For i.i.d. exp$(\lambda)$ random variables $T_{1},T_{2},T_{3},\cdots$, let $N_{t}=\max\left\{k:T_{1}+\cdots+T_{k}\leq t\right\}$, where we take the value to be zero if the set under the max is empty. It can be shown that $\left(N_{t},t\geq 0\right)$ is a Poisson process.
Let $X^{\lambda}_{t}=N_{t}-\lambda t$, $t\geq 0$, $X^{\lambda}_{0}=0$, denote the centered Poisson process. This process increases by jumps of size one but decreases continuously. Using CLT can show that $\frac {1}{\sqrt{\lambda}}X^{\lambda}_{1}$ converges to a standard normal random variable as $\lambda\to \infty$. I wonder how to extend this to find the limit of the finite dimensional distributions: $0\leq t_{1}<t_{2}<\cdots<t_{k}$, $k\in \mathbb {N}$, $$ \lim_{\lambda\to \infty}\frac {1}{\sqrt{\lambda}}\left(X^{\lambda}_{t_{1}},\cdots,X^{\lambda}_{t_{k}}\right), $$ where the above limit is the sense of weak convergence. What about the limit of the entire process $\left(\lambda^{-1/2}X^{\lambda}_{t},t\geq 0\right)$?
Let $Y^{(\lambda)}_t=\frac{X_t^{\lambda}}{\sqrt{\lambda}} $. In above problem, you mention following fact already, \begin{equation*} d-\lim_{\lambda\to\infty} Y^{(\lambda)}_1=N(0,1)\overset{d}{=}W_1. \end{equation*} where $d-\lim$ denote the limt of convergence in distribution, $W=\{W_t, t\ge 0\}$ is Brownian Motion.
By same way, it could prove, also, \begin{equation*} d-\lim_{\lambda\to\infty} (Y^{(\lambda)}_{t_j}-Y^{(\lambda)}_{t_{j-1}}) = N(0,t_j-t_{j-1})\overset{d}{=}W_{t_j}-W_{t_{j-1}},\quad 0\le t_{j-1}<t_j. \tag{1} \end{equation*} Since both $W$ and $Y^{(\lambda)}=\{Y^{(\lambda)}_t,t\ge0\}$ are processes with independet increments, from (1) get \begin{gather*} d-\lim_{\lambda\to\infty}(Y^{(\lambda)}_{t_j}-Y^{(\lambda)}_{t_{j-1}}, 1\le j\le k) = (W_{t_j}-W_{t_{j-1}}, 1\le j\le k), \\ 0= t_{0}\le t_{j-1}< t_{j},\quad 1\le j\le k. \end{gather*} Furthhermore, \begin{equation*} d-\lim_{\lambda\to\infty}(Y^{(\lambda)}_{t_1},\cdots Y^{(\lambda)}_{t_k})=(W_{t_1},\cdots,W_{t_k}). \tag{2} \end{equation*} This means, as $\lambda\to+\infty$, the finite-dimension distributions of $Y^{(\lambda)}$ converge to the finite-dimension distributions of $W$. Meanwhile, the predictable variation $\langle Y^{(\lambda)}\rangle_t=t$, this also means that the distributions of $\{Y^{(\lambda)}, \lambda>0\} $ are tight(cf., J. Jacod, and A. N. Shiryaev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, Theorem 6.4.13, p.358.) Hence, the distributions of process $Y^{(\lambda)}$ weak convergence to $W$ also.