Visual Guide to Statistics. Part III: Asymptotic Properties of Estimators

A minimal condition for a good estimator is that it is getting closer to estimated parameter with growing size of sample vector. In this post we will focus on asymptotic properties of estimators.

Consistency of estimators

Berore talking about estimators convergence, let’s recall that there exist several different notions of convergence of random variables. Let $(X_n)$ be sequence of real-valued random variables, then we say

  • $X_n$ converges in distribution towards the random variable $X$ if
\[\lim\limits_{n \to \infty} F_{n}(x) = F(x),\]

for every $x \in \mathbb{R}$, at which $F$ is continuous. $F_n(x)$ and $F(x)$ are the cumulative distribution functions for $X_n$ and $X$ respectively. We denote convergence in distribution as $X_n \xrightarrow[]{\mathcal{L}} X$.

  • $X_n$ converges in probability to random variable $X$ if
\[\lim\limits_{n \to \infty} P(|X_n-X|>\varepsilon)=0 \quad \forall \varepsilon > 0.\]

Convergence in probability implies convergence in distribution. In the opposite direction, convergence in distribution implies convergence in probability when the limiting random variable $X$ is a constant. We denote convergence in probability as $X_n \xrightarrow[]{\mathbb{P}} X$.

  • $X_n$ converges almost surely towards $X$ if
\[P(\omega \in \Omega: \lim\limits_{n \to \infty} X_n(\omega) = X(\omega)) = 1.\]

Almost sure convergence implies convergence in probability, and hence implies convergence in distribution. Notation: $X_n \xrightarrow[]{\text{a.s.}} X$.

The similar logic can be applied to a sequence of $d$-dimensional random variables. Also, recall continuous mapping theorem, which states that for a continuous function $f$ we have

\[\begin{aligned} &X_n \xrightarrow[]{\mathcal{L}} X \quad \Rightarrow \quad f(X_n) \xrightarrow[]{\mathcal{L}} f(X), \\ &X_n \xrightarrow[]{\mathbb{P}} X \quad \Rightarrow \quad f(X_n) \xrightarrow[]{\mathbb{P}} f(X), \\ &X_n \xrightarrow[]{\text{a.s.}} X \quad \Rightarrow \quad f(X_n) \xrightarrow[]{\text{a.s.}} f(X). \end{aligned}\]

Now let $g_n$ be an estimator of $\gamma(\vartheta)$ with values in metric space. Assume that all experiments are defined on a joint probability space $P_\vartheta$ for all $n$. We say that

  • $g_n$ is (weakly) consistent if
\[g_n \xrightarrow[]{\mathbb{P}}\gamma(\vartheta) \quad \forall \vartheta \in \Theta.\]
  • $g_n$ is strongly constistent if
\[g_n \xrightarrow[]{\text{a.s.}} \gamma(\vartheta) \quad \forall \vartheta \in \Theta.\]

Recall the method of moments from Part I: $X_1, \dots, X_n$ i.i.d. $\sim P_\vartheta$, $\vartheta \in \Theta \subset \mathbb{R}^k$ and $\gamma: \Theta \rightarrow \Gamma \subset \mathbb{R}^l$. Also

\[m_j = \mathbb{E}_\vartheta[X_1^j] = \int x^j P_\vartheta(dx)\]

for $j = 1, \dots, k$, and

\[\gamma(\vartheta) = f(m_1, \dots, m_k).\]

Then choose

\[\hat{\gamma}(X) = f(\hat{m}_1, \dots, \hat{m}_k),\]


\[\hat{m}_j = \frac{1}{n} \sum_{i=1}^{n}X_k^j.\]

By Law of Large Numbers $\hat{m}_j \rightarrow m_j$ a.s. Since $f$ is continuous, we obtain

\[\hat{\gamma}(X) \xrightarrow[]{\text{a.s.}} \gamma(\vartheta).\]

Hence, $\hat{\gamma}(X)$ is a strongly consistent estimator.

