# Note(5)--Extension of CLT

Posted by Gwan Siu on February 9, 2018

In this blog, we just talk about some extension of CLT.

1. Lyapunov CLT: just requires only independent but not identically idstributed.
2. Multivariate CLT: Original CLT is applied to one dimension, what happens to high dimension space.
3. CLT with estimated variance: when we use CLT, we should know about the variance of random variables. Could we replace estimated variance with true variance when we use CLT?
4. How about the Convergenvce rate in CLT? Berry Esseen
5. The delta method, we use a continuous function to construct another limiting distribution.

#### 1. Berry-Esseen Theorem.

The central limit theorem is from the asymptotic view. It just states that the average of i.i.d random variables converges in distribution to standard normal distribution as $n\rightarrow \infty$, but it does not answer the question that how close to this standard normal distribution. The berry-esseen theorem answer this question.

Berry-Esseen: Suppose that $X_{1},…,X_{n}\sim P$. Let $\mu=\mathbb{E}[X_{1}],\sigma^{2}=\mathbb{E}[\arrowvert X_{1}-\mu\arrowvert^{2}]$, and $\mu_{3}=\mathbb{E}[\arrowvert X_{1}-\mu\arrowvert^{3}]$. Let

$$$F_{n}(x)=\mathbb{P}(\frac{\sqrt{n}(\widetilde{\mu}-\mu)}{\sigma}\leq x),$$$

denote the CDF of the normalized sample average. If $\mu_{3}<\infty$ then,

$$$\sup_{x}\arrowvert F_{n}(x)-\Phi(x)\arrowvert\leq \frac{9\mu_{3}}{\sigma^{3}\sqrt{n}}$$$

This bound is roughly saying that if $\mu/sigma^{3}$ is small then the convergence to normality in distribution happens quite fast.

#### 2.Multivariate CLT-CLT in high dimensional space

Multivariate CLT: If $X_{1},…,X_{n}$ are i.i.d with mean $\mu\in \mathbb{R}^{d}$, and covariance matrix $\Sigma\in \mathbb{R}^{d\times d}$(with finite entries) then,

$$$\sqrt{n}(\widetilde{\mu}-\mu)\overset{d}{\rightarrow} N(0,\Sigma)$$$

Notes:

1. You might wonder what convergence in distribution means for random vectors. A random vector still has a CDF, typically we define this case:
$$$F_{X}(x_{1},...,x_{d})=\mathbb{P}(X_{1}\leq x_{1},...,X_{d}\leq x_{d})$$$

so we can still define convergence in distribution via pointwise convergence of the CDF. In order to define points of continuity it turns out that the correct definition is that a point of continuity of the CDF if the boundary of the rectangle whose upper right corner is (x_{1},…,x_{d}) has probability 0.

1. Although $d$ can be larger than 1, it is taken to be fixed as $n\rightarrow \infty$. Central limit theorems, when $d$ is allowed to grow, i.e. high-dimensional CLTs are rare and are an active topic of research.

2. The proof of this result follows directly from the proof of the univariate CLT and a powerful result in asymptotic known as the Cramer-World device. The Cramer-World device roughly asserts that if $a^{T}X_{n}\overset{d}{\rightarrow}a^{T}X$ for all vectors $a\in \mathbb{R}^{d}$ then $X_{n}^{d}\overset{d}{\rightarrow}X$.

#### 3. CLT with estimated variance

In the typical case of the CLT, we need to know the variance $\sigma$. In practics, we can use estimated variance to replace with variance. The estimated variance is defined as:

$$$\widetilde{\sigma}_{n}^{2} = \frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\widetilde{\mu})^{2}$$$

It turns out that we can replace the standard deviation in the CLT by $\widetilde{\sigma}$ and still have the same convergence in distribution, i.e.

$$$\frac{\sqrt{n}(\widetilde{\mu}-\mu)}{\widetilde{\sigma}_{n}}\overset{d}{\rightarrow}N(0,1)$$$

This prood can be derived from a sequence of applications of Slutsky’s theorem and the continuous mapping theorem.

Proof:First observe that if we can show that $\frac{\sigma}{\widetilde{\sigma}_{n}}\overset{d}{\rightarrow}1$, then an application of Slutsky’s theorem and the CLT gives us the desired result.

