Show that two random vectors have the same distribution

Question

Let $W_1,W_2,...$ be independent identically distributed random variables on $[0,\infty).$ Define $T_0=0,T_n=\sum_{k=1}^{n}W_k (n\ge 1).$ Show that, for any $n,m\in\mathbb{Z}^{+}$ and $0\le t_1<\cdots<t_{m},$ both $\left(\sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{1}\right\}}, \cdots, \sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{m}\right\}}\right)$ and $\left(\sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{1}\right\}}, \cdots, \sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{m}\right\}}\right)$ have the same distribution $$\text{i.e. }\left(\sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{1}\right\}}, \cdots, \sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{m}\right\}}\right)\stackrel{\textbf {d}}=\left(\sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{1}\right\}}, \cdots, \sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{m}\right\}}\right).$$

Applying characteristic function,I can deduce that for any positive integer $k\ge n+1$,$T_k-T_n\stackrel{\textbf {d}}=T_{k-n}$. Then for any $c\ge 0 $, $\mathbf{1}_{\left\{T_{k}-T_{n} \leq c \right\}}\stackrel{\textbf {d}}=\mathbf{1}_{\left\{T_{k-n}\leq c\right\}}$.The random variables $X_1,X_2,Y_1,Y_2,$ satisfying $X_1\stackrel{\textbf {d}}=Y_{1}$ and $X_2\stackrel{\textbf {d}}=Y_{2}$, can not get $X_1+X_2\stackrel{\textbf {d}}=Y_{1}+Y_2$ ,unless $X_1\perp\!\!\!\!\perp Y_1$,$X_2\perp\!\!\!\!\perp Y_2$ ($X \perp\!\!\!\!\perp Y$ means $X$ is independent of $Y$).

---

As @jwhite's comment,define $$X:=\left(\sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{1}\right\}}, \cdots, \sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{m}\right\}}\right),Y:=\left(\sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{1}\right\}}, \cdots, \sum_{\ell=1}^{\infty} \mathbf{1}_{\left\{T_{\ell} \leq t_{m}\right\}}\right),$$

directly show that for any $m$-tuple of non-negative integers $p=(p_1,...,p_m),$ $$\mathbb{P}(X=p)=\mathbb{P}(Y=p).$$ Considering the $\mathbb{P}(X=p)$,since $X=\left(\sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{1}\right\}}, \cdots, \sum_{k=n+1}^{\infty} \mathbf{1}_{\left\{T_{k}-T_{n} \leq t_{m}\right\}}\right)$ and $W_{n}$ are non-negative,$X=p\Rightarrow \bigcap_{i=1}^{m}S_{i}$,where $$S_i=\left\{\sum_{k=1}^{p_i}W_{k+n}\le t_{i}\right\}\bigcap \left\{\sum_{k=1}^{p_i+1}W_{k+n}> t_{i}\right\}.$$ So $\mathbb{P}(X=p)=\mathbb{P}(\bigcap_{i=1}^{m}S_{i}).$ Similarly,$\mathbb{P}(Y=p)=\mathbb{P}(\bigcap_{i=1}^{m}R_{i})$,where $$R_i=\left\{\sum_{k=1}^{p_i}W_{k}\le t_{i}\right\}\bigcap \left\{\sum_{k=1}^{p_i+1}W_{k}> t_{i}\right\}.$$ Note those are not independent,how to prove that $\mathbb{P}(X=p)=\mathbb{P}(Y=p)?$(this is my key issue)

Answer 1

Step $1$: For any $q\in\mathbb{N}$, $$(W_1,\ldots, W_q)\sim (W_{n+1},\ldots, W_{n+q}),$$ where $\sim$ denotes having the same distribution.

