Convergence of Types Theorem

Question

(Convergence of Types Theorem) Suppose that $F_n(u_nx+v_n) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $u_n>0, a_n>0$ and $F$ an $G$ are non-degenerate. Then there exist $a>0$ an $b \in\mathbb R$ such that $a_n/u_n \to a$ and $(b_n-v_n)/u_n \to b$, and $F(ax+b) = G(x).$

In Billingsley's textbook, he proves the above theorem using the following lemmas.

1. If $F_n \Rightarrow F$, $a_n \to a$ and $b_n \to b$, then $F_n(a_nx +b_n) \Rightarrow F(ax+b)$.
2. If $F_n \Rightarrow F$ and $a_n \to \infty$, then $F_n(a_nx) \Rightarrow \Delta(x)$, where $\Delta(x)$ is the degenerate distribution at $x$.
3. If $F_n \Rightarrow F$ and $b_n$ is unbounded, then $F_n(x+b_n)$ cannot converge weakly.
4. If $F_n(x) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $F$ and $G$ are non-degenerate, then $$ 0<\inf_n a_n \leq \sup_n a_n < \infty;\; \sup_n |b_n| < \infty .$$

I have difficulty in understanding the proof of the forth lemma (highlighted parts). The argument in the book is as follows. Suppose that $a_n$ is not bounded above. Arrange by passing to a sub-sequence that $a_n \to \infty$. Then by lemma 2, $$F_n(a_nx) \Rightarrow \Delta(x).(*)$$ Then since $$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] = F_n(a_nx+b_n)\Rightarrow G(x),(**)$$ it follows by lemma 3 that $\frac{b_n}{a_n}$ is bounded. [Note that in lemma 3, we did not have $a_n$ in front of $(x+b_n)$. But we do now. My question is how to use lemma 3 to get the desired boundedness of $b_n/a_n$, please?] By passing to a further sub-sequence, arrange that $b_n/a_n$ converges to some $c$. By $(*)$ and lemma 1,$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] \Rightarrow \Delta(x+c)$ along this sub-sequence. But $(**)$ now implies that $G$ is degenerate, contrary to hypothesis. Thus $a_n$ is bounded above. If $G_n(x) = F_n(a_nx+b_n)$, then $G_n(x) \Rightarrow G(x)$ and $G_n(a_n^{-1}x - a_n^{-1}b_n) = F_n(x) \Rightarrow F(x)$. The result just proved shows that $a_n^{-1}$ is bounded. Thus $a_n$ is bounded away from $0$ and $\infty$. My question here is how to know $a_n$ is positive, please? How come $a_n$ cannot be negative? Thank you!

Answer 1

It seems that Lemma 3 is applied with $\widetilde{F_n}(t):=F_n(a_nt)$ instead of $F_n$ (which is allowed by Lemma 2) and the sequence $(b_n/a_n)_{n\geqslant 1}$ instead of $(b_n)_{n\geqslant 1}$.

Answer 2

This is an old question but I think it is worth while to have a rather detail solution in MSE.

For each $n$, let $X_n$ be a random variable with distribution $F_n(u_nx+b)$. Then $Y_n=\frac{u_n}{a_n}X_n+\frac{v_n-b_n}{a_m}$ has distribution $F_n(a_nx+b_n)$: $$\mathbb{P}\big[\frac{u_n}{a_n}X_n+\frac{v_b-b_n}{a_n}\leq x\big]=\mathbb{P}\big[X_n\leq \frac{a_n}{u_n}x+\frac{b_n-v_n}{u_n}\big]=F[a_nx+b_n] $$ By assumption $X_n\Longrightarrow X$ and $Y_n\Longrightarrow Y$ where $\mathbb{P}[X\leq x]=F(x)$ and $\mathbb{P}[Y\leq y]=G(y)$ are bot non-degenerate (i.e., neither $X$ nor $Y$ are constants). The problem in the OP reduces to show that:

If $X_n\Longrightarrow X$, and for numeric sequences $\alpha_n>0$, $\beta_n\in\mathbb{R}$, $Y_n:=\alpha_nX_n+\beta_n\Rightarrow Y$, where $X$ and $Y$ are non-degenerate random variables, then there are $\alpha>0$ and $\beta\in\mathbb{R}$ such that $\alpha_n\xrightarrow{n\rightarrow\infty}\alpha$ and $\beta_n\xrightarrow{n\rightarrow\infty}\beta$. As a consequence, $Y=\alpha X+\beta$ in law.

Following Breiman, L. Probability, Addison-Wesley Publishing Company, 1968, the proof presented here is based on characteristic functions. Denote by $\widehat{F_n}$ the characteristic function of $X_n$, and by $\widehat{G_n}$ the characteristic function of $Y_n$. Similarly, let $\widehat{F}$ and $\widehat{G}$ be the characteristic functions of $X$ and $Y$ respectively. Then $$ \widehat{G_n}(t)=\int_{\mathbb{R}}e^{-itx}\,G_n(dx)=\int_{\mathbb{R}}e^{it(\alpha_n x+\beta_n)}\,F_n(dx)=e^{-it\beta_n}\int_{\mathbb{R}}e^{-it\alpha_nx}\,F_n(dx)=e^{-it\beta_n}\widehat{F_n}(\alpha_nt)$$ The assumptions imply that

1. $\widehat{F_n}(t)\xrightarrow{n\rightarrow\infty}\widehat{F}(t)$ uniformly in compact sets in $\mathbb{R}$,
2. $e^{-it\beta_n}\widehat{F_n}(\alpha_nt)=\widehat{G_n}(t)\xrightarrow{n\rightarrow\infty}G(t)$ uniformly in compact sets in $\mathbb{R}$.

