Finding limiting distribution with delta method

Question

Let $X_1, X_2, ..., X_n$ be iid $N \sim (\theta, \sigma^2)$, Let $\delta_{n} = \bar{X}^2 - \frac{1}{n(n-1)}S^2$, where $S^2 = \sum (X_i - \bar{X})^2$ . Find the limiting distribution of

(i) $\sqrt{n} (\delta_{n} - \theta^2)$ for $\theta \neq 0$

(ii) $n(\delta_{n} - \theta^2)$ for $\theta = 0$

(i) I know that $\sqrt{n} (\bar{X^2} - \theta^2) \overset{L}{\to} N(0, 4\sigma^2 \theta^2)$ using delta method. For $\sqrt{n} (\bar{X}^2 - \frac{1}{n(n-1)}S^2 -\theta^2)$, i can write it as $\sqrt{n} (\bar{X}^2 -\theta^2) - \sqrt{n}(\frac{1}{n(n-1)}S^2)$, what's the limiting distribution for $\sqrt{n}(\frac{1}{n(n-1)}S^2)$?

(ii) I know that $n (\bar{X^2} ) \overset{L}{\to} \sigma^2 \chi_{1}^2$ , but I don't know how to find $n (\bar{X^2} - \frac{1}{n(n-1)}S^2)$

Answer 1

The following relies only on the fact that $X_1,\ldots,X_n$ are i.i.d with finite mean $\theta$, finite variance $\sigma^2(>0)$ and $E|\delta_n|<\infty$ for every $n\ge 1$.

Your $\delta_n$ is a U-statistic, since it can be written in the form

$$\delta_n=\frac1{\binom{n}{2}}\sum_{1\le \alpha<\beta\le n}X_\alpha X_\beta$$

Asymptotic distributions of U-statistics in general can be derived using projections.

If $\theta\ne 0$, the general result gives

$$\sqrt n(\delta_n-\theta^2)\stackrel{d}\longrightarrow N(0,4\theta^2\sigma^2) \tag{1}$$

And if $\theta=0$,

$$n(\delta_n-\theta^2)\stackrel{d}\longrightarrow \sigma^2(\chi^2_1-1) \tag{2}$$

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To prove $(1)$ using projection theorem, let $T_n=\sqrt n(\delta_n-\theta^2)$, so that $E(T_n)=0$.

Define $K_P(X_i)=E(T_n\mid X_i)$, which simplifies to

$$K_P(X_i)=\frac{\sqrt n}{\binom{n}{2}}(n-1)(\phi_1(X_i)-\theta^2)=\frac2{\sqrt n}(\phi_1(X_i)-\theta^2)\,,$$

with $\phi_1(X_1)=E[X_1X_2\mid X_1]=\theta X_1$.

Note that $E(\phi_1)=\theta^2$ and $\operatorname{Var}(\phi_1)=\theta^2\sigma^2=\sigma_1^2$ (say).

So by classical CLT, $$V_P=\sum_{i=1}^n K_P(X_i) \stackrel{d}\longrightarrow N(0,4\sigma_1^2)$$

It can be shown that $\operatorname{Var}(\delta_n)=\frac{4\sigma_1^2}{n}+O\left(\frac1{n^2}\right)$. And clearly, $\operatorname{Var}(V_P)=4\sigma_1^2$.

Therefore, as $n\to \infty$, $$\operatorname{Var}(T_n)-\operatorname{Var}(V_P)=O\left(\frac1n\right)\longrightarrow 0$$

Projection theorem (which is another application of Slutsky's theorem) then says that limiting distributions of $T_n$ and $V_P$ are identical.

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Edit:

I might have complicated things before. Ignore the proof above as it is missing some details.

For $\theta \ne 0$, simply write

$$\sqrt n(\delta_n-\theta^2)=\sqrt n\left(\overline X^2 -\theta^2\right)-\frac1{\sqrt n}\left(\frac{S^2}{n-1}\right)$$

Using delta method, OP can show that

$$\sqrt n\left(\overline X^2 -\theta^2\right)\stackrel{d}\longrightarrow N(0,4\theta^2\sigma^2)$$

And we know that

$$\frac{S^2}{n-1} \stackrel{P}\longrightarrow \sigma^2\,,$$

which implies $$\frac1{\sqrt n}\left(\frac{S^2}{n-1}\right)\stackrel{P}\longrightarrow 0$$

Hence $(1)$ follows directly from Slutsky's theorem.

Now when $\theta=0$,

$$n(\delta_n-\theta^2)=n\overline X^2 - \frac{S^2}{n-1}$$

We have from CLT and continuous mapping theorem that

$$n\overline X^2\stackrel{d}\longrightarrow \sigma^2\chi^2_1$$

And as before, $$\frac{S^2}{n-1} \stackrel{P}\longrightarrow \sigma^2$$

Therefore, Slutsky's theorem again proves $(2)$.