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 $\hat{r} = (\hat{\mu})^2$

This time using the Delta method:
$$E(y)=\mu$$
$$V(y)=\mu^3$$
$$r=\mu^2$$
$$\hat{r}=\left(\sum_{i=1}^n y_i/n\right)^2$$
Now the function of interest is
$$f=\sqrt{n} (\hat{r}-r)$$
The asymptotic variance using the Delta method is found as follows (under the assumption that the $y_i$'s are all independent:
$$\sum_{i=1}^n \left(\frac{\partial f}{\partial y_i}\bigg|_{y_i=\mu}\right)^2 V(y)=\sum_{i=1}^n \frac{4 \left(\sum _{j=1}^n y_j\right)^2}{n^3}\bigg|_{y_i=\mu} \mu^3=\sum_{i=1}^n \frac{4(n\mu)^2}{n^3} \mu^3=4\mu^5=4r^{5/2}$$