Suppose $a\in[0,\infty]$ is a sub sequential limit of $\alpha_n$, i.e., for some subsequence $\alpha_{n_k}\xrightarrow{k\rightarrow\infty}a$.

Claim I: $0<a<\infty$. Suppose $a=\infty$. Then, from $|\widehat{F_n}(\alpha_n x)|=|\widehat{G_n}(t)|$ it follows that for any $t$ $$|\widehat{F}(t)|=\lim_k|\widehat{F_{n_k}}(t)|=\lim_k|\widehat{G_{n_k}}\big(\tfrac{t}{a_{n_k}}\big)|=|G(0)|=1$$ where in the last part we have used the fact that the convergence of characteristic functions is uniform in compact sets. However $|F(t)|=1$ for all $t$ implies that $F$ is the characteristic function of a degenerate random variable, which is a contradiction to the assumption on $X$. Similarly, if $a=0$, then $$|G(t)|=\lim_k|\widehat{G_{n_k}}(t)|=\lim_k|\widehat{F_{n_k}}(a_{n_k}t)|=|F(0)|=1$$ in which case, $Y$ is a degenerate random variable. This is a contradiction to the assumption on $Y$.

Claim II: $\alpha_n$ converges. Suppose $\alpha_n$ admits two different sub sequential limits $0<a'<a<\infty$. Then $$ |\widehat{F}(a't)|=|\widehat{F}(at)|$$ for all $t$ since \begin{align} |\widehat{F}(a't)|&=\lim_{n'}|\widehat{F}(ta_{n'})|=\lim_{n'}|\widehat{G_{n'}}(t)|=|G(t)|\\ |\widehat{F}(at)|&=\lim_{n''}|\widehat{F}(ta_{n''})|=\lim_{n''}|\widehat{G_{n''}}(t)|=|G(t)| \end{align} Consequently $$|\widehat{F}(t)|=\big|F\big(a\tfrac{t}{a}\big)\big|=\Big|\widehat{F}\big(\frac{a'}{a} t\big)\Big|=\ldots \Big|\widehat{F}\Big(\big(\tfrac{a'}{a}\big)^nt\Big)\Big|\xrightarrow{n\rightarrow\infty} |F(0)|=1$$ implying that $X$ is degenerate, a contradiction! As the set of sub sequential limits of a sequence in not empty in $[-\infty,\infty]$, $\alpha_n$ has a unique sequential limit in $(0,\infty)$; hence $\alpha_n\xrightarrow{n\rightarrow\infty}\alpha$ for some $\alpha>0$.

Now we turn our attention to $\beta_n$. Notice that on a neighborhood $(-\delta,\delta)$ of $0$ where $\widehat{F}(at)\neq0$ $$ e^{it\beta_n}=\frac{\widehat{G_n}(t)}{\widehat{F_n}(\alpha_n t)}\xrightarrow{n\rightarrow\infty}\frac{\widehat{G}(t)}{\widehat{F}(\alpha t)}$$ Claim III: $\beta_n$ converges in $\mathbb{R}$. If $\beta_n$ admits a subsequential limit $|\beta_{n_k}|\xrightarrow{k\rightarrow\infty}\infty$, then for all $k$ large enough $|\beta^{-1}_{n_k}|<\delta$ and so $$e^{-i}=\frac{\widehat{G_{n_k}}(\beta^{-1}_{n_k})}{\widehat{F_{n_k}}(\alpha_{n_k}\beta^{-1}_{n_k}t)}\xrightarrow{k\rightarrow\infty}\frac{\widehat{G}(0)}{\widehat{F}(0)}=1$$ which is absurd. Hence all sub sequential limits of $\beta_n$ are finite. If there were two different subsequential limits $b$, $b'$ of $\beta_n$, then $$e^{-ib't}=\frac{\widehat{G}(t)}{\widehat{F}(\alpha t)}=e^{-ibt},\qquad -\delta < t< \delta$$ Then $h(t)=\frac{t(b-b')}{2\pi}$ is a continuos map on $(-\delta,\delta)$ taking integer values. Hence $h$ is constant and $h(t)=h(0)=0$ for all $|t|<\delta$. This shows that $b=b'$. We conclude that $\beta_n\xrightarrow{n\rightarrow\infty}\beta$ for some $\beta\in\mathbb{R}$.

As $\widehat{G_n}(t)=e^{-it\beta_n}\widehat{F_n}(\alpha_nt)\xrightarrow{n\rightarrow\infty}e^{-i\beta t} \widehat{F}(\alpha t)$ uniformly in compacts in $\mathbb{R}$, we also conclude that $Y=\alpha X+\beta$ in distribution. $\Box$.

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In the course of the proof we use the following result from analysis and Probability

1. If $f_n:\mathbb{R}\rightarrow\mathbb{C}$ converges uniformly in compacts to $f$, then for any convergent real sequence $\lambda_n\xrightarrow{n\rightarrow\infty}\lambda$, then $f_n(\lambda_n t)\xrightarrow{n\rightarrow\infty}f(\lambda t)$ uniformly in compacts.

2. If $X$ is a random variable with characteristic function $\widehat{F}$ and $|\widehat{F}(t)|=1$ for all $t$ in a neighborhood of $0$, then $X$ is a degenerate random variable (i.e., for some $b\in\mathbb{R}$, $\mathbb{P}[X=b]=1$).