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

• Did you mean $r=Y^2$ or perhaps $r=\bar Y^2$? If $\mu$ is fixed in the definition of the distribution of $Y$ than $\mu^2$ is also fixed Commented Nov 28, 2022 at 0:16
• If $\hat{r}=\frac{1}{n}\sum_{i=1}^n Y^2$, should you have $r=\mu^2 (\mu+1)$ as $r$ is the mean of $Y^2$?
– JimB
Commented Nov 28, 2022 at 5:28
• OK, but what is the definition of $\hat{r}$?
– JimB
Commented Nov 28, 2022 at 16:23
• Thanks. That should go into the question rather than in the comments as folks don't always read through the commens.
– JimB
Commented Nov 28, 2022 at 16:53
• I guess the question is: what is the asymptotic distribution of $\hat\mu^2$; $\mu^2$ has degenerate distribution at $\mu^2$. Commented Nov 29, 2022 at 11:57

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:

$$\sum_{i=1}^n \left(\frac{\partial f}{\partial y_i}\bigg|_{y_i=\mu}\right)^2 V(y)=\sum_{i=1}^n \frac{4 \left(\sum _{j=1}^n y_j\right)^2}{n^3}\bigg|_{y_i=\mu} \mu^3=\sum_{i=1}^n \frac{4(n\mu)^2}{n^3} \mu^3=4\mu^5=4r^{5/2}$$

• Are you using that version of the delta method? If so, you should additionally justify the asymptotic normality. Commented Nov 30, 2022 at 10:04
• @statmerkur Feel free to modify the above answer.
– JimB
Commented Nov 30, 2022 at 16:25
• After the modifications I have in mind there would be no difference to my answer since the OP has already established asymptotic normality of $\sqrt{n}(\hat{\mu} - \mu)$, where $\hat\mu$ is the arithmetic mean of the $y_i$s. Commented Nov 30, 2022 at 16:56

Consider $$\mathop{g}\left(x\right)\mathrel{:=x^2}$$ with derivative $$\mathop{g'}\left(x\right)=2x$$. What does the Wikipedia article on the delta method tell you about the asymptotic distribution of $$\sqrt{n}\left[\mathop{g}\left(\hat\mu\right)-\mathop{g}\left(\mu\right)\right]$$?1

1 You can ignore the assumption $$g'(x) \neq 0$$ if you allow for singular normal limits.

• Thank you for the answer. Sorry but I do not understand. I have then $\sqrt{n}\left[\mathop{g}\left(\hat\mu\right)-\mathop{g}\left(\mu\right)\right] \rightarrow N(0, \sigma^2 \cdot [g'(\mu)]^2)$ I do not follow how this will end up being $\sqrt{n}(\hat{r} - r) \sim N(0, 4r^{5/2})$. Since $[g'(\mu)]^2 = 4\mu^2 = 4r$, how should I get $\sigma^2$?
– 0xcc
Commented Nov 29, 2022 at 13:31
• @0xcc From the Wikipedia article you can see that $\sigma^2$ is the variance of the asymptotic normal distribution of $\sqrt{n}\left[\hat\mu-\mu\right]$. With that you will get the desired result. Commented Nov 29, 2022 at 15:58
• Ok thank you, I think i see the connection now, since $V[X] = \mu^3$ this leads to $\sigma^2 = \mu^3$ which is $r \cdot \sqrt{r}$. Is this the correct interpretation?
– 0xcc
Commented Nov 29, 2022 at 16:44
• @0xcc almost. $\mu^3$ is the asymptotic variance of $\sqrt{n}\left[\hat\mu-\mu\right]$ which corresponds to $\sigma^2$ being the asymptotic variance of $\sqrt{n}\left[X_n-\theta\right]$ in the Wikipedia article. Commented Nov 29, 2022 at 22:03

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).