观察到的信息矩阵是否是预期信息矩阵的一致估计？

我试图证明在弱一致性最大似然估计器（MLE）处评估的观测信息矩阵是预期信息矩阵的弱一致性估计器。这是被广泛引用的结果，但没有人提供参考或证明（我已经用尽我认为Google搜索结果的前20页和我的统计资料教科书）！

使用MLE的弱一致序列，我可以使用大数弱定律（WLLN）和连续映射定理来获得所需的结果。但是，我相信不能使用连续映射定理。相反，我认为需要使用统一的大数定律（ULLN）。有人知道有证明这一点的参考文献吗？我尝试了ULLN，但为简洁起见，现在省略。

对于这个问题的冗长，我深表歉意，但必须引入一些符号。表示法如下（我的证明在结尾）。

假设我们有随机变量的IID样本$\{Y_1,\ldots,Y_N\}$与密度$f(\tilde{Y}|\theta)$，其中$\theta\in\Theta\subseteq\mathbb{R}^{k}$（这里$\tilde{Y}$是具有相同密度的只是一般随机变量作为样本的任何成员）。向量$Y=(Y_1,\ldots,Y_N)^{T}$是所有样本向量的向量，其中$Y_{i}\in\mathbb{R}^{n}$所有$i=1,\ldots,N$。密度的真实参数值是$\theta_{0}$和 θ Ñ（Ý）是的弱一致最大似然估计（MLE） θ 0。根据规律性条件，Fisher信息矩阵可以写为$\hat{\theta}_{N}(Y)$$\theta_{0}$

其中${H}_{\theta}$是Hessian矩阵。等效样本为

其中$I_{y_i}=-E_\theta \left[H_{\theta}(\log f(Y_{i}|\theta)\right]$。所观察到的信息矩阵是;

$J(\theta) = -H_\theta(\log f(y|\theta)$，

（有些人的需求矩阵在评估θ，但有些却没有）。样本观察信息矩阵为：$\hat{\theta}$

$J_N(\theta)=\sum_{i=1}^N J_{y_i}(\theta)$

其中$J_{y_i}(\theta)=-H_\theta(\log f(y_{i}|\theta)$。

我可以证明在所述估计的概率收敛到我（θ ），但不ñ - 1 Ĵ Ñ（θ Ñ（Ý ））到我（θ 0$N^{-1}J_N(\theta)$$I(\theta)$$N^{-1}J_{N}(\hat{\theta}_N(Y))$$I(\theta_{0})$。到目前为止，这是我的证明；

Now $(J_{N}(\theta))_{rs}=-\sum_{i=1}^N (H_\theta(\log f(Y_i|\theta))_{rs}$ is element $(r,s)$ of $J_N(\theta)$, for any $r,s=1,\ldots,k$. If the sample is iid, then by the weak law of large numbers (WLLN), the average of these summands converges in probability to $-E_{\theta}[(H_\theta(\log f(Y_{1}|\theta))_{rs}]=(I_{Y_1}(\theta))_{rs}=(I(\theta))_{rs}$. Thus $N^{-1}(J_N(\theta))_{rs}\overset{P}{\rightarrow}(I(\theta))_{rs}$ for all $r,s=1,\ldots,k$, and so $N^{-1}J_N(\theta)\overset{P}{\rightarrow}I(\theta)$. Unfortunately we cannot simply conclude $N^{-1}J_{N}(\hat{\theta}_N(Y))\overset{P}{\rightarrow}I(\theta_0)$ by using the continuous mapping theorem since $N^{-1}J_{N}(\cdot)$ is not the same function as $I(\cdot)$.

Any help on this would be greatly appreciated.

$\newcommand{\convp}{\stackrel{P}{\longrightarrow}}$

I guess directly establishing some sort of uniform law of large numbers is one possible approach.

Here is another.

We want to show that $\frac{J^N(\theta_{MLE})}{N} \convp I(\theta^*)$.

(As you said, we have by the WLLN that $\frac{J^N(\theta)}{N} \convp I(\theta)$. But this doesn't directly help us.)

One possible strategy is to show that

and

If both of the results are true, then we can combine them to get

which is exactly what we want to show.

The first equation follows from the weak law of large numbers.

The second almost follows from the continuous mapping theorem, but unfortunately our function $g()$ that we want to apply the CMT to changes with $N$: our $g$ is really $g_N(\theta) := \frac{J^N(\theta)}{N}$. So we cannot use the CMT.

(Comment: If you examine the proof of the CMT on Wikipedia, notice that the set $B_\delta$ they define in their proof for us now also depends on $n$. We essentially need some sort of equicontinuity at $\theta^*$ over our functions $g_N(\theta)$.)

Fortunately, if you assume that the family $\mathcal{G} = \{g_N | N=1,2,\ldots\}$ is stochastically equicontinuous at $\theta^*$, then it immediately follows that for $\theta_{MLE} \convp \theta^*$,

(See here: http://www.cs.berkeley.edu/~jordan/courses/210B-spring07/lectures/stat210b_lecture_12.pdf for a definition of stochastic equicontinuity at $\theta^*$, and a proof of the above fact.)

Therefore, assuming that $\mathcal{G}$ is SE at $\theta^*$, your desired result holds true and the empirical Fisher information converges to the population Fisher information.

Now, the key question of course is, what sort of conditions do you need to impose on $\mathcal{G}$ to get SE? It looks like one way to do this is to establish a Lipshitz condition on the entire class of functions $\mathcal{G}$ (see here: http://econ.duke.edu/uploads/media_items/uniform-convergence-and-stochastic-equicontinuity.original.pdf ).

