2
$\begingroup$

To keep the question short: Let $C([0,1], \mathbb{R}^d)$, $d \geq 2$ be the space of all $\mathbb{R}^d$-valued continuous processes. $X$ and $Y$ are two $C([0,1],\mathbb{R}^d)$-valued random variates, and we know that $\lambda' X \stackrel{d}{=} \lambda' Y$ for all $\lambda \in \mathbb{R}^d$. Is it necessary the case that $X \stackrel{d}{=} Y$?

More Details:

The problem is reformulated to be short. I came across the problem upon reading this paper (proposition 4.1), in which the authors seek to establish an analogue of Cramer-Wold device so as to reduce the proof of invariance principle for multivariate stochastic process to the univariate case.

The proposition they state is

Let $\{W_n\}$ be a sequence in $\mathbb{D}_{[0,1]}(\mathbb{R}^d)$, the Skorohod space of $\mathbb{R}^d$-valued stochastic processes. Then $W_n \stackrel{d}{\longrightarrow} W$ if and only if $\lambda' W_n \stackrel{d}{\longrightarrow} \lambda' W$ for any $\lambda \in \mathbb{R}^d$.

Some other papers have pointed out that this proposition is incorrect as stated, and continuity of $W$ should be imposed. But this is just a minor problem.

I was puzzled by the question posed above. Let's consider the if part. We could easily show tightness of each coordinate sequence of $\{W_n\}$ and so tightness of $\{W_n\}$ itself. Prokhorov's theorem tells us that now it suffices to show that $W$ is the only possible limit. This would be done once we show finite dimensional convergence along continuities of $W$. But is the "Cramer-Wold"-like condition really sufficient to entail finite-dimensional convergence? I really can't see how.

More$\times 2$ Details:

In some other book, I see a proof of the (if part of the) same proposition, which goes like this:

Since $\lambda' W_n \stackrel{d}{\longrightarrow} \lambda' W$, by continuous mapping theorem $$ (\lambda' W_n(t_1), ..., \lambda' W_n(t_k) ) \stackrel{d}{\longrightarrow} (\lambda' W(t_1), ..., \lambda' W(t_k)) $$ along continuities of $W$. Then by Cramer-Wold device, $$ (W_n(t_1), ..., W_n(t_k)) \stackrel{d}{\longrightarrow} (W(t_1), ..., W(t_k)) $$

Unfortunately, I fail to see why Cramer-Wold device applies here: $(W_n(t_1), ..., W_n(t_k))$ is a $d \times k$-matrix, so it seems we need to show $$ \alpha' \mathrm{vec}(W_n(t_1), ..., W_n(t_k)) \stackrel{d}{\longrightarrow} \alpha' \mathrm{vec}(W(t_1), ..., W(t_k)) $$ for any $dk \times 1$-vector $\alpha$, which is more stringent than the assumption.

Any help would be appreciated!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ If it is true, the continuity will be essential. For instance, if you take $d=2$ and replace $[0,1]$ by $\{0,1\}$, the question becomes: if $(X,Y)$, $(X', Y')$ are two pairs of $\mathbb{R}^2$-valued random vectors, and $(\lambda \cdot X, \lambda \cdot Y) \overset{d}{=} (\lambda \cdot X', \lambda \cdot Y')$ for all $\lambda \in \mathbb{R}^2$, does $(X,Y) \overset{d}{=} (X', Y')$? The answer is no. Let $(\eta, \xi)$ be iid standard 1-D normal random variables, $X=(\eta, \xi)$, $Y=(-\xi, \eta)$, and let $(X',Y')$ be iid standard 2-D normals (so $(X', Y')$ is standard 4-D normal). $\endgroup$
    – Nate Eldredge
    Apr 20, 2018 at 20:41
  • 1
    $\begingroup$ You may then check that $\lambda \cdot X, \lambda \cdot Y$ are independent for any $\lambda$, just like $\lambda \cdot X', \lambda \cdot Y'$. $\endgroup$
    – Nate Eldredge
    Apr 20, 2018 at 20:43
  • $\begingroup$ @NateEldredge Thanks, that's an enlightening example! Extending your example, I think a counter-example could be constructed for the stated analogue of Cramer-Wold device: Let $(\xi, \eta)'$ be 2d standard Wiener process. Define $X(t) \equiv \sqrt(t - t^2) (\xi(t), \eta(t)) + t (-\eta(t), \xi(t))$. Then $\lambda' X$ agrees with a Wiener process multiplied by a constant, yet $X$ is obviously not Wiener. $\endgroup$
    – Dormire
    Apr 25, 2018 at 15:29

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.