TITLE: Convergence of Random Vector to Gaussian Process in Distribution
QUESTION [5 upvotes]: I am trying to show the following:
Let $\{N_{c}(t)\}$ be a poisson process with parameter $c \ge 0$.  Define the process $X_{c}(t)=N_{c}(t)-ct$ for $t \ge 0.$
Show for every $k \in \mathbb{N}$ and $(t_1,...,t_k) \in \mathbb{R}^+$ the random vector $\frac{1}{c^{1/2}}(X_{c}(t_1),...,X_{c}(t_k))$ converges in distribution to $(Y(t_1),...,Y(t_k))$ where $\{Y(t)\}$ is a Gaussian process. Describe this process.
My thoughts are this can be shown via characteristic functions similarly to how the CLT is proven, that is $\phi_{\frac{1}{c^{1/2}}N_{c}-ct}(t)$ converges to a characteristic function of a multivariate normal for each $t$ and $k$.  This would be sufficient to show the resulting limit is a Gaussian process.  I am struggling to actually show this algebraically.  
As far as describing this limit, is it interpreted similarly to the CLT?  Since the poisson process is a counting process, does the limit Gaussian process imply that the arithmetic mean of poisson process after sufficiently many time steps will follow a multivariate normal distribution?

REPLY [4 votes]: This answer requires some background knowledge on Lévy processes.

Theorem Let $(L_t^{(n)})_{t \geq 0}$ be a sequence of Lévy processes with characteristic exponents $\psi_n$. If there exists a Lévy process $(L_t)_{t \geq 0}$ with characteristic exponent $\psi$ such that $$\lim_{n \to \infty} \psi_n(\xi) = \psi(\xi)  \qquad \text{for all} \, \, \xi \in \mathbb{R}^d, \tag{1}$$ then $L^{(n)} \to L$ in finite-dimensional distribution.

Proof: We prove by induction that $$(L_{t_1}^{(n)},\ldots,L_{t_k}^{(n)}) \xrightarrow[d]{n \to \infty} (L_{t_1},\ldots,L_{t_k}) \tag{2}$$ for any $k \in \mathbb{N}$ and $0 \leq t_1 < \ldots <t_k$.

$k=1$: By the Lévy-Khintchine formula, we have $$\mathbb{E} e^{i \xi L^{(n)}_t} = e^{ -t \psi_n(\xi)} \xrightarrow[]{n \to \infty} e^{-t \psi(\xi)} = \mathbb{E} e^{i \xi L_t},$$ and therefore it follows from Lévy's continuity theorem that $L_t^{(n)} \to L_t$ in distribution.
induction hypothesis: $(2)$ holds
inductive step ($k \to k+1$): By the independence of the increments, we have $$\begin{align*} &\quad \mathbb{E} \exp \left( i \sum_{j=1}^{k+1} \xi_j L_{t_j}^{(n)} \right) \\ &= \mathbb{E} \left[ \exp \left( i \sum_{j=1}^{k} \xi_j L_{t_j}^{(n)} + i \xi_{k+1} L_{t_k}^{(n)} \right) \mathbb{E} \left( \exp \left( i \xi_{k+1} (L_{t_{k+1}}^{(n)}-L_{t_k}^{(n)}) \right) \mid \mathcal{F}_{t_k} \right) \right] \\ &=  \mathbb{E}\exp\left(i \xi_{k+1} (L_{t_{k+1}}^{(n)}-L_{t_k}^{(n)})\right) \mathbb{E} \exp \left( i \sum_{j=1}^{k} \xi_j L_{t_j}^{(n)} + i \xi_{k+1} L_{t_k}^{(n)} \right). \tag{3} \end{align*}$$ Using the induction hypothesis (for the second term) and the Lévy-Khintchine formula (for the first one), we get
$$\begin{align*} \lim_{n \to \infty} \mathbb{E} \exp \left( i \sum_{j=1}^{k+1} \xi_j L_{t_j}^{(n)} \right) &= e^{-(t_{k+1}-t_k) \psi(\xi_{k+1})} \mathbb{E} \exp \left( i \sum_{j=1}^{k} \xi_j L_{t_j} + i \xi_{k+1} L_{t_k} \right) \\ &= \mathbb{E}e^{i \xi_{k+1} (L_{t_{k+1}}-L_{t_k})} \mathbb{E} \exp \left( i \sum_{j=1}^{k} \xi_j L_{t_j} + i \xi_{k+1} L_{t_k} \right). \end{align*}$$ By performing exactly the same calculations as in $(3)$ (but now backwards), we find that the right-hand side equals $$\mathbb{E} \exp \left( \sum_{j=1}^{k+1} \xi_j L_{t_j} \right).$$ Applying Lévy's continuity theorem, we conclude $(L_{t_1}^{(n)},\ldots,L_{t_{k+1}}^{(n)}) \to (L_{t_1},\ldots,L_{t_{k+1}})$ in distribution. This finishes the proof.

Remark: The above theorem holds, under slightly stronger assumptions, in a similar fashion for Markov processes. The reason is that the finite dimensional distributions of a Markov process $(X_t)_{t \geq 0}$ are uniquely determined by the one-dimensional distributions $\mathbb{P}^x(X_t \in \cdot)$, $t \geq 0$, $x \in \mathbb{R}^d$.

In order to solve your problem, it just remains to verify that

$X_c(t) =N_c(t)-ct$ is a Lévy process
the characteristic exponent of $X_c$ converges to $\psi(\xi) = |\xi|^2/2$ as $c \to \infty$.