Deriving the asymptotic distribution using delta method I have the density function:
$P_Y(y) = \sqrt{\frac{1}{2\pi y^3}} \exp\left(-\frac{(y-\mu)^2}{2\mu^2y}\right)$
If we define $r := \mu^2$ what is its asymptotic distribution?
The right answer is $\sqrt{n}(\hat{r} - r) \sim N(0, 4r^{5/2})$ but I don't understand how i should get there.
I know that $E[Y] = \mu$ and $V[Y] = \mu^3$
And for $P_Y$ I know that the ML-estimation for $\mu$ is just $\hat{\mu} = \bar{y}$ and the asymptotic distribution for that ML-estimation is $\sqrt{n}(\hat{\mu} - \mu) \sim N(0, \mu^3)$
In this case I've calculated the Fisher information to be: $I(\theta_n) = -E[\ell_n''(\mu)] = \frac{n}{\mu^3}$ and this seems to be right.
I've used the formula $\sqrt{n}(\hat{\theta} - \theta) \sim N(0, \frac{1}{I(\theta)})$, where $I(\theta)$ is the Fisher information. But this does not yield the right answer, at least not how I try to use it when we define $r$ to be $r := \mu^2$
But I am now stuck. I think I've written down the relevant information.
EDIT 1
The ML-estimation for $r$ is $\hat{r} = (\hat{\mu})^2$

If from the comments that knowing the variance of $\hat{\mu}^2$ is desired, then that can be determined exactly for any sample size.
The mean of $\hat{r}=\hat{\mu}^2$ is $\mu^2 (1+\mu/n)$ and the variance is
$$V(\hat{r})=\frac{\mu ^5 \left(15 \mu ^2+4 n^2+14 \mu  n\right)}{n^3}$$
The makes the variance of $\sqrt{n}(\hat{r}-r)$ equal to $n V(\hat{r}-r)=n V(\hat{r})=4 \mu ^5+\frac{15 \mu ^7}{n^2}+\frac{14 \mu ^6}{n}$.
One can see that the variance of $\sqrt{n}(\hat{r}-r)$ approaches $4\mu^5=4r^{5/2}$ as $n$ approaches $\infty$ (as you noted in your question).
