Central Limit Theorems

Abstract

Let P be a Markov kernel on $\mathsf {X}\times \mathscr {X}$ that admits an invariant probability measure $\pi$ and let be the canonical Markov chain.

21.A A Covariance Inequality

Lemma 21.A.1

Let $(\varOmega ,\mathscr {A},\mathbb {P})$ be a probability space and X, Y two square-integrable random variables defined on $(\varOmega ,\mathscr {A},\mathbb {P})$. Define

$$\begin{aligned} \alpha = \alpha (X, Y) = 2 \sup _{(x, y) \in \mathbb {R}^2} |\mathbb {P}(X> x, Y> y) - \mathbb {P}(X> x) \mathbb {P}(Y > y)|\;. \end{aligned}$$

(21.A.1)

Then

Proof

For X, Y two square-integrable random variables defined on a probability space $(\varOmega ,\mathscr {A},\mathbb {P})$, one has

(21.A.2)

Note indeed that every random variable X can be written as

$$ X = X^+ - X^- = \int _0^\infty \left[ \mathbbm {1}_{\{X > x\}} - \mathbbm {1}_{\{X < -x\}} \right] \mathrm {d}x \;. $$

Writing Y similarly and applying Fubini’s theorem yields (21.A.2). For $x\in \mathbb {R}$, set . Since the functions $I_x$ are uniformly bounded by 1, we obtain

On the other hand, using that ${\mathbb E}\left[ |I_x(X)| \right] = \mathbb {P}(|X|>x)$, we get

Plugging these bounds into (21.A.2), we obtain

The proof is concluded by applying Hölder’s inequality. ${\Box }$

Lemma 21.A.2

Let $(\varOmega ,\mathscr {A},\mathbb {P})$ be a probability space. Let X be a real-valued random variable and V a uniformly distributed random variable independent of X defined on $(\varOmega ,\mathscr {A},\mathbb {P})$. Define $F_X(x^-)=\lim _{y\rightarrow x \atop y<x} F_X(y)$, $\varDelta F_X(x) = F_X(x) -F_X(x^-)$, where $F_X$ is the cumulative distribution function and

$$\begin{aligned} U&= 1-F_X(X^-) - V\varDelta F_X(X) \; . \end{aligned}$$

Then U is uniformly distributed and $Q_X(U)=X$ , where $Q_X$ is the tail quantile function.

Proof

That $Q_X(U)=X$ is straightforward, since by definition, $Q_X(v)=x$ for all $v\in \left[ 1-F_X(x^-), 1-F_X(x)\right]$, whether there is a jump at x or not. To check that U is uniformly distributed over $\left[ 0,1\right]$, note that $\mathbb {P}(X>x)>u$ if and only if $Q_X(u)>x$. Since V is uniformly distributed on [0, 1], this yields

$$\begin{aligned}&\mathbb {P}(U> u)\\&= \mathbb {P}(1-F_X(X)>u) + \mathbb {P}\left( X=Q_X(u), F_X(F_X^\leftarrow (u)^-) + V \varDelta F_X(F_X(F_X^\leftarrow (u)^-)) \le u \right) \\&= F_X(Q_X(u)^-) + \mathbb {P}(X=Q_X(u)) \frac{1-F_X(Q_X(u)^-)-u}{F_X(Q_X(u)^-) -F_X(Q_X(u))} = 1-u \; . \end{aligned}$$

${\Box }$

Lemma 21.A.3

Let $(\varOmega ,\mathscr {A},\mathbb {P})$ be a probability space and $\mathscr {B}$ a sub-$\sigma$-algebra of $\mathscr {A}$. Let X be a square-integrable random variable and . Then for all $a\in \left[ 0,1\right]$,

$$\begin{aligned} \int _0^a Q_Y^2(u)\mathrm {d}u \le \int _0^a Q_X^2(u)\mathrm {d}u \;. \end{aligned}$$

Proof

By Lemma 21.A.2, let V be a uniformly distributed random variable, independent of $\mathscr {B}$ and X, and define $U = 1-F_Y(Y^-) - V \{F_Y(Y)-F_Y(Y^-)\}$. Set $\mathscr {G}=\mathscr {B}\vee \sigma (V)$. Then $Q_Y(U) = Y$ is $\mathscr {B}$-measurable and Applying Jensen’s inequality, we obtain

Noting that $\mathbb {P}(X^2>x) > u$ if and only if $Q_X^2(u) > x$ and applying Fubini’s theorem, we obtain

$$\begin{aligned} \int _0^\infty [\mathbb {P}(X^2>x) \wedge a] \mathrm {d}x&= \int _0^\infty \left( \int _0^a\mathbbm {1}\left\{ \mathbb {P}(X^2>x)>u\right\} \mathrm {d}u\right) \mathrm {d}x \\&= \int _0^\infty \left( \int _0^a\mathbbm {1}\left\{ Q_X^2(u)>x\right\} \mathrm {d}u\right) \mathrm {d}x \\&= \int _0^a \left( \int _0^\infty \mathbbm {1}\left\{ Q_X^2(u)>x\right\} \mathrm {d}x \right) \mathrm {d}u = \int _0^a Q_X^2(u) \mathrm {d}u \;. \end{aligned}$$

${\Box }$