Since square-root is a continuous map, by the continuous mapping theorem, it suffices to show that $\frac{\sigma^{2}}{\widetilde{\sigma}^{2}_{n}}\overset{d}{\rightarrow}1$. We will instead show the stronger statement that,

$$$\widetilde{\sigma}^{2}_{n}\overset{p}{\rightarrow}\sigma^{2}$$$

which implies that the desired statement via the continuous mapping theorem. Note that:

\begin{align} \widetilde{\sigma}^{2}_{n} &= \frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\widetilde{\mu})^{2}\\ &\overset{p}{\rightarrow}\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\widetilde{\mu})^{2} \end{align}

using the fact that $\frac{n-1}{n}\rightarrow 1$. Now,

$$$\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\widetilde{\mu})^{2}=\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}-(\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2})^{2}\overset{p}{\rightarrow}\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2}$$$

using the WLLN. This concludes the proof.

#### 4. The Delta Method

Delta Method states that suppose we have a sequence of random variables $X_{n}$ that converges in distribution to a Gaussian distribution then can we characterize the limiting distribution of $g(X_{n})$ where $g$ is a smooth function. (Continuous mapping theorem).

Delta Method: Suppose that,

$$$\frac{\sqrt{n}(X_{n}-\mu)}{\sigma}\overset{d}{\rightarrow}N(0,1)$$$

and that $g$ is a continuously differentiable function such that $g^{}(\mu)\neq 0$. Then,

$$$\frac{\sqrt{n}(g(X_{n})-g(\mu))}{\sigma}\overset{d}{\rightarrow}N(0,[g^{}(\mu)]^{2})$$$

Proof: The basic idea is simply to use Taylor’s approximation. We know that

$$$g(X_{n})\approx g(\mu) + g^{}(\mu)(X_{n}-\mu)$$$

so that,

$$$\frac{\sqrt{n}(g(X_{n})-g(\mu))}{\mu}\approx g^{}(\mu)\frac{\sqrt{n}(X_{n}-\mu)}{\sigma}\overset{d}{\rightarrow}N(0,[g^{}(\mu)]^{2})$$$

Alternatively, here is a more formal proof. By a rigorous application of Taylor’s theorem we obtain

$$$\frac{\sqrt{n}(g(X_{n})-g(\mu)}{\sigma}\approx g^{}(\mu)\frac{\sqrt{n}(X_{n}-\mu)}{\sigma}\overset{d}{\rightarrow}N(0,[g^{}(\mu)]^{2})$$$

where $\widetilde{\mu}$ is on the line joining $\mu$ to $\widetilde{\mu}$. We know by the WLLN that $\widetilde{\mu}\overset{p}{\rightarrow}\mu$ and so $\widetilde{\mu}\overset{p}{\rightarrow}\mu$ Since $g$ is continuously differentiable， we can use the continuous mapping theorem to conclude that,

$$$g^{}(\widetilde{\mu})\overset{p}{\rightarrow}g^{}(\mu)$$$

Now, we apply Slutsky’s theorem to obtain that,

$$$g^{}(\widetilde{\mu})\frac{\sqrt{n}(X_{n}-\mu)}{\sigma}\overset{d}{\rightarrow} g^{}(\mu)N(0,1)\overset{d}{=}N(0,[g^{}(\mu)]^{2})$$$

#### 5. Multivariate Delta Method: How about delta method in high dimensional space?

Suppose random vector $X_{1},…,X_{n}\in \mathbb{R}^{d}$, and $g:\mathbb{R}^{d}\mapsto \mathbb{R}$ is a continuously differentiable function, then

$$$\sqrt{n}(g(\widetilde{\mu}_{n})-g(\mu))\overset{d}{\rightarrrow}N(0,\bigtriangleup_{\mu}(g)^{T}\Sigma\bigtriangleup_{\mu}(g))$$$

where

$$$\bigtriangleup_{g}(\mu)=\pmatrix{\frac{\partial g(x)}{\partial x_{1}}\cr ...\cr \frac{\partial g(x)}{\partial x_{d}}}_{x=\mu}$$$

is the gradient of $g$ evaluated at $\mu$.