Central Limit Theorem

Let $(X_n)$ be a sequence of $d$-dimensional random variables. Lévy’s continuity theorem states that

\[X_n \xrightarrow[]{\mathcal{L}} X \quad \Longleftrightarrow \quad \mathbb{E}[\exp(iu^TX_n)] \rightarrow \mathbb{E}[\exp(iu^TX)] \quad \forall u \in \mathbb{R}^d.\]

If we write $u=ty$ for $t \in \mathbb{R}$, $y \in \mathbb{R}^d$, then we can say that $X_n \xrightarrow[]{\mathcal{L}} X$ if and only if

\[y^TX_n \xrightarrow[]{\mathcal{L}} y^TX \quad \forall y \in \mathbb{R}^d.\]

This statement is called Cramér–Wold theorem.

If $X_1, \dots, X_n$ are i.i.d. with $\mathbb{E}[X_j]=\mu \in \mathbb{R}^d$ and $\operatorname{Cov}(X_j)=\Sigma \in \mathbb{R}^{d \times d}$ (positive-definite, $\Sigma > 0$), then for random vector

\[X^{(n)} = \frac{1}{n}\sum_{j=1}^n X_j \in \mathbb{R}^d\]

we know from one-dimensional Central Limit Theorem (CLT) that

\[\sqrt{n}(y^TX^{(n)} -y^T\mu) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, y^T\Sigma y) \quad \forall y \in \mathbb{R}^d.\]

Applying Cramér–Wold theorem we get

\[\sqrt{n}(X^{(n)}-\mu) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \Sigma).\]

This statement is known as Multidimensional Central Limit Theorem.

Fig. 1. Visualization of multidimensional CLT for two-dimensional case. On the left-hand side there is random vector of two uniformly distributed random variables: $X_1, X_2 \sim \mathcal{U}(-1, 1)$ with mean $\mu=(0, 0)^T$ and correlation $\rho$. On the right-hand side is $\sqrt{n} X^{(n)}$ which for large $n$ has approximately normal distribution with zero mean and the same covariance as $X$.


Let $(X_n)$ and $(Y_n)$ be sequences of $d$-dimensional random variables, such that

\[X_n \xrightarrow[]{\mathcal{L}} X \quad \text{and} \quad Y_n \xrightarrow[]{\mathbb{P}} c\]

for some constant vector $c$. Then we can apply the continuous mapping theorem, recognizing the functions $f(x, y)=x+y$ and $f(x, y)=xy$ are continuous, and conclude that

  • $X_n+Y_n \xrightarrow[]{\mathcal{L}} X + c,$
  • $Y_n^TX_n \xrightarrow[]{\mathcal{L}} c^TX.$

This statement is called Slutsky’s lemma and it can be extremely useful in estimating approximate distribution of estimators. For example, let $X_1, \dots X_n$ i.i.d. $\sim \operatorname{Bin}(1, p)$. Estimator of $p$ $g_n(X) = \overline{X}_n$ is unbiased and we know from CLT that

\[\sqrt{\overline{X}_n(1-\overline{X}_n)} \xrightarrow[]{\mathbb{P}} \sqrt{p(1-p)}.\]

By Slutsky’s lemma,

\[\frac{\sqrt{n}(\overline{X}_n-p)}{\sqrt{\overline{X}_n(1-\overline{X}_n)}} \xrightarrow[]{\mathcal{L}} \mathcal{N}(0,1)\]

and for large $n$ we have

\[P_p(|\overline{X}_n-p|<\varepsilon) \approx 2 \Phi\Bigg(\varepsilon\sqrt{\frac{n}{\overline{X}_n(1-\overline{X}_n)}}\Bigg) -1 \quad \forall p \in (0, 1),\]

where $\Phi$ is cumulative distribution function for $\mathcal{N}(0,1)$.

Slutsky’s lemma also leads to important asymptotic property of estimator $g_n$, called Delta-method. Let $(X_n)$ be sequence of $d$-dimensional random variables, such that

\[\frac{X_n-\mu}{c_n} \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \Sigma),\]

where $c_n \rightarrow 0$, $\mu \in \mathbb{R}^d$ и $\Sigma \geq 0 \in \mathbb{R}^{d \times d}$. Let also $g:\mathbb{R}^d \rightarrow \mathbb{R}^m$ be continuously differentiable in $\mu$ with Jacobian matrix $D \in \mathbb{R}^{m \times d}$. Then:

