Deriving the limiting distribution of the Hodges-Le Cam estimator in Bickel and Doksum (2015) I am trying to better understand the Hodges-Le Cam estimator, and am having difficulty rendering explicit some of the asymptotic arguments in the derivation of the estimator's limiting distribution. I would appreciate if someone could make certain aspects of the argument explicit by addressing the queries further below.
Context.
The following presentation is extracted from Mathematical Statistics: Basic Ideas and Selected Topics Volume I (2nd ed.) by Bickel and Doksum (2015).

Example 5.4.2. Hodges' Example. Let $X_1, \dots X_n$ be i.i.d. $\mathcal{N}(\theta, 1)$. Then $\overline{X}$ is the MLE of $\theta$ and it is trivial to calculate $I(\theta) \equiv 1$. Consider the following competitor to $\overline{X}$:
\begin{align}
\tilde{\theta}_n = 
\begin{cases}
0 \quad \text{if} \quad \vert \overline {X}\vert \leq n^{-1/4} \\
\overline{X} \quad \text{if} \quad \vert \overline {X} \vert > n^{-1/4}.
\end{cases}
\tag{5.4.37}
\end{align}
[...]
We next compute the limiting distribution of $\sqrt{n}(\tilde{\theta} - \theta)$. Let $Z \sim \mathcal{N}(0, 1)$. Then
\begin{align}
P_{\theta} [\vert \overline{X} \vert \leq n^{-1/4}] &= P_{\theta}[ \vert Z + \sqrt{n} \theta \vert \leq n^{1/4}] \\
&= \Phi(n^{1/4} - \sqrt{n} \theta) - \Phi(-n^{1/4} - \sqrt{n} \theta).
\tag{5.4.38}
\end{align}
Therefore if $\theta \neq 0$, $P_{\theta}[\vert \overline{X} \vert \leq n^{-1/4}] \longrightarrow 0$ because $n^{1/4} - \sqrt{n} \theta \longrightarrow -\infty$, and, thus, $P_{\theta}[\tilde{\theta}_n = \overline{X}] \longrightarrow 1$. If $\theta = 0$, $P_{\theta}[\vert \overline{X} \vert \leq n^{1/4}] \longrightarrow 1$, and $P_{\theta}[\tilde{\theta}_n = 0] \longrightarrow 1$. Therefore,
\begin{align}
\mathcal{L}_{\theta}(\sqrt{n}(\tilde{\theta}_n - \theta)) \longrightarrow \mathcal{N}(0, \sigma^2(\theta)) \\ \tag{5.4.39}
\end{align}
where $\sigma^2(\theta) = 1 = \frac{1}{I(\theta)}$, $\theta \neq 0, \sigma^2(0) = 0 < \frac{1}{I(\theta)}$.

Queries.
1. Where it stated for the case $\theta \neq 0$, $n^{1/4} - \sqrt{n} \theta \longrightarrow - \infty$, wouldn't this only hold if $\theta > 0$? If that is the case, I am unable to understand why the author has omitted dealing with the case where $\theta < 0$?
2. What results are being invoked when we go from both $P_{\theta}[\tilde{\theta} = \overline{X}] \longrightarrow 1$ for $\theta \neq 0$ and $P_{\theta}[\tilde{\theta}_n = 0] \longrightarrow 1$ for $\theta = 0$ to the two limiting distributions in $(5.4.39)$, that is, a standard Normal and a degenerate distribution?
Intuitively, I am partly comfortable with the ideas that:

*

*As we collect more observations $n \rightarrow \infty$, then in the regime where $\theta \neq 0$, the Hodges estimator $\tilde{\theta}_n$ behaves likes the MLE with increasingly high probability because $P_{\theta}[\tilde{\theta}_n = \overline{X}] \rightarrow 1$. And because of the asymptotic normality of MLE, this means that the limiting distribution of $\sqrt{n}(\tilde{\theta}_n - \theta)$ is standard normal.


*As we collect more observations $n \rightarrow \infty$, then in the regime where $\theta = 0$, the Hodges estimator $\tilde{\theta}_n$ has increasingly high probability of being set to exactly 0 because $P_{\theta}[\tilde{\theta} = 0] \rightarrow 1$. And this means that the limiting distribution of $\sqrt{n} \tilde{\theta}_n$ is that of a degenerate CDF at 0.
However, I am having difficulty seeing what results are being used here.
3. Is $\tilde{\theta}_n$ a discrete or continuous random variable; or some hybrid that depends on whether $\theta = 0$ or $\theta \neq 0$?
I am experiencing doubt over this because I am not entirely comfortable with how much sense the statements $P_{\theta}[\tilde{\theta}_n = \overline{X}]$ and $P_{\theta}[\tilde{\theta}_n = 0]$ make. Because if $\tilde{\theta}_n$ is continuous, both probabilities on $\tilde{\theta}_n$ taking a value would be 0?
I tried rewriting the Hodges estimator using indicator functions $\tilde{\theta}_n = \overline{X}  \cdot \mathbb{I}(\vert \overline{X} \vert > n^{-1/4})$, which suggests to me that the intention is that $\tilde{\theta}_n$ is continuous when $\theta \neq 0$, but I'm not sure about whether $\tilde{\theta}_n$ is discrete or continuous when $\theta = 0$.
 A: *

*Yes, that's for $\theta>0$, but the argument for $\theta<0$ is exactly symmetric.


*The key random variable is the indicator $A_n=\{|\bar X_n|\leq n^{-1/4}\}$. Since $\bar X_n\sim N(0,1/n)$, we know $P(A_n)$ exactly. Asymptotically, if $\theta=0$, $P(A_n)\to 1$. If $\theta\neq 0$, $P(A_n)\to 0$.
Now,
$$\tilde\theta_n= A_n\times 0 + (1-A_n)\times \bar X_n$$
and so
$$\sqrt{n}(\tilde\theta-\theta)=A_n\times 0+(1-A_n)\sqrt{n}(\bar X_n-\theta)$$
and by the Continuous Mapping Theorem you get
$$\sqrt{n}(\tilde\theta-\theta)\stackrel{p}{\to}0$$
if $\theta=0$ and
$$\sqrt{n}(\tilde\theta-\theta)\stackrel{p}{\to}N(0,1)$$
if $\theta=1$.
So, for any fixed $\theta$, the asymptotic distribution is either identical to that of $\bar X_n$ (if $\theta\neq 0$) or better (if $\theta=0$).


*$\tilde\theta$ has a mixed discrete and continuous distribution: it has a point mass at zero, it has zero mass on $[-n^{-1/4},0)$ and $(0,n^{-1/4}]$, and otherwise it has a strictly positive density.

*

*$P(\tilde\theta=0)$ is well-defined and non-zero; it's just $P(A_n)$. It is non-zero because $\tilde\theta$ has point mass at zero


*$P(\tilde\theta=\bar X_n)$ is well-defined and non-zero; it's just $1-P(A_n)$. It is non-zero even though $\bar X_n$ is continuous, but that's fine because $\tilde\theta-\bar X_n$ has point mass at zero.




*While you didn't ask, the other interesting part about the Hodges estimator is how it breaks down. If you take $\theta=n^{-1/4}$, then $P(A_n)\approx 1/2$. When $A_n=1$, $\sqrt{n}(\tilde\theta_n-\theta)\approx n^{1/2}n^{-1/4}=n^{1/4}$, which is very large for large $n$.
