Strong consistency in quantum estimation problem

Question

I'm reading the paper: Strong consistency and asymptotic efficiency for adaptive quantum estimation problems by Akio Fujiwara.

In this paper, describes the next adaptive scheme of estimation:

"Consider a sequential design problem that allows us at each stage an experiment $E$ to be taken from an experiment space $\mathcal{E}$. The observed data $x_n \in \mathcal{X}$ at time $n$ have probability density $f(x_n ; \theta, E_n )$, with respect to some $\sigma$ -finite measure $\mu$ on $\mathcal{X}$, which depends on both the parameter $\theta \in \Theta$ and the experiment $E_n \in \mathcal{E}$ selected at stage $n$. It is assumed that $E_n$ is measurable with respect to the natural filtration $\mathcal{F}_{n−1} := \sigma(X_1 , . . . , X_{n−1} )$, that is, $E_n$ is chosen according to the information of the past data $X_1 , . . . , X_{n−1}$ . The likelihood function is therefore given by $$L_n(\theta) := \prod_{i=1}^{n} f (x_i ; \theta, E_i ).$$

In a quantum estimation problem, $\mathcal{X}$ is a finite set (with $\mu$ the counting measure), and $\Theta, \mathcal{E}$ are both compact. "

Also, they claim that under the next regularity conditions they can obtain the strong consistency for the Maximum likelihood estimator. The regularity conditions are

1. $f(x; \theta, e)$ is positive for all $(x, \theta, e)$, and is continous in $(x, \theta,e)$
2. $\mu(\left\{x; f(x; \theta, e) \neq f(x; \theta', e) \right\})>0 $ for any $\theta \neq \theta'$ and $e \in \mathcal{E}$.

However, I don't have a good measure theory background. Then I have a couple of questions, My questions are:

¿What means the second regularity condition, is it the identifiability for each probability density? ¿If any of the conditions is not holds, the Maximum likelihood estimator could be consistent?

Thanks for any help.

Answer 1

The definition of identifiability is this: a family of densities or p.m.f. $f_\theta$ is identifiable if and only if $\theta_1\neq\theta_2$ implies $f_{\theta_1} \neq f_{\theta_1}$; that is, saying that a family is identifiable is the same as saying that different parameters cannot give you the same density.

In your case, for each fixed experiment $e$, the family $f(\cdot;\theta,e)$ is identifiable if and only if $\theta_1\neq\theta_2$ implies that $f(\cdot;\theta_1,e)$ and $f(\cdot;\theta_2,e)$ are different functions, i.e., it cannot be true that for all $x\in\mathcal{X}$ $f(x;\theta_1,e) = f(x;\theta_2,e)$. When $\mathcal{X}$ is finite and discrete, the equation \begin{equation} \mu(\{x|f(x;\theta_1,e)\neq f(x;\theta_2,e)\}) > 0 \end{equation} reads literally as "there must be at least one element in $\mathcal{X}$ that can make $f(\cdot;\theta_1,e)$ and $f(\cdot;\theta_2,e)$ evaluate to different values". This is the same as saying that $f(\cdot;\theta_1,e)$ and $f(\cdot;\theta_2,e)$ must be different functions. So yes, the second condition is equivalent to saying that the family of densities for each experiment $e$ must be identifiable.

Without the identifiability condition, there cannot be any guarantees that the MLE $\hat{\theta}_n$ will be (weakly) consistent. To see this without going too technical, for simplicity let all the experiments $e_n$ be the same, say, $e_n=e$ for all $n$ and let's imagine $\theta_1$ and $\theta_2$ are two different parameters such that $f(\cdot;\theta_1,e)$ and $f(\cdot;\theta_2,e)$ are the same function. Notice that whichever one of $\theta_1$ and $\theta_2$ were the truth, your observations $x_1\cdots x_n$ will be drawn from the same density; therefore $\hat{\theta}_n$, which is a function of $x_1\cdots x_n$, would also be drawn from the same density no matter which one of them were the truth. If $\hat{\theta}_n$ were consistent, by the definition of weak consistency, the density of $\hat{\theta}_n$ needs to become more and more concentrated around the true $\theta$; but both possible $\theta$'s give rise to the same density for $\hat{\theta}_n$, so now which one of $\theta_1$ or $\theta_2$ should density of $\hat{\theta}_n$ converge in probability to? It cannot converge in probability to both, so the consistency idea is broken in this situation.