\[\frac{g(X_n)-g(\mu)}{c_n} \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, D\Sigma D^T).\]
Proof By Slutsky's Lemma $$X_n-\mu = \frac{X_n-\mu}{c_n}c_n \xrightarrow[]{\mathcal{L}} 0.$$ Convergence in distribution to a constant implies convergence in probability: $X_n \xrightarrow[]{\mathbb{P}} \mu$. Then $$\frac{g(X_n)-g(\mu)}{c_n}=g'(\mu)\frac{X_n-\mu}{c_n}+(g'(\xi_n)-g'(\mu))\frac{X_n-\mu}{c_n},$$ for some intermediate point $\xi_n$, such that $\|\xi_n-\mu \| \leq \|X_n-\mu \|$. From $X_n \xrightarrow[]{\mathbb{P}} \mu$ we have $\xi_n \xrightarrow[]{\mathbb{P}} \mu$ and $g'(\xi_n) \xrightarrow[]{\mathbb{P}} g'(\mu)$ (because $g$ is continuously differentiable). Applying again Slutsky's Lemma: $$ g'(\mu) \frac{X_n-\mu}{c_n} \xrightarrow[]{\mathcal{L}} g'(\mu) \cdot \mathcal{N}(0, \Sigma) $$ finishes the proof.
  • Recall example with method of moments, but now with additional conditions on $\mathbb{E}[X_1^{2k}] < \infty$ for all $\vartheta \in \Theta$ and $\gamma$ being continuously differentiable with Jacobian matrix $D$. We know from CLT that
\[\sqrt{n}((\hat{m}_1, \dots, \hat{m}_k)^T - (m_1, \dots, m_k)^T) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \Sigma),\]


\[\Sigma = (\Sigma)_{i,j=1}^k = (m_{i+j} - m_i m_j)_{i,j=1}^k.\]


\[\sqrt{n}(\gamma(\hat{m}_1, \dots, \hat{m}_k) - \gamma(m_1, \dots, m_k)) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, D \Sigma D^T).\]
  • Take another example: let $X_1, \dots X_n$ be i.i.d. with
\[\mathbb{E}_\vartheta[X_i] = \mu \quad \text{and} \quad \operatorname{Var}_\vartheta(X_i) = \sigma^2.\]

From CLT we have

\[\sqrt{n}(\overline{X}_n - \mu) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \sigma^2).\]

Choose $\overline{X}_n^2$ as an estimator for $\mu^2$. Applying Delta-method we get

\[\sqrt{n}(\overline{X}_n^2-\mu^2) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, 4\mu^2\sigma^2).\]
  • Let
\[(X_i, Y_i)^T \sim \mathcal{N} \begin{pmatrix} \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix}, \begin{pmatrix} \sigma^2 & \rho \sigma \tau \\ \rho \sigma \tau & \tau^2 \end{pmatrix} \end{pmatrix}, \quad i = 1, \dots, n,\]

be i.i.d with parameter $\vartheta = (\mu_1, \mu_2, \sigma^2, \tau^2, \rho)^T$. The estimator

\[\hat{\rho}_n = \frac{SQ_{xy}}{\sqrt{SQ_{xx} SQ_{yy}}},\]


\[SQ_{xy} = \frac{1}{n} \sum_{i=1}^{n}(X_i-\overline{X}_n)(Y_i - \overline{Y}_n),\]

$SQ_{xx}, SQ_{yy}$ - likewise, is called the Pearson correlation coefficient. Without loss of generality, assume $\mu_1=\mu_2=0$, $\sigma=\tau=1$, because $\hat{\rho}_n$ is invariant under affine transformation.

