Portmanteau Theorem

Statement: $X_n\stackrel{D}{\to }X$ iff $E[g(X_n)]\to E[g(X)]$ for any bounded, continuous Lipschitz g. We prove this in two parts:

Part 1

Here we show $X_n\stackrel{D}{\to }X$ implies $E[g(X_n)]\to E[g(X)]$ for any continuous bounded g. Let $F_n$ denote the cdf for $X_n$, and likewise let $F$ be the cdf for $X$.

Pick $ϵ$, and a continuous g with M as its bound. By assuming convergence, we have $X_n=O_P(1)$, hence for some $R>0$, we know $P(|X_n|>R)<ϵ$, and also $P(|X|>R)<ϵ$. Thus, ${\displaystyle |{\int }_{|X_n|>R}g(X_n)d{F}_n|<ϵM}$, and likewise for X. Since this can be made arbitrarily small, we need only look at convergence on the interval $[-R,R]$.

Let’s approximate g on $[-R,R]$. Since g is continuous, it’s uniformly continuous on this interval, so we can pick $N+1$ points $t_i$ such that $-R=t_0<t_1<\dots <t_N=R$ and also $x,y\in [t_k,t_{k+1}]$ implies $|g(x)–g(y)|<ϵ$. Furthermore, since F is continuous outside a countable set, we can choose the points such that F is continuous at each $t_i$. Finally, let’s define h, a step function approximation to g, by $x\in [t_k,t_{k+1})$ implies $h(x)=g(t_k)$. (And of course $h(t_N)=g(t_N)$, $h(x)=0$ everywhere else.)

We need two lemmas:

Lemma 1: $|E[g(Y)]–E[h(Y)]|<ϵ+ϵM$ where $Y\in \{X,X_1,X_2,\dots \}$.

Proof:

$$\begin{array}{rcl}|E[g(Y)]–E[h(Y)]|& =& |E[g(Y)1_{|Y|>R}]+(E[g(Y)1_{|Y|\le R}]–E[h(Y)])|\\ & \le & |E[g(Y)1_{|Y|>R}]|+|{\int }_{[-R,R]}g(y)–h(y)dY|\\ \text{(1)}& & <& ϵM+ϵP(|Y|\le R)& \le & ϵM+ϵ\end{array}$$

where (1) follows since we’ve constructed h such that $|g(x)–h(x)|<ϵ$ for $x\in [-R,R]$. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

Lemma 2: $E[h(X_n)]\to E[h(X)]$.

Proof: It’s really easy to integrate h, since it’s a step function. We have:

$$\begin{array}{rcl}|E[h(X_n)]–E[h(X)]|& =& |{\int }_{[-R,R]}h(x)d{F}_n–{\int }_{[-R,R]}h(x)dF|\\ & =& \left|(\sum _{k=0}^{N}g(t_k)(F_n(t_{k+1})–F_n(t_k)))–(\sum _{k=0}^{N}g(t_k)(F(t_{k+1})–F(t_k)))\right|\\ & \le & \sum _{k=0}^N|g(t_k)(F_n(t_{k+1})–F(t_{k+1}))+g(t_k)(F_n(t_k)–F(t_k))|.\end{array}$$

Since there are only finitely many $t_i$’s, and since each $t_i$ is a continuity point of F, and since $F_n\to F$ at each continuity point, it follows that this last quantity tends to 0 as n approaches infinity. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

Finally, note that

$$\begin{gathered} |E[g(X_n)–E[g(X)]|\le |E[g(X_n)]–E[h(X_n)]|+|E[h(X_n)]–E[h(X)]| \\ +|E[h(X)]–E[g(X)]| \end{gathered}$$

and our two lemmas show that each of the three terms on the right can be made arbitrarily small for large n by picking suitably close approximations h. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

Part 2

Let $F_n$ and $F$ be the cdf’s as above. Suppose $E[g(X_n)]\to E[g(X)]$ for any bounded, Lipschitz g. We must show that $F_n\to F$ at any continuity point of F. Let x be such a continuity point.

Pick $ϵ>0$. Let’s define two Lipschitz continuous functions $a$ and $b$ as follows:

• Let $a(y)=1$ for $y\le x–ϵ$, $a(y)=0$ for $y\ge x$, and $a$ is linear on $[x–ϵ,x]$.
• Let $b(y)=1$ for $y\le x$, $b(y)=0$ for $y\ge x+ϵ$, and $b$ is linear on $[x,x+ϵ]$.

So we have

$$\int ad{F}_n\le \int 1_{X_n\le x}d{F}_n=F_n(x)\le \int bd{F}_n.$$

Using the second and third parts here and the convergence of the third part, we get ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{F}_n(x)–F(x)\le {\int }_{[x,x+ϵ]}dF=F(x+ϵ)–F(x)}$. Likewise, using $a$, we get ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}{F}_n(x)–F(x)\ge F(x–e)–F(x)}$. But since F is continuous at x, these can both be made arbitrarily small as $ϵ$ decreases. Hence $F_n(x)\to F(x)$. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

Notes: Why is this called the Portmanteau Theorem? I can’t find any references.