On finding the asymptotic distribution of the sample variance using the delta method

This is an exercise I am stuck on.

Given an IID sample $X_1, \dots, X_n \sim N(\mu, \sigma^2)$ with $\mu \ne 0$ let

$$S_n^2 = \frac{1}{n} \sum_{i = 1}^n ( X_i - \overline{X_n})^2$$

be the sample variance and $\overline{X_n}$ the sample mean. I am tasked in finding the asymptotic distribution of $S_n^2$ using the second order delta method. This says that given a continuous and doubly differentiable function $\phi$ with $\phi'(\theta) = 0$ and an estimator $T_n$ of a parameter $\theta$ such that $$\sqrt{n} (T_n - \theta) \rightarrow_d T$$ where $T$ is some probability density function then $$\sqrt{n} (\phi (T_n) - \phi(\theta)) \rightarrow_d \frac{1}{2} \phi''(\theta) T^2$$.

In my exercise I think I have to utilize this method taking as $T_n$ the sample mean, the problem I run into is that I don't know how to define the function $\phi$. I know that

$$\sum_{i = 1}^n ( X_i - \overline{X_n})^2 = \sum_{i = 1}^n X_i^2-n \overline{X}_n$$ but this does not help me in defining $\phi$. Moreover as a hint I am given to work with $$\frac{S^2_n}{\overline{X_n}}$$ and I can't see why this would help. Any ideas?

EDIT: My theorem was wrong as pointed out by Chaconne in his answer, I correct it for the sake of clarity. I am still searching for the appropriate $\phi$ to utilize to apply the delta method to my exercise.

The second order delta method that you have is incorrect. To see why, take $X_1, \dots \stackrel {\text{iid}}\sim \mathcal N(0,1)$ so that $$\sqrt n \bar X_n \to_d \mathcal N(0, 1)$$ and consider $g(z)=z^2$ so $g''(z) = 2$. By the statement as you have it, we should find $$\sqrt ng(\bar X_n) \stackrel ? \to_d \frac 12 \left(2\right) Y$$ where $Y \sim \mathcal N(0, 1)$. But clearly this isn't true as $\sqrt n g(\bar X_n) \geq 0$ always.

The full theorem that you need is rather unsightly at first glance but tells us exactly what to do:

(source: Jun Shao's Mathematical Statistics)

Let's apply this to our example. We have $k=1$ and $\frac{\partial^2 g}{\partial x^2}(0) \neq 0$ (so we'll have $m=2$) so this simplifies dramatically. In particular, we have $$(\sqrt n)^2g(\bar X_n)^2 \to_d \frac 1{2!} \sum_{i_1=1}^1 \sum_{i_2=1}^1 \frac{\partial^2 g}{\partial x_{i_1}\partial x_{i_2}}\big\vert_{x = 0} Y_{i_1} Y_{i_2}.$$ Now since $Y$ has only one component (i.e. it's a scalar) we have $Y = Y_{i_1} = Y_{i_2}$ so this reduces to $$ng(\bar X_n)^2 \to_d \frac 12 2 Y^2 =_d Y^2.$$ Let's check this by using our knowledge of the exact distribution of $\bar X_n$.

We know that $\bar X_n \sim \mathcal N(0, \frac 1n)$ therefore in actuality $$\left(\sqrt n \bar X_n \right) \sim \mathcal N(0,1) \implies \left(\sqrt n \bar X_n \right)^2 \sim \chi^2_1.$$

Via the delta method we just showed that $$a_n^2 (\bar X_n)^2 = n \bar X_n^2 \to_d \left[\mathcal N(0,1)\right]^2 =_d \chi^2_1.$$ so we have indeed gotten the correct answer.

Does this give you any directions to try with your particular problem? I'll add more steps in that direction if this doesn't help much.

• First off thank you greatly! I will read carefully through the answer and see if I manage to solve my problem. – Monolite Feb 12 '18 at 18:23
• @Monolite you're very welcome, I'll be happy to update if you have further questions – jld Feb 12 '18 at 18:42
• I wrote down the theorem wrong in my notes (I called a friend so I know), I am still not fully sure what $g$ I should use in my case though... – Monolite Feb 12 '18 at 18:47
• Could you tell me what $g$ should be in my case? – Monolite Feb 12 '18 at 19:39
• @Monolite sorry I won't be able to respond until this evening, hopefully someone else will jump in – jld Feb 12 '18 at 20:48