Prove first that $S_n = (SQ_{xx}, SQ_{yy}, SQ_{xy})^T$ satisifies

\[\sqrt{n}(S_n - m) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, V),\]

where $m=(1, 1, \rho)^T$ and

\[V = 2 \begin{pmatrix} 1 & \rho^2 & \rho \\ \rho^2 & 1 & \rho \\ \rho & \rho & (1 + \rho^2)/2 \end{pmatrix}.\]
Sketch of the proof We use Slutsky's Lemma and CLT to show that $$\sqrt{n}(\overline{X}_n \overline{Y}_n) \xrightarrow[]{\mathbb{P}} 0, \quad \sqrt{n}(\overline{X}_n)^2 \xrightarrow[]{\mathbb{P}} 0, \quad \sqrt{n}(\overline{Y}_n)^2 \xrightarrow[]{\mathbb{P}} 0. $$ Then it is simple to conclude $$\sqrt{n}(S_n - m) - \sqrt{n}\Big(\frac{1}{n}\sum_{i=1}^{n}Z_i - m \Big) \xrightarrow[]{\mathbb{P}} 0,$$ with $Z_i = (X_i^2, Y_i^2, X_iY_i)^T$. Then prove that $$\operatorname{Cov}(Z_i) = \mathbb{E}[Z_i Z_i^T]-\mathbb{E}[Z_i]\mathbb{E}[Z_i]^T = V. $$ The rest follows from multidimensional CLT.

Then take $g(S_n)=\hat{\rho}_n$ with $g(x_1, x_2, x_3) = \frac{x_3}{\sqrt{x_1 x_2}}$. Jacobian matrix of $g$ at $m$:

\[D = (-\rho/2, -\rho/2, 1).\]

In total,

\[\sqrt{n}(\hat{\rho}_n - \rho) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, DVD^T) = \mathcal{N}(0, (1-\rho^2)^2).\]

Fig. 2. Visualization of asymptotic normality for Pearson correlation coefficient. Drag sample dots to observe how it affects $SQ$ coefficients and $\hat{\rho}_n$.

Asympotic efficiency

Let $g_n \subset \mathbb{R}^l$ be a sequence of estimators with

\[\mu_n(\vartheta)=\mathbb{E}_\vartheta[g_n] \in \mathbb{R}^l \quad \text{and} \quad \Sigma_n(\vartheta)=\operatorname{Cov}(\vartheta) \in \mathbb{R}^{l \times l},\]

such that $\lVert \Sigma_n(\vartheta) \rVert \rightarrow 0$. Then

  • $g_n$ is called asymptotically unbiased for $\gamma(\vartheta)$ if
\[\mu_n(\vartheta) \rightarrow \gamma(\vartheta),\]
  • $g_n$ is called asymptotically normal if
\[\Sigma_n^{-\frac{1}{2}}(\vartheta)(g_n-\mu_n(\vartheta)) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \mathbb{I}_l),\]

where $\mathbb{I}_l$ is identity matrix.

Let $f_n: \mathcal{X} \rightarrow \mathbb{R}^l$ be asymptotically unbiased and asymptotically normal sequence of estimators. Under regularity conditions from Cramér–Rao theorem we call $g_n$ asymptotically efficient, if

\[\lim\limits_{n \rightarrow \infty} \Sigma_n(\vartheta) \mathcal{I}(f_n(\cdot, \vartheta))=\mathbb{I}_l \quad \forall \vartheta \in \Theta,\]

where $\mathcal{I}(f_n(\cdot, \vartheta))$ is Fisher information.

The intuition behind definition above is the following: if $g_n$ is unbiased, then by Cramér–Rao theorem $\operatorname{Cov}_\vartheta(g_n) \geq \mathcal{I}^{-1}(f_n(\cdot, \vartheta))$. Due to asymptotic normality:

\[\Sigma_n^{-\frac{1}{2}}(\vartheta)(g_n-\mu_n(\vartheta)) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \mathbb{I}_l)\]

we have approximately

\[\operatorname{Cov}_\vartheta(g_n) \approx \Sigma_n(\vartheta) \approx \mathcal{I}^{-1}(f_n(\cdot, \vartheta))\]

and $g_n$ is asymptotically unbiased and asymptotically efficient.

Recall example from Part I: for $X_1, \dots X_n$ i.i.d. $\sim \mathcal{N}(\mu, \sigma^2)$ estimator

\[g_n(X) = \begin{pmatrix} \overline{X}_n \\ \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X}_n)^2 \end{pmatrix}\]

satisfies the equality

\[\operatorname{Cov}_\vartheta(g_n) = \begin{pmatrix} \sigma^2/n & 0 \\ 0 & 2\sigma^4 / (n - 1) \end{pmatrix} = \Sigma_n(\vartheta).\]

But Fisher information is

\[\mathcal{I}^{-1}(f_n(\cdot, \vartheta)) = \begin{pmatrix} \sigma^2/n & 0 \\ 0 & 2\sigma^4 / n \end{pmatrix}\]

and $g_n$ is not efficient, but asymptotically efficient.