The answer above using stochastic equicontinuity works very well, but here I am answering my own question by using a uniform law of large numbers to show that the observed information matrix is a strongly consistent estimator of the information matrix , i.e. $N^{-1}J_{N}(\hat{\theta}_{N}(Y))\overset{a.s.}{\longrightarrow}I(\theta_{0})$ if we plug-in a strongly consistent sequence of estimators. I hope it is correct in all details.

We will use $I_{N}=\{1,2,...,N\}$ to be an index set, and let us temporarily adopt the notation $J(\tilde{Y},\theta):=J(\theta)$ in order to be explicit about the dependence of $J(\theta)$ on the random vector $\tilde{Y}$. We shall also work elementwise with $(J(\tilde{Y},\theta))_{rs}$ and $(J_{N}(\theta))_{rs}=\sum\nolimits_{i=1}^{N}(J(Y_{i},\theta))_{rs}$, $r,s=1,...,k$, for this discussion. The function $(J(\cdot,\theta))_{rs}$ is real-valued on the set $\mathbb{R}^{n}\times\Theta^{\circ}$, and we will suppose that it is Lebesgue measurable for every $\theta\in\Theta^{\circ}$. A uniform (strong) law of large numbers defines a set of conditions under which

$\underset{\theta\in\Theta}{\text{sup}}\left|N^{-1}(J_{N}(\theta))_{rs}-E_{\theta}\left[(J(Y_{1},\theta))_{rs}\right]\right|=\nonumber\\ \hspace{60pt}\underset{\theta\in\Theta}{\text{sup}}\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(Y_{i},\theta))_{rs}-(I(\theta))_{rs}\right|\overset{a.s}{\longrightarrow}0\hspace{100pt}(1)$

The conditions that must be satisfied in order that (1) holds are (a) $\Theta^{\circ}$ is a compact set; (b) $(J(\tilde{Y},\theta))_{rs}$ is a continuous function on $\Theta^{\circ}$ with probability 1; (c) for each $\theta\in \Theta^{\circ}$ $(J(\tilde{Y},\theta))_{rs}$ is dominated by a function $h(\tilde{Y})$, i.e. $|(J(\tilde{Y},\theta))_{rs}|<h(\tilde{Y})$; and (d) for each $\theta\in \Theta^{\circ}$ $E_{\theta}[h(\tilde{Y})]<\infty$;. These conditions come from Jennrich (1969, Theorem 2).

Now for any $y_{i}\in\mathbb{R}^{n}$, $i\in I_{N}$ and $\theta'\in S\subseteq\Theta^{\circ}$, the following inequality obviously holds

$\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(y_{i},\theta'))_{rs}-(I(\theta'))_{rs}\right|\leq\underset{\theta\in S}{\text{sup}}\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(y_{i},\theta))_{rs}-(I(\theta))_{rs}\right|.\hspace{50pt}(2)$

Suppose that $\{\hat{\theta}_{N}(Y)\}$ is a strongly consistent sequence of estimators for $\theta_{0}$, and let $\Theta_{N_{1}}=B_{\delta_{N_{1}}}(\theta_{0})\subseteq K\subseteq \Theta^{\circ}$ be an open ball in $\mathbb{R}^{k}$ with radius $\delta_{N_{1}}\rightarrow 0$ as $N_{1}\rightarrow\infty$, and suppose $K$ is compact. Then since $\hat{\theta}_{N}(Y)\in \Theta_{N_{1}}$ for $N$ sufficiently large enough we have $P[\underset{N}{\text{lim}}\{\hat{\theta}_{N}(Y)\in\Theta_{N_{1}}\}]=1$ for sufficiently large $N$. Together with (2) this implies

$P\left[\underset{N\rightarrow\infty}{\text{lim}}\left\{\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(Y_{i},\hat{\theta}_{N}(Y)))_{rs}-(I(\hat{\theta}_{N}(Y)))_{rs}\right|\leq\right.\right.\nonumber\\ \hspace{40pt}\left.\left.\underset{\theta\in\Theta_{N_{1}}}{\text{sup}}\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(Y_{i},\theta))_{rs}-(I(\theta))_{rs}\right|\right\}\right]=1.\hspace{100pt}(3)$

Now $\Theta_{N_{1}}\subseteq\Theta^{\circ}$ implies conditions (a)-(d) of Jennrich (1969, Theorem 2) apply to $\Theta_{N_{1}}$. Thus (1) and (3) imply

$P\left[\underset{N\rightarrow\infty}{\text{lim}}\left\{\left|N^{-1}\sum\nolimits_{i=1}^{N}(J(Y_{i},\hat{\theta}_{N}(Y)))_{rs}-(I(\hat{\theta}_{N}(Y)))_{rs}\right|=0\right\}\right]=1.\hspace{100pt}(4)$

Since $(I(\hat{\theta}_{N}(Y)))_{rs}\overset{a.s.}{\longrightarrow}I(\theta_{0})$ then (4) implies that $N^{-1}(J_{N}(\hat{\theta}_{N}(Y)))_{rs}\overset{a.s.}{\longrightarrow}(I(\theta_{0}))_{rs}$. Note that (3) holds however small $\Theta_{N_{1}}$ is, and so the result in (4) is independent of the choice of $N_{1}$ other than $N_{1}$ must be chosen such that $\Theta_{N_{1}}\subseteq \Theta^{\circ}$. This result holds for all $r,s=1,...,k$, and so in terms of matrices we have $N^{-1}J_{N}(\hat{\theta}_{N}(Y))\overset{a.s.}{\longrightarrow}I(\theta_{0})$.