温莎日记 28

Let's try to understand why MLEs are 'good':

If I get more and more data, I can uncover the truth.

Law of Large Numbers:

TT the U.S. physician: chenpengaizhongguo@126.com.

If distribution of the i.i.d. sample $X_1,...,X_n$ is such that $X_1$ has a finite expectation, i.e. $\vert EX_1\vert < ∞$ , then the sample average

$\tilde{X_n} =\frac{X_1+...+X_n}{n} \rightarrow EX_1$

converges to its expectation in probability, which means that for any arbitrarily small $\varepsilon >0$ .

$P(\vert \tilde{X} -EX_1 \vert >\epsilon )\rightarrow 0$ as $n\rightarrow ∞$ .

Note. Whenever we will use the LLN below we will simply say that the average converges to its expectation and will not mention in what sense. More mathematically inclined clients are welcome to carry out these steps more rigorously, especially when we use LLN in combination with the Central Limit Theorem.

Central Limit Theorem:

If distribution of the i.i.d. sample $X_1,...,X_n$ is such that $X_1$ has finite expectation and variance, i.e. $\vert EX_1 \vert <∞$ and $\sigma ^2=Var(X)<∞$ , then

$\sqrt{n} (\tilde{X_n} -EX_1 ) {\to d} N(0,\sigma ^2)$

converges in distribution to normal distribution with zero mean and variance $\sigma ^2$ , which means that for any interval [a,b],

$P(\sqrt{n}(\tilde{X}_n - EX_1 ) \in [a,b])\rightarrow \int_{a}^{b} \frac{1}{\sqrt{2\pi } \sigma } e^{-\frac{x^2}{2\sigma ^2} } dx$ .

In other words, the random variable $\sqrt{n} (\tilde{X}_n -EX_1 )$ will behave like a random variable from normal distribution when n gets large.

We will prove that MLE satisfies usually the following two properties called consistency and asymptotic normality.

1. Consistency. We say that an estimate $\hat{\theta }$ is consistent if $\hat{\theta } \rightarrow \theta _0$ in probability as $n \rightarrow ∞$ , where $\theta _0$ is the 'true' unknown parameter of the distribution of the sample.

2. Asymptotic Normality. We say that $\hat{\theta }$ is asymptotically normal if

$\sqrt{n} (\hat{\theta }-\theta _0 ){\to d} N(0,\sigma _{\theta _0}^2 )$

where $\sigma _{\theta _0}^2$ is called the asymptotic variance of the estimate $\hat{\theta }$ . Asymptotic normality says that the estimator not only converges to the unknown parameter but also converges fast enough at rate $1/\sqrt{n}$ .

Consistency of MLE:

Suppose that the data $X_{1:n}$ is generated from a distribution with unknown parameter $\theta _0$ and $\hat{\theta }$ converges to the unknown parameter $\theta _0$ ? This is not immediately obvious and we will give a sketch of why this happens.

First of all, MLE $\hat{\theta }$ is the maximizer of $L_n(\theta )=\frac{1}{n} \sum_{i=1}^n logf(X_i|\theta )$ which is a log-likelihood function normalized by $\frac{1}{n}$ . Notice that function $L_n(\theta )$ depends on data. Let us consider a function $l(X|\theta )=logf(X|\theta )$ and define $L(\theta )=E_{\theta _0}l(X|\theta )$ , where $E_{\theta _0}$ denotes the expectation with respect to the true unknown parameter $\theta _0$ of the sample $X_{1:n}$ .

If we deal with continuous distributions then $L(\theta )=\int(logf(x|\theta ))f(x|\theta _0)dx$ .

By law of large numbers, for any $\theta$ , $L_n(\theta )\rightarrow E_{\theta _0}l(X|\theta )=L(\theta )$ . Note that $L(\theta )$ this does not depend on the sample, it only depends on $\theta$ . We will need the following.

