# Billingsley Ch: $4$ Asymptotically Independent random elements

A NEW DOUBT arose in a previous question, so I am posting it again since the previous edits did not draw views.

Let $X_n$ be random elements of $D$ (space of cad lag functions on $[0,1]$ as domain). $X_n$ has asymptotically independents if $0\leq s_1 \leq t_1 \leq s_2 \leq \ldots < s_r \leq t_r \leq 1$, then for all linear Borel sets $H_1,\ldots,H_r$ we have

$P\{X_n(t_i)-X_n(s_i)\in H_i, i = 1,\ldots,r \}-\prod_{i=1}^{r}P\{X_n(t_i)-X_n(s_i)\in H_i\} \star$

converges to zero as $n\to\infty$. Now Theorem $19.2$ in Billingsley's book on convergence of probability measures reads

Let $X_n$ have the following properties,

1. asymptotically independent increments,

2. $\{X_n^{2}(t)\}$ is uniformly integrable for each $t$,

3. $E\{X_n(t)\}\to 0$, $E\{X_n^{2}(t)\}\to t$ as $n\to\infty$,

4. Also for each positive $\epsilon$ and $\eta$, there exists a positive $\delta$ such that $P\{w(X_n,\delta)\geq \epsilon\}\leq \eta$ for all sufficiently large $n$.

Then $X_n \to W$ (Wiener measure/process).

Its proof seems pretty standard, I had one doubt in it. Tightness of $\{X_n\}$ follows from condition $4$, which also implies that if $X$ is a limit point, then $P(X\in C) = 1$. It remains to show that $X$ has distribution of $W$.

Conditions $2,3$ ensure that $E\{X(t)\} = 0$ and $E\{X^{2}(t)\} = t$. Next is written that Condition $1$ implies that increments of $X(t)$ are independent.

NEW DOUBT

The following doubt in relaxing the condition $t_i <s_{i+1}$ using the fact $\mathbb{P}(X \in C) = 1$. Let us consider the case of only $3$ points in time, i.e. $0,s,t$, where assume that $0<s<t$. My intuition is that this case will entail the main idea. We would like to show that

$\mathbb{P}(X_s \in H_1, X_t-X_s \in H_2) = \mathbb{P}(X_s \in H_1)\mathbb{P}(X_t-X_s \in H_2) \star$.

Also we are given that if $s_1 <s$, then

$\mathbb{P}(X_{s_{1}} \in A, X_t-X_s \in B) = \mathbb{P}(X_{s_{1}} \in A)\mathbb{P}(X_t-X_s \in B) \star\star$.

Using the fact that $\mathbb{P}(X\in C) = 1$, we can write

$\{X_s \in H_1, X_t-X_s \in H_2 \}\equiv \{X_{s^{-}} \in H_1, X_t-X_s \in H_2 \}\equiv \cap_{k=1}^{\infty}\cup_{m=1}^{\infty}E_{k,m}\cap\{X_t-X_s \in H_2\}$

where $E_{k,m} := \{X_l \in B_{\frac{1}{k},H_1} 1-\frac{1}{m}\leq l<s \}$,and $B_{\frac{1}{k},H_1}$ denotes an open ball of radius $\frac{1}{k}$ around the set $H_1$. Now I know that I have to use $\star \star$ somehow, but cannot proceed.

Take $s_n:=s-1/n$: then $(X_{s_n},X_t-X_s)\to (X_s,X_t-X_s)$ almost surely (by the property $\mathbb P\{X\in C\}=1$). Now we can check, uisng characteristic functions, that $X_s$ and $X_{t}-X_s$ are independent.
• Take $s_n:=s-1/n$: then $Y_n : = (X_{s_n},X_t-X_s)\to (X_s,X_t-X_s):=Y$ almost surely (by the property $\mathbb P\{X\in C\}=1$). No doubt in this part. But then regarding the comment on characteristic function, did you mean to say that $p_n(t) \to p(t)$ (they are characteristic functions of $Y_n$ and $Y$), so that the characteristic function of $Y$ factorizes? – user24367 Nov 6 '13 at 2:55
• Yes: for each $n$ the characteristic function factorizes, and $Y_n\to Y$ in distribution. – Davide Giraudo Nov 7 '13 at 13:06