Joint Convergence and Donsker's Theorem I have a question about joint convergence results derived from an FCLT (i.e., a Functional Central Limit Theorem). To motivate my question, consider the following setup:
Let $y_t$ be a random walk
$$y_t = \rho y_t + \epsilon_t ,\quad \rho = 1$$ where $(\epsilon_t)$ satisfies a FCLT, i.e.,
$$ \frac{1}{\sqrt T} \sum_{t=1}^{[rT]}\epsilon_t\Longrightarrow \sigma W(r),$$
where $W$ denotes a standard Brownian motion on $[0,1]$, $\sigma>0$ is a constant, and $\Longrightarrow$ denotes weak convergence. Then I can follow standard textbook arguments (using the FCLT and the Continuous Mapping Theorem) to derive the following convergence results:


*

*$$T^{-1} \sum_{t=1}^T y_{t-1}\epsilon_t\Longrightarrow \sigma^2\int_0^1W(t)dW(t)$$

*$$T^{-3/2} \sum_{t=1}^T y_{t-1}\Longrightarrow \sigma\int_0^1W(t)dt$$

*$$T^{-2} \sum_{t=1}^T y_{t-1}^2\Longrightarrow \sigma^2\int_0^1W(t)^2dt$$


So no problem so far. But very often (in research papers and textbooks), you see the following result:
$$\left(\begin{array}{c} T^{-1} \sum_{t=1}^T y_{t-1}\epsilon_t \\
 T^{-3/2} \sum_{t=1}^T y_{t-1}  \\
 T^{-2} \sum_{t=1}^T y_{t-1}^2  \end{array}\right) \Longrightarrow \left(\begin{array}{c} \sigma^2\int_0^1W(t)dW(t) \\ \sigma\int_0^1W(t)dt \\ \sigma^2\int_0^1W(t)^2dt \end{array} \right)$$
And this I do not understand (i.e., here is my question): HOW DO I DERIVE THE JOINT CONVERGENCE, provided I have the convergence of the marginals. I do know that convergence of the marginals in distribution (or weakly) is not sufficient for convergence of the vector in distribution (or weakly). Am I missing something? What tools, theorems, insights can I use to derive the above joint convergence? Many thanks for any help, I really appreciate everything you can give me! 
Cheers!
 A: For each positive $T$, consider the process $x_T=(x_T(s))_{0\leqslant s\leqslant 1}$ defined by 
$$
x_T(s)=\frac{y_{\lfloor sT\rfloor}}{\sigma\sqrt{T}},
$$ for every $0\leqslant s\leqslant 1$.
The hypothesis is that $x_T$ converges in distribution to a standard Brownian motion $W=(W(s))_{0\leqslant s\leqslant 1}$ when $T\to\infty$. Hence, for every suitable functional $G$, $G(x_T)$ converges in distribution to $G(W)$.
Our goal is to use this for some specific three-dimensional functional $G$. The two last components of $G$ are obvious but the first one requires some more massaging. To this end, note that
$$
\frac2T\sum_{t=1}^Ty_{t-1}\epsilon_t=\frac1Ty_T^2-\eta_T=\sigma^2x_T(1)^2-\eta_T,\qquad\eta_T=\frac1T\sum_{t=1}^T\epsilon_t^2.
$$
Assume that $\eta_T\to\sigma^2$ almost surely (this holds, for example, if $(\epsilon_t)_t$ is i.i.d. centered with variance $\sigma^2$). Then the case which interests you is
$$
G(x)=\left(\frac{x(1)^2-1}2,\int_0^1x(s)\mathrm ds,\int_0^1x(s)^2\mathrm ds\right).
$$
Indeed, note for example that
$$
\int_0^1x_T(s)^2\mathrm ds=\frac1T\frac1T\sum_{t=0}^{T-1}y_t^2.
$$
This proves the convergence in distribution of the three-dimensional vector considered since, finally,
$$
\frac{W(1)^2-1}2=\int_0^1W(s)\mathrm dW(s).
$$
Nota: There exists simple counterexamples to the hypothesis that $\eta_T\to\sigma^2$. For example, if $\epsilon_t=a_t+(-1)^t$ where $(a_t)_t$ is i.i.d. and the distribution of $a_t$ is symmetric, then the FCLT holds with $\sigma^2=\mathbb E(a_1^2)$ but $\eta_T\to1+\mathbb E(a_1^2)\ne\sigma^2$.
