ and variance $\sigma_{y}^{2}$ unknown. I could also prove that $\frac{1}{n}\left(\frac{1}{n}\sum_{i=1}^{n}Y_{i}^{2}-\sigma_{y}^{2}\right)\overset{D}{\to}N(0,2\sigma^{4})$ using the CLT, but my question is:

Question: How can I find a function $f(t)$ such that the variance of the limiting distribution does not depend on the parameter of interest $\sigma_{y}^{2}$ in the sense of the Delta method: $$\sqrt{n}\left(f\left(\frac{1}{n}\sum_{i=1}^{n}Y_{n}\right)-f(\mathbb{E}X_{i})\right)\overset{D}{\to} N(0,\sigma^{2}(\mu)f'(\mu)^{2}) , \quad Y_{i}\overset{i.i.d}{\sim}N(\mu,\sigma^{2})$$

For example, I could see that $f$ might satisfy $2(f'(\sigma^{2}))^{2}\sigma^{4}$ shouldn't depend on $\sigma^{2}$.

Answer 1

I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean $$ \frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0. $$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, i.e. $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them.

This means the derivative is just $$ f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}. $$ But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.