# Portmanteau Theorem

We recall definition of convergence in distribution and its related Portmanteau Theorem here. This is based on [Bil99] ([Patrick Billingsley 1999]).

Definition 1 ${X}$ is called random element if it satisfies the following conditions:

1. ${(\Omega, \mathcal F, \mathbb P)}$ is a probability space;
2. ${X: \Omega \mapsto S}$ for some metric space ${S}$.

Random elements ${X_{n}}$ are said to be convergent to an random element ${X}$ in distribution, and denoted by ${X_{n} \Rightarrow X}$, if $\displaystyle \lim_{n\rightarrow \infty} \mathbb E[ f(X_{n}) ] = \mathbb E[ f(X) ], \quad \forall f\in C_{b}(S).$

– QED. –

The following theorem can be found in Page 26 of [Bil99].

Theorem 2 (Portmanteau Theorem)

The following five statements are equivalent:

1. ${X_{n} \Rightarrow X}$;
2. ${\mathbb E f(X_{n}) \rightarrow \mathbb E f(X)}$ for all bounded uniformly continuous ${f}$;
3. ${\lim\sup_{n} \mathbb P(X_{n} \in F) \le \mathbb P(X \in F)}$ for all closed ${F}$;
4. ${\lim\inf_{n} \mathbb P(X_{n} \in G) \ge \mathbb P(X \in G)}$ for all open ${G}$;
5. ${\mathbb P(X_{n} \in A) = \mathbb P(X\in A)}$ for all ${X}$-continuity sets ${A}$.

– QED –

No. 3 of Pormanteau Theorem can be written by $\displaystyle \lim\sup_{n} \mathbb E I_{F}(X_{n}) \le \mathbb E I_{F}(X),$

which is indeed to be still true for all bounded upper semicontinuous functions in place of ${I_{F}}$. Similarly, No. 4 is equivalent to ${\lim\inf_{n} \mathbb E I_{G}(X_{n}) \ge \mathbb E I_{G}(X),}$ which can be extended to all bounded lower semicontinuous functions.

Example 1

If ${X_{n} \Rightarrow X}$, then

1. ${\lim\sup_{n} \mathbb E f(X_{n}) \le \mathbb E f(X)}$ for all bounded upper semicontinuous ${f}$;
2. ${\lim\inf_{n} \mathbb E f(X_{n}) \ge \mathbb E f(X)}$ for all bounded lower semicontinuous ${f}$.

Proof: If ${f}$ is bounded upper semicontinuous, then there exists bounded continuous functions ${f_{\epsilon}}$ such that ${f_{\epsilon} \downarrow f}$. Thus, $\displaystyle \lim\sup_{n\rightarrow \infty} \mathbb E [f(X_{n})] \le \lim\sup_{n} \mathbb E[ f_{\epsilon}(X) ] = \mathbb E [f_{\epsilon}(X)].$

Note that, ${ \mathbb E [f_{\epsilon}(X)] \downarrow \mathbb E [f(X)]}$ by bounded convergence theorem, and it yields the first conclusion. The second one is the consequence of the first conclusion applying to upper semicontinuous function ${(-f)}$. $\Box$

– QED –

We can also extend the above example even further to ${\mathbb PX^{-1}}$-almost semicontinuous functions.

Example 2

Let ${X_{n} \Rightarrow X}$.

1. If ${f}$ is bounded upper semicontinuous ${\mathbb PX^{-1}}$-almost surely, then ${\lim\sup_{n} \mathbb E f(X_{n}) \le \mathbb E f(X)}$;
2. If ${f}$ is bounded lower semicontinuous ${\mathbb PX^{-1}}$-almost surely, then ${\lim\inf_{n} \mathbb E f(X_{n}) \ge \mathbb E f(X)}$.

Proof: We will only show the first result. The condition ${f}$ is upper semicontinuous ${\mathbb PX^{-1}}$-almost surely

means that ${\exists N \in \mathcal B(S)}$ such that ${\mathbb P(X\in N) = 0}$ and ${f}$ is USC in ${S\setminus N}$.

Denote ${K = \sup_{S} |f(x)| <\infty}$, and define ${g = f (1 - I_{N}) + K I_{N}}$. Then ${g}$ is bounded USC and this yields by Example 1 $\displaystyle \lim\sup_{n} \mathbb E g(X_{n}) \le \mathbb E g(X).$

One can conclude the result by easily checking that $\displaystyle \mathbb E g(X) = \mathbb E f(X), \hbox{ and } \mathbb E g(X_{n}) \ge \mathbb E f(X_{n}).$ $\Box$