The Portmanteau Theorem

It is important to consider criteria of weak convergence not only on Euclidean spaces, but on more general setting of an abstract metric space. For a random element ξ\xi with probability PP in metric space (S,ρ)(S, \rho) with Borel σ\sigma-field S\mathcal{S}.
The following theorem provides useful conditions equivalent to weak convergence; any of them could serve as the definition. A set AA in S\mathcal{S} whose boundary A\partial A satisfies P(ξA)=0P( \xi \in \partial A)=0 is called a PP-continuity set (note that A\partial A is closed and hence lies in S\mathcal{S} ).
Theorem (Portmanteau Theorem)
Let ξ,ξ1,ξ2,ξ3,\xi, \xi_1, \xi_2, \xi_3, \dots be random elements in metric space (S,S)(S, \mathcal{S}). Then these five conditions are equivalent:
  • A.\mathbf{A.}

    \quad ξndξ\xi_n \xrightarrow{d} \xi.

  • B.\mathbf{B.}

    \quad lim supnP{ξnF}P{ξF}\limsup_{n\to\infty} P\{ \xi_n \in F \} \leq P\{ \xi \in F \} for all closed FF.

  • C.\mathbf{C.}

    \quad lim infnP{ξnG}P{ξG}\liminf_{n\to\infty} P\{ \xi_n \in G \} \geq P\{ \xi \in G \} for all open GG.

  • D.\mathbf{D.}

    \quad P{ξnA}P{ξA}P\{ \xi_n \in A \} \rightarrow P\{ \xi \in A \} for all PP-continuity sets AA.

Proof of Portmanteau Theorem

\quad Proof. Assume and fix closed set FF. Let's introduce the approximation of the indicator 1F\mathbb{1}_{F} by function f(x)=max(0,1ρ(x,F) ϵ)f(x) = \max(0, 1 - \rho(x, F) \ \epsilon). It is bounded and continuous so it can be used in the definition of convergence of distribution. Moreover we have the following inequalities:1F(x)f(x)=max(0,1ρ(x,F) ϵ)1Fϵ(x)\begin{equation*}\mathbb{1}_{F}(x) \leq f(x) = \max(0, 1 - \rho(x, F) \ \epsilon) \leq \mathbb{1}_{F^{\epsilon}}(x)\end{equation*}
 portmanteauTheorem indicators.png
Example of approximation
Therefore we getlim supnE1F(ξn)lim supnEf(ξn)=Ef(ξ)E1Fϵ(ξ)=P{ξFϵ}\begin{equation*}\limsup_{n\to\infty} E \mathbb{1}_{F} (\xi_n) \leq \limsup_{n\to\infty} E f(\xi_n) = E f(\xi ) \leq E \mathbb{1}_{F^{\epsilon}}(\xi) = P\{ \xi \in F^{\epsilon} \}\end{equation*}Letting ϵ0\epsilon \downarrow 0 gives the inequality in B.\mathbf{B.}
The equivalence of B.\mathbf{B.} and C.\mathbf{C.} follows easily by complementation.
Proof that B.&C.\mathbf{B.} \& \mathbf{C.} \rightarrow D\mathbf{D}. If AA^{\circ} and Aˉ\bar{A} are the interior and closure of AA, then conditions B.\mathbf{B.} and C.\mathbf{C.} together implyP{ξAˉ}lim supnP{ξnAˉ}lim supnP{ξnA}lim infnP{ξnA}lim infnP{ξnA}P{ξA}.\begin{align*}P \{ \xi \in \bar{A} \} & \geq \limsup_{n\to\infty} P \{ \xi_n \in \bar{A} \} \geq \limsup_{n\to\infty} P \{ \xi_n \in A \} \\& \geq \liminf_{n\to\infty} P \{ \xi_n \in A \} \geq \liminf_{n\to\infty} P \{ \xi_n \in A^{\circ} \} \geq P \{ \xi \in A^{\circ} \} .\end{align*}If AA is a PP-continuity set, then the extreme terms here coincide with PAP A, and D.\mathbf{D.} follows.
Proof that D.A.\mathbf{D.} \rightarrow \mathbf{A.}. By linearity of the mathematical expectation we may consider the bounded ff which satisfies 0<f<10<f<1. Then Ef(ξ)=0P{f(ξ)>t}dt=01P{f(ξ)>t}dt E f(\xi) =\int_0^{\infty} P \{ f(\xi)>t \} d t=\int_0^1 P \{ f(\xi)>t \} dt. If ff is continuous, then {x ⁣:f(x)>t}{x ⁣:f(x)=t}\partial \{x \colon f(x)>t \} \subset \{ x\colon f(x)=t \}, and hence {x ⁣:f(x)>t}\{x \colon f(x)>t \} is a PP-continuity set except for countably many tt. By condition D.\mathbf{D.} and the bounded convergence theorem,Ef(ξn)=01P{f(ξn)>t}dt01P{f(ξ)>t}dt=Ef(ξ)\begin{equation*}E f(\xi_n) = \int_0^1 P \{ f(\xi_n) > t \} \textrm{d} t \rightarrow \int_0^1 P \{ f(\xi) > t \} \textrm{d} t = E f(\xi)\end{equation*}which proves A.\mathbf{A.}

References