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

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    $\begingroup$ 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 $\endgroup$
    – Henry
    Commented Nov 28, 2022 at 0:16
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    $\begingroup$ 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 $Y^2$? $\endgroup$
    – JimB
    Commented Nov 28, 2022 at 5:28
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    $\begingroup$ OK, but what is the definition of $\hat{r}$? $\endgroup$
    – JimB
    Commented Nov 28, 2022 at 16:23
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    $\begingroup$ Thanks. That should go into the question rather than in the comments as folks don't always read through the commens. $\endgroup$
    – JimB
    Commented Nov 28, 2022 at 16:53
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    $\begingroup$ I guess the question is: what is the asymptotic distribution of $\hat\mu^2$; $\mu^2$ has degenerate distribution at $\mu^2$. $\endgroup$
    – utobi
    Commented Nov 29, 2022 at 11:57

3 Answers 3

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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:

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

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  • $\begingroup$ Are you using that version of the delta method? If so, you should additionally justify the asymptotic normality. $\endgroup$
    – statmerkur
    Commented Nov 30, 2022 at 10:04
  • $\begingroup$ @statmerkur Feel free to modify the above answer. $\endgroup$
    – JimB
    Commented Nov 30, 2022 at 16:25
  • $\begingroup$ 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. $\endgroup$
    – statmerkur
    Commented Nov 30, 2022 at 16:56
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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.

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  • $\begingroup$ 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 $\sigma^2$? $\endgroup$
    – 0xcc
    Commented Nov 29, 2022 at 13:31
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    $\begingroup$ @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. $\endgroup$
    – statmerkur
    Commented Nov 29, 2022 at 15:58
  • $\begingroup$ 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? $\endgroup$
    – 0xcc
    Commented Nov 29, 2022 at 16:44
  • $\begingroup$ @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. $\endgroup$
    – statmerkur
    Commented Nov 29, 2022 at 22:03
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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).

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