Lemma. We have that for any $\theta$ , $L(\theta )\leq L(\theta _0)$ . Moreover, the inequality is strict, $L(\theta )<L(\theta _0)$ , unless $P_{\theta _0}(f(X|\theta )=f(X|\theta _0))=1$ , which means that $P_\theta =P_{\theta_0}$ .

Proof. Let us consider the difference

$L(\theta )-L(\theta_0)=E_{\theta _0}(logf(X|\theta )-logf(X|\theta _0))=E_{\theta _0}log\frac{f(X|\theta )}{f(X|\theta _0)}$ .

Since $logt\leq t-1$ , we can write

$E_{\theta _0}log\frac{f(X|\theta )}{f(X|\theta _0)} \leq E_{\theta _0}(\frac{f(X|\theta )}{f(X|\theta _0)} -1)=\int(\frac{f(x|\theta )}{f(x|\theta _0)} -1)f(x|\theta _0)dx$

$=\int f(x|\theta)dx-\int f(x|\theta _0)dx=1-1=0$ .

Both integrals are equal to 1 because we are integrating the probability density function. This proves that $L(\theta )-L(\theta _0) \leq 0$ . The second statement of Lemma is also clear. We will use this Lemma to sketch the consistency of the MLE.

Theorem. Under some regularity conditions on the family of distributions, MLE $\hat{\theta }$ is consistent, i.e. $\hat{\theta } \rightarrow \theta _0$ as $n\rightarrow ∞$ .

Proof. We have the following facts:

1) $\hat{\theta }$ is the maximizer of $L_n(\theta )$ by definition.

2) $\theta _0$ is the maximizer of $L(\theta )$ by Lemma.

3) $\forall \theta$ we have $L_\theta (\theta ) \rightarrow L(\theta )$ by LLN.

Asymptotic normality of MLE, Fisher information.

We want to show the asymptotic normality of MLE, i.e. to show that

$\sqrt{n}(\hat{\theta } -\theta _0) {\to d} N(0,\sigma ^2_{MLE})$ for some $\sigma ^2_{MLE}$ and compute it.

This asymptotic variance in some sense measures the quality of MLE. First, we need to introduce the notion called Fisher Information.

Definition. (Fisher Information) Fisher information of a random variable X with distribution $P_{\theta _0}$ from the family $\left\{ P_\theta : \theta \in \Theta \right\}$ is defined by

$I(\theta _0)=E_{\theta _0}(l$ ' $(X|\theta _0)^2\equiv E_{\theta _0}(\frac{\partial}{\partial \theta } logf(X|\theta )|_{\theta =\theta _0})^2$ .

Theorem. (Asymptotic normality of MLE) We have, $\sqrt{n}(\hat{\theta }-\theta _0 ) \rightarrow N(0,\frac{1}{I(\theta _0)} )$ .

Example. The family of Bernoulli distributions $B(p)$ has p.f. $f(x|p)=p^x(1-p)^{1-x}$ and taking the logarithm $logf(x|p)=xlogp + (1-x)log(1-p)$ . The second derivative with respect to parameter $p$ is

$\frac{\partial}{\partial p} logf(x|p)=\frac{x}{p}-\frac{1-x}{1-p}$ , $\frac{\partial^2}{\partial p^2} logf(x|p)=-\frac{x}{p^2}-\frac{1-x}{(1-p)^2}$ .

Then the Fisher information can be computed as

$I(p)=-E\frac{\partial^2}{\partial p^2} logf(X|p)=\frac{EX}{p^2} +\frac{1-EX}{(1-p)^2} =\frac{p}{p^2} +\frac{1-p}{(1-p)^2} =\frac{1}{p(1-p)}$ .

The MLE of $p$ is $\hat{p}=\tilde{X}$ and the asymptotic normality result states that

$\sqrt{n}(\hat{p} -p_0) \rightarrow N(0,p_0(1-p_0))$ which, of course, also follows directly from the CLT.

最后编辑于：2022.09.09 21:18:05

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温莎日记 28

Asymptotic normality of MLE, Fisher information.

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