asymptotic normality of z-estimator I'm working on Problem 5.4.1 in Bickel and Docksum's Mathematical Statistics
Let $X_1, \dots, X_n$ be i.i.d. random variables distributed according to $P\in\mathcal{P}$. Suppose $\psi:\mathbb{R}\to\mathbb{R}$:
(i) is monotone nondecreasing
(ii)$\psi(-\infty)<0<\psi(\infty)$
(iii)$|\psi|(x)\le M<\infty$ for all $x$
And suppose for all $P\in\mathcal{P}$, $\theta(P)$ is the unique solution of $\mathbb{E}_P\psi(X_1-\theta) = 0$. Let $\hat{\theta}_n = \theta(\hat{P})$, where $\hat{P}$ is the empirical distribution of $X_1,\dots,X_n$. Now set $\lambda(\theta) = \mathbb{E}_P\psi(X_1-\theta)$ and $\tau^2(\theta) = \text{var}\psi(X_1-\theta)$. Assume $\lambda'(\theta)<0$ exists and that
$$\frac{1}{\sqrt{n}\tau(\theta)}\sum_{i=1}^n\left[\psi(X_i-\theta_n)-\lambda(\theta_n)\right] \xrightarrow{\mathcal{L}} N(0,1)$$
for every sequence $\{\theta_n\}$ with $\theta_n = \theta = t/\sqrt{n}$ for $t\in\mathbb{R}$.\
$\textbf{The problem is to show that}$
$$\sqrt{n}(\hat{\theta}_n-\theta) \xrightarrow{\mathcal{L}} N(0,\frac{\tau^2(\theta)}{[\lambda'(\theta)]^2})$$
Hint: $P(\sqrt{n}(\hat{\theta}_n-\theta)<t) = P(\hat{\theta}_n<\theta_n) = P(-\sum_{i=1}^n\psi(X_i-\theta_n)<0)$.
$\textbf{My question}$ is that I don't understand the last equality in the hint, and also how this could lead to the conclusion. Any help will be appreciated! (I manage to prove the consistency of this z-estimator $\hat{\theta}_n$, but not sure whether this could help this problem or not)
 A: Here's how to understand the last equality. The $z$-estimator $\hat\theta_n$ by definition solves the equation $$\mathbb{E}_{\hat P}\psi(X-\hat\theta_n)=0.\tag1$$ The empirical distribution $\hat P$ assigns mass $\frac1n$ to each of $X_1,\ldots,X_n$, so the above translates to
$$\frac1n\sum_{i=1}^n \psi(X_i-\hat\theta_n)=0.\tag2$$
The mapping $t\mapsto \sum_{i=1}^n\psi(X_i-t)$ is decreasing, since the function $\psi$ is increasing. Therefore the statement $$\hat\theta_n<\theta_n\tag3$$ is equivalent to the statement $$\sum \psi(X_i-\hat\theta_n)>\sum \psi(X_i-\theta_n)\tag4$$ (all sums run from $i=1$ to $n$). By (2), the LHS of (4) equals zero, so (4) is equivalent to
$$\sum \psi(X_i-\theta_n)<0.\tag5$$ (The negative sign in your version is a typo?) To compute the probability of (5), the natural manipulation is to write the algebraically equivalent statement
$$\frac1{\sqrt n\tau(\theta)}\sum\left[\psi(X_i-\theta_n)-\lambda(\theta_n)\right]<-\sqrt n\frac{\lambda(\theta_n)}{\tau(\theta)}\tag6
$$
since by assumption, the LHS of (5) converges in distribution to a standard normal as $n\to\infty$. What about the RHS? Recall that $\theta_n:=\theta + \frac t{\sqrt n}$ where $\theta$ solves $\lambda(\theta)=0$. Hence we can rearrange the RHS of (6) to the form
$$-\sqrt n\frac{\lambda(\theta_n)}{\tau(\theta)}=-\frac t{\tau(\theta)}\left(\frac{\lambda\left(\theta+\frac t{\sqrt n}\right)-\lambda(\theta)}{\frac t{\sqrt n}}\right)\tag7
$$
which by the definition of derivative converges to $\displaystyle-\frac t{\tau(\theta)}\lambda'(\theta)$ as $n\to\infty$.
