convergence in distribution for $N(\theta, 1)$ sample statistic

Question

${X_1,...,X_n}$ is a random sample of size $n$ from a $N(\theta,1)$ distribution. Find the limiting distribution of $\sqrt{n} (|\bar{X}|−|\theta|)$ when $\theta$ is not equal to $0$. Does it hold if $\theta$ is zero?

So far I have by CLT that $\sqrt{n} (\bar{X}−\theta) \sim N(\theta,1)$.

So let $g(\theta) = \text{abs}(\theta)$ then if the MLE of $\theta = \bar{X}$, the MLE of $g(\theta)$ will be $g(\bar{X})$.

I want to use the delta method to show that this should have a limiting distribution of $N(0, [g'(\theta)]^2\,\sigma^2)$ but I am not sure how to treat the absolute value. Any advice would be greatly appreciated!

Answer 1

If $\theta \neq 0$, you can directly apply Delta method as $g$ is differentiable at $\theta$. The limiting distribution is the same as $\sqrt{n}(\bar{X} - \theta)$, i.e., $N(0, \sigma^2)$.

On the other hand, when $\theta = 0$, we cannot use Delta method since $g$ is not differentiable at $0$. But a direct calculation is expedient: For $x \geq 0$, \begin{align} & P[\sqrt{n}|\bar{X}| \leq x] = P[-x \leq \sqrt{n}\bar{X} \leq x] \\ = & P[\sqrt{n}\bar{X} \leq x] - P[\sqrt{n}\bar{X} < -x] \\ \to & \Phi(\sigma^{-1}x) - \Phi(-\sigma^{-1}x) = 2\Phi(\sigma^{-1}x) - 1, \tag{1} \end{align} where $\Phi(\cdot)$ denotes the distribution function of the standard normal random variable. In $(1)$, we used the result $\sqrt{n}\bar{X} \Rightarrow N(0, \sigma^2)$ and the definition of convergence in distribution.

If $x < 0$, then clearly $P[\sqrt{n}|\bar{X}| \leq x] = 0$. Therefore, $\sqrt{n}\bar{|X|}$ converges in distribution to $$F(x) = \begin{cases} 0 & x < 0, \\ 2\Phi(\sigma^{-1}x) - 1 & x \geq 0. \end{cases}$$

When $\sigma = 1$, the plot of $F$ looks like as follows: