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!



1 Answer 1


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 $\eta_T\to1+\mathbb E(a_1^2)\ne\sigma^2$.

  • $\begingroup$ many thanks for your very well explained answer. just one question (regarding the appearing $x(1)$ term in the first coordinate of $G$): are coordinate mappings in the Skorohod space continuous? $\endgroup$
    – s_2
    Jan 15, 2013 at 15:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .