5
$\begingroup$

Say that $Z_t = (X_t, Y_t)$ is a 2-dimensional Gaussian process (by definition, it means that the random vector $(X_{t_1},Y_{t_1},...,X_{t_n},Y_{t_n})$ is a Gaussian random vector for all $t_1 ,...,t_k$.

I want to prove that if $E(X_s Y_t)=0$ for all $s,t$, then two processes $X_t$ and $Y_t$ are independent.

For this, it seems likely that if a random variable $X$ is independent with some family (can be uncountable) of random variables $Y_t$, $X$ is also independent with the $\sigma$-algebra generated by all $Y_t$.

Is this necessarily true? If not, how do we prove the above statement?

$\endgroup$
4
$\begingroup$

By definition, two stochastic processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are independent iff their corresponding canonical filtrations are independent, i.e.

$$\mathcal{F}^X := \sigma(X_s; s \geq 0) \quad \text{and} \quad \sigma(Y_s; s \geq 0) =: \mathcal{F}^Y .$$

are independent. From

$$\mathcal{F}^X = \sigma \bigg( \underbrace{\bigcup_{n \in \mathbb{N}} \bigcup_{0<t_1<\ldots<t_n} \sigma(X_{t_1},\ldots,X_{t_n})}_{=:\mathcal{G}^X} \bigg)$$

and the fact that $\mathcal{G}^X$ is a generator stable under intersections, we conclude that it suffices to show that $X:=(X_{t_