Asymptotic properties of maximum-likelihood estimators

In Part I we briefly mentioned maximum-likelihood estimators as one of the most common estimation methods in statistic. It is worth knowing what their asymptotic properties are. Let’s rewrite the definition here: let $X_1, \dots X_n$ be i.i.d. $\sim P_\vartheta$, $\vartheta \in \Theta$ with densities $f(\cdot, \vartheta)$. We call

\[\ell(\cdot, \vartheta) = \log f(\cdot, \vartheta)\]

the log-likelihood function and set

\[\begin{aligned}\hat{\theta}_n(X) &= \arg \sup_{\vartheta \in \Theta} f(X, \vartheta) \\&= \arg \sup_{\vartheta \in \Theta} \ell (X, \vartheta) \\&= \arg \sup_{\vartheta \in \Theta} \frac{1}{n} \sum_{i=1}^{n} \ell (X_i, \vartheta) \end{aligned}\]

as the maximum-likelihood estimator for $\vartheta$.

Now, say the following conditions are satisfied:

  1. $\Theta \subset \mathbb{R}^k$ is compact space
  2. $L(\eta, \vartheta) = \mathbb{E}[\ell(X_i, \eta)]$ and $L_n(\eta) = \frac{1}{n}\sum_{i=1}^n\ell(X_i, \eta)$ are a.s. continuous functions over $\eta$.
  3. \[\sup_{\eta \in \Theta} | L_n(\eta)-L(\eta, \vartheta)|\xrightarrow{\mathcal{L}}0.\]

Then maximum-likelihood estimator $\hat{\theta}_n$ is consistent.

Proof: for any $\eta \in \Theta$:

\[L(\eta, \vartheta) = \int \ell(x, \eta) f(x,\vartheta) dx = \int \ell(x,\vartheta)f(x,\vartheta)dx - KL(\vartheta | \eta),\]

where $KL$ is Kullback-Leibler divergence:

\[KL(\vartheta | \eta) = \int_{\mathcal{X}} \log\Big(\frac{f(x,\vartheta)}{f(x,\eta)}\Big) f(x,\vartheta)dx.\]

It can be shown that

\[\begin{aligned} KL(\vartheta | \eta) & = \int_{\mathcal{X}} -\log\Big(\frac{f(x,\eta)}{f(x,\vartheta)}\Big) f(x,\vartheta)dx \\ \color{\Salmon}{\text{Jensen inequality} \rightarrow } & \geq -\log\int_{\mathcal{X}} \frac{f(x,\eta)}{f(x,\vartheta)} f(x,\vartheta)dx \\ & = 0, \end{aligned}\]

and it turns into equality only when $f(x,\vartheta) = f(x,\eta)$ for almost every $x$. Therefore we conclude that $L(\eta, \vartheta)$ reaches maximum at $\eta = \vartheta$.

Using the fact that function $m_f = \arg\max_{\eta \in \Theta} f(\eta)$ is continuous if $m_f$ is unique, we finish the proof from

\[\vartheta = \arg \max L(\eta, \vartheta)\quad \text{and} \quad \hat{\theta}_n=\arg \max L_n(\eta)\]

and condition 3.

Asymptotic efficiency of maximum-likelihood estimators

If the following conditions are satisfied:

  • $\Theta \subset \mathbb{R}^k$ is compact and $\vartheta \subset \operatorname{int}(\Theta)$.
  • $\ell(x, \eta)$ is continuous $\forall \eta \in \Theta$ and twice continuously differentiable over $\vartheta$ for almost every $x \in \mathcal{X}$.
  • $\ell(x, \eta)$ is Lipschitz: there exist functions $H_0, H_2 \in L^1(P_\vartheta)$ and $H_1 \in L^2(P_\vartheta)$, such that:
\[\sup_{\eta \in \Theta} \|\ell(x, \eta)\| \leq H_0(x), \quad \sup_{\eta \in \Theta} \|\dot{\ell}(x, \eta)\| \leq H_1(x), \quad \sup_{\eta \in \Theta} \|\ddot{\ell}(x, \eta)\| \leq H_2(x) \quad \forall x \in \mathcal{X}.\]
  • Fisher information
\[\mathcal{I}(f(\cdot, \vartheta))=\mathbb{E}_\vartheta[\dot{\ell}(X,\vartheta)\dot{\ell}(X,\vartheta)^T]\]