Proof: Let $F=(W_1,\ldots, W_q)$ and $G=(W_{n+1},\ldots, W_{n+q})$. Since the Borel $\sigma$-algebra on $\mathbb{R}^q$ is generated by Cartesian products of the form $\prod_{i=1}^q B_i$, $B_i\subset \mathbb{R}$ Borel, it is sufficient to show that for any Borel $B_1,\ldots, B_q\subset \mathbb{R}$, $$\mathbb{P}\Bigl(F\in\prod_{i=1}^q B_i\Bigr)=\mathbb{P}\Bigl(G\in \prod_{i=1}^q\mathbb{B}_i\Bigr).$$ But \begin{align*}\mathbb{P}\Bigl(F\in\prod_{i=1}^q B_i\Bigr) & =\mathbb{P}\Bigl(\bigcap_{i=1}^q (W_i\in B_i)\Bigr)=\prod_{i=1}^q \mathbb{P}(W_i\in B_i)=\prod_{i=1}^q\mathbb{P}(W_{n+i}\in B_i) \\ & = \mathbb{P}\Bigl(\bigcap_{i=1}^q (W_{n+i}\in B_i)\Bigr)=\mathbb{P}\Bigl(G\in\prod_{i=1}^q B_i\Bigr).\end{align*} This proves that $F\sim G$.

Step $2$: Fix $0\leqslant t_1<\ldots < t_m$. For each $n\in\{0,1,\ldots\}$, let $$X_n=\Bigl(\sum_{j=1}^\infty 1_{\{T_{n+j}-T_n\leqslant t_1\}}, \sum_{j=1}^\infty 1_{\{T_{n+j}-T_n\leqslant t_2\}},\ldots, \sum_{j=1}^\infty 1_{\{T_{n+j}-T_n\leqslant t_m\}}\Bigr).$$ Then for any $p=(p_1,\ldots,p_m)\in \{0,1,\ldots\}^m$, $$\mathbb{P}(X_n=p)$$ does not depend on $n$.

Proof of Step $2$: As noted in the edit, since $$T_{n+1}-T_n\leqslant T_{n+2}-T_n\leqslant \ldots,$$ $$\sum_{j=1}^\infty 1_{\{T_{n+j}-T_n\leqslant t_i\}}=p_i$$ if and only if $1_{\{T_{n+j}-T_n\leqslant t_i\}}=1$ for all $j\leqslant p_i$ and $1_{\{T_{n+j}-T_n\leqslant t_i\}}=0$ for all $j>p_i$. Of course, this is equivalent to the conditions that $$\sum_{j=1}^{p_i} W_{n+j}=T_{n+p_i}-T_n\leqslant t_i\tag{$i$}$$ and $$\sum_{j=1}^{p_i+1}W_{n+j}=T_{n+p_i+1}-T_n>t_i.\tag{$ii$}$$ Let $q=1+\max\{p_1,\ldots,p_m\}$. For $1\leqslant i\leqslant m$, let $$Q_i=\{(x_1,\ldots, x_q)\in \mathbb{R}^q: x_1+\ldots+x_{p_i}\leqslant t_i\}\cap \{(x_1,\ldots, x_q)\in\mathbb{R}^q:x_1+\ldots+x_{p_i+1}>t_i\}.$$ Note that $Q_i$ does not depend on $n$. Let $Q=\bigcap_{i=1}^m Q_i$. Another way of stating conditions $(i)$ and $(ii)$ is that $$X_n=p\Leftrightarrow (W_{n+1},\ldots, W_{n+q})\in Q.$$ A special case of this is that $X_0=p$ if and only if $(W_1,\ldots,W_q)\in Q$. But by Step $1$, $F:=(W_1,\ldots, W_q)$ and $G:=(W_{n+1},\ldots, W_{n+q})$ have the same distribution, so $$\mathbb{P}(X_n=p)=\mathbb{P}(G\in Q)=\mathbb{P}(F\in Q)=\mathbb{P}(X_0=p),$$ and this last quantity does not depend on $n$.

Step $3$: $X_n$ and $X_0$ have the same distribution.

Proof of Step $3$: Since $X_0,X_n$ are discrete, it follows that for any subset $P$ of $\{0,1,\ldots\}^m$, $$\mathbb{P}(X_0\in P)=\sum_{p\in P}\mathbb{P}(X_0=p)=\sum_{p\in P}\mathbb{P}(X_n=p)=\mathbb{P}(X_n\in P).$$ Since these sets generate the full $\sigma$-algebra on $(\{0,1,\ldots\}\cup \{\infty\})^m$, where $X_n$ takes its values, this is enough to show that $X_0,X_n$ have the same distribution.