Portmanteau Theorem

μ-continuity set

First, we need a definition.

Definition: μ-continuity set

A set $A$ is a $\mu$-continuity set if $\mu(\partial A) = 0$ (the boundary of set $A$ in the topological sense has probability zero).

Portmanteau theorem

The Portmanteau Theorem provides several equivalent definitions of weak convergence. It connects the convergence of distribution functions to the convergence of expectations of bounded continuous functions and probabilities of $\mu$-continuity sets.

Theorem

The following conditions are equivalent:

1. $X_n \xrightarrow{d} X$ (recall weak convergence).
2. $\mathbb{E}[f(X_n)] \to \mathbb{E}[f(X)]$ for all bounded, continuous functions $f$.
3. $\mu_n(A) \to \mu(A)$ for every $\mu$-continuity set $A$, where:
   • $\mu_n$ is the distribution of $X_n$, and
   • $\mu$ is the distribution of $X$.

Proofs

1 ⟹ 2

Given $X_n \xrightarrow{d} X$, by Skorohod’s Representation Theorem, there exist random variables $Y_n$ and $Y$ on a common probability space such that:

$$Y_n \stackrel{d}{=} X_n, \quad Y \stackrel{d}{=} X, \quad \text{and} \quad Y_n \xrightarrow{a.s.} Y$$

1. Since $Y_n \stackrel{d}{=} X_n$ and $Y \stackrel{d}{=} X$, we have: $$\mathbb{E}[f(X_n)] = \mathbb{E}[f(Y_n)] \quad \text{and} \quad \mathbb{E}[f(X)] = \mathbb{E}[f(Y)]$$
2. Since $f$ is continuous and $Y_n \xrightarrow{a.s.} Y$, we have: $$f(Y_n) \xrightarrow{a.s.} f(Y)$$
3. Moreover, since $f$ is bounded, by the DCT: $$\mathbb{E}[f(Y_n)] \to \mathbb{E}[f(Y)]$$

Combining everything, we obtain $\mathbb{E}[f(X_n)] \to \mathbb{E}[f(X)]$.

2 ⟹ 1

Assume condition 2 holds: $\mathbb{E}[f(X_n)] \to \mathbb{E}[f(X)]$ for all bounded continuous $f$. We want to show $F_n(x) \to F(x)$ for any point $x$ where $F$ is continuous.

Note that $F_n(x) = \mathbb{E}[\mathbb{1}_{(-\infty, x]}(X_n)]$. The indicator $\mathbb{1}_{(-\infty, x]}$ is not continuous (jump at $x$), so we cannot apply condition 2 directly. Instead, we approximate the indicator function with continuous functions.

For any $x$ and any $y > x$, define a continuous function $f_{x,y}(t)$:

$$f(t) = \begin{cases} 1 & t \le x \\ \frac{y-t}{y-x} & x < t < y \\ 0 & t \ge y \end{cases}$$

Observe that $\mathbb{1}_{(-\infty, x]} \le f \le \mathbb{1}_{(-\infty, y]}$.

1. Upper Bound

   $$F_n(x) = \mathbb{E}[\mathbb{1}_{(-\infty, x]}(X_n)] \le \mathbb{E}[f(X_n)]$$

   Taking limit superior as $n \to \infty$:

   $$\limsup_{n \to \infty} F_n(x) \le \lim_{n \to \infty} \mathbb{E}[f(X_n)] = \mathbb{E}[f(X)]$$

   (The limit exists by condition 2). Also, $\mathbb{E}[f(X)] \le \mathbb{E}[\mathbb{1}_{(-\infty, y]}(X)] = F(y)$. So:

   $$\limsup_{n \to \infty} F_n(x) \le F(y) \quad \text{for any } y > x$$

   Since $F$ is a CDF, it is right-continuous. Taking $y \downarrow x$, we get:

   $$\limsup_{n \to \infty} F_n(x) \le F(x)$$

2. Lower Bound
   By a symmetric argument, construct a function $g$ such that $\mathbb{1}_{(-\infty, z]} \le g \le \mathbb{1}_{(-\infty, x]}$ for $z < x$. Then:

   $$F(z) \le \liminf_{n \to \infty} F_n(x) \quad \text{for any } z < x$$

   Taking $z \uparrow x$:

   $$F(x-) \le \liminf_{n \to \infty} F_n(x)$$

   (where $F(x-)$ is the limit from the left).

3. Conclusion
   Combining the bounds:

   $$F(x-) \le \liminf_{n \to \infty} F_n(x) \le \limsup_{n \to \infty} F_n(x) \le F(x)$$

   If $F$ is continuous at $x$, then $F(x-) = F(x)$. The inequalities collapse to an equality:

   $$\lim_{n \to \infty} F_n(x) = F(x)$$

1 ⟹ 3

Similarly, given $X_n \xrightarrow{d} X$, we use the Skorohod construction ($Y_n \xrightarrow{a.s.} Y$). Let $A$ be a $\mu$-continuity set, meaning $\mu(\partial A) = 0$.

Consider the indicator function $f = \mathbb{1}_A$. Note that $f$ is continuous everywhere except on the boundary $\partial A$. Since $\mu(\partial A) = 0$, $Y$ falls in the continuity set of $f$ almost surely. Thus:

$$\mathbb{1}_A(Y_n) \xrightarrow{a.s.} \mathbb{1}_A(Y)$$

By DCT (bounded by 1):

$$\mathbb{P}(X_n \in A) = \mathbb{E}[\mathbb{1}_A(Y_n)] \to \mathbb{E}[\mathbb{1}_A(Y)] = \mathbb{P}(X \in A)$$

3 ⟹ 1

Assume condition 3 holds: $\mathbb{P}(X_n \in A) \to \mathbb{P}(X \in A)$ for all sets $A$ such that $\mathbb{P}(X \in \partial A) = 0$.

We want to show convergence of the CDF at continuity points. Let $A = (-\infty, x]$. The boundary of this set is the single point set $\partial A = \{x\}$.

If $x$ is a continuity point of $F$, then by definition the probability mass at $x$ is zero:

$$\mathbb{P}(X \in \partial A) = \mathbb{P}(X = x) = F(x) - F(x-) = 0$$

Since the boundary has measure zero, $A$ is a $\mu$-continuity set. By condition 3, the probabilities of these sets converge:

$$\mu_n((-\infty, x]) \to \mu((-\infty, x])$$

In terms of CDFs, this is exactly:

$$F_n(x) \to F(x)$$

Thus, $F_n(x)$ converges to $F(x)$ at every point $x$ where $F$ is continuous.