is positive definite (and therefore invertible),

then $\hat{\theta}_n$ is asymptotically normal:

\[\sqrt{n}(\hat{\theta}_n-\vartheta) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \mathcal{I}(f(\cdot, \vartheta))^{-1}).\]

We will prove it in 4 steps:

Step 1. Prove the constistency of $\hat{\theta}_n$. For this we need to verify that all conditions from theorem about consistency of maximum-likelihood estimator are satisfied:

  1. Satisfied by the assumption.
  2. $L_n(\eta)$ is a.s. continuous. Using 2-3 conditions and dominated convergence we get \(|L(\eta_1, \vartheta) - L(\eta_2, \vartheta)| \leq \int_{\mathcal{X}} |\ell(x, \eta_1) - \ell(x,\eta_2)| f(x,\vartheta) \mu(dx) \rightarrow 0,\) for $\eta_1 \rightarrow \eta_2$.
  3. By Law of Large Numbers: \(\begin{aligned} \limsup_{n \rightarrow \infty} \sup_{\| \eta_1 - \eta_2 \| < \delta} | L_n(\eta_1) - L_n(\eta_2)| & \leq \limsup_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \sup_{\| \eta_1 - \eta_2 \| < \delta} |\ell(X_i, \eta_1) - \ell(X_i, \eta_2) |\\ & = \mathbb{E}_\vartheta[\sup_{\| \eta_1 - \eta_2 \| < \delta}|\ell(X,\eta_1) - \ell(X, \eta_2)|] \end{aligned}\)

Because $\Theta$ is compact, function $\ell(X, \eta)$ is a.s. uniformly continuous in $\eta$. As a consequence, the last statement converges to zero for $\delta \rightarrow 0$ (using again dominated convergence).

Step 2. Let

\[\dot{L}_n(\vartheta) := \frac{1}{n} \sum_{i=1}^{n} \dot{\ell}(X_i, \vartheta).\]

Prove that $\sqrt{n}\dot{L}_n(\vartheta) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \mathcal{I}(f(\cdot, \vartheta))).$

Let $A_n$ be $k$-dimensional rectangle with vertices in $\hat{\theta}_n$ and $\vartheta$. Because $\hat{\theta}_n \xrightarrow{\mathcal{L}} \vartheta$ and $\vartheta \in \operatorname{int}(\Theta)$, we have

\[P_\vartheta(A_n \subset \operatorname{int}(\Theta)) \rightarrow 1.\]


\[\dot{L}_n(\hat{\theta}_n) = \frac{1}{n} \sum_{i=1}^{n} \dot{\ell}(X_i, \hat{\theta}_n) = 0\]

by definition of $\hat{\theta}_n$, and

\[\begin{aligned} \mathbb{E}[\dot{\ell}(X_i, \vartheta)] & = \int_{\mathcal{X}} \dot{\ell}(x, \vartheta) f(x, \vartheta) dx \\ & = \int_{\mathcal{X}} \dot{f}(x, \vartheta) dx \\ & =\frac{\partial}{\partial \vartheta}\int_{\mathcal{X}} f(x, \vartheta) dx = 0. \end{aligned}\]

By definition

\[\operatorname{Cov}(\dot{\ell}(X_i, \vartheta)) = \mathcal{I}(f(\cdot, \vartheta)).\]

Then by CLT:

\[\sqrt{n} \dot{L}_n(\vartheta) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \mathcal{I}(f(\cdot, \vartheta))).\]

Step 3. By Mean Theorem:

\[-\dot{L}_n(\vartheta) = \dot{L}_n(\hat{\theta}_n) - \dot{L}_n(\vartheta) = \ddot{L}_n(\widetilde{\theta}_n)(\hat{\theta}_n - \vartheta)\]

for some $\widetilde{\theta}_n \in A_n$. Prove that $\ddot{L}_n(\widetilde{\theta}_n) \xrightarrow{\mathcal{L}} -\mathcal{I}(f(\cdot, \vartheta))$.

We use the equation

\[\ddot{\ell}(x, \vartheta) = \frac{\ddot{f}(x, \vartheta)}{f(x,\vartheta)} - \dot{\ell}(x, \vartheta)\dot{\ell}(x, \vartheta)^T.\]

to show that

\[\mathbb{E}_\vartheta[\ddot{\ell}(X, \vartheta)] + \mathcal{I}(f(\cdot, \vartheta)) = \mathbb{E}_\vartheta\Big[ \frac{\ddot{f}(X, \vartheta)}{f(X,\vartheta)} \Big] = 0,\]

From Law of Large Numbers it follows that

\[\ddot{L}_n(\vartheta) \xrightarrow{\mathcal{L}} - \mathcal{I}(f(\cdot, \vartheta)).\]

Finally, we use the equality

\[\lim\limits_{\delta \rightarrow 0} \lim\limits_{n \rightarrow \infty} P_\vartheta(\| \widetilde{\theta}_n - \vartheta \| < \delta) = 1\]

and continuity of $\ddot{\ell}$ over $\vartheta$ to finish the proof.

Step 4. Now we conclude that

\[\lim\limits_{n \rightarrow \infty} P_\vartheta(\ddot{L}_n(\widetilde{\theta}_n) \text{ is invertible}) = 1.\]

and applying Slutsky’s lemma we get

\[\begin{aligned} \sqrt{n}(\hat{\theta}_n - \vartheta) & = -\ddot{L}_n(\widetilde{\theta}_n)^{-1} \sqrt{n} \dot{L}_n(\vartheta) \\ & \rightarrow \mathcal{I}(f(\cdot, \vartheta))^{-1} \mathcal{N}(0, \mathcal{I}(f(\cdot, \vartheta))) \\&= \mathcal{N}(0, \mathcal{I}(f(\cdot, \vartheta))^{-1}). \color{Salmon}{\square} \end{aligned}\]

Take an example: let $X_1, \dots X_n$ be i.i.d. $\sim \operatorname{Exp}(\lambda)$ with joint density

\[f_n(X, \lambda) = \lambda^n \exp \Big(-\lambda \sum_{i=1}^n X_i \Big) \quad \forall x \in \mathbb{R}^+.\]

To find maximum-likelihood estimator one must maximize log-density

\[\ell_n(X, \lambda) = n \log(\lambda) - \lambda \sum_{i=1}^n X_i \quad \forall x \in \mathbb{R}^+\]

with respect to $\lambda$. Taking the derivative and equating it to zero we get

\[\frac{n}{\lambda} = \sum_{i=1}^{n} X_i,\]

and estimator is

\[\hat{\lambda}_n = \frac{1}{\overline{X}_n}.\]

Next, using the fact that

\[\mathbb{E}[X] = \lambda^{-1} \quad \text{and} \quad \operatorname{Var}(X) = \lambda^{-2},\]

and $\dot{\ell}_1(X, \lambda) = -(X - \lambda^{-1})$, we calculate Fisher information:

\[\mathcal{I}(f(\cdot, \lambda)) = \mathbb{E}\Big[\Big(X - \frac{1}{\lambda}\Big)^2\Big]=\frac{1}{\lambda^2}.\]

By theorem of asymptotic efficiency of maximum-likelihood estimators we get

\[\sqrt{n}(\hat{\lambda}_n - \lambda) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \lambda^2),\]

On the other hand by CLT

\[\sqrt{n}\Big(\overline{X}_n - \frac{1}{\lambda}\Big) \xrightarrow[]{\mathcal{L}} \mathcal{N}\Big(0, \frac{1}{\lambda^2}\Big).\]

Using Delta-method for $g(x) = x^{-1}$ we get the same result:

\[\sqrt{n}(\overline{X}_n^{-1} - \lambda) \xrightarrow[]{\mathcal{L}} \mathcal{N}(0, \lambda^2).\]
This post is licensed under CC BY 4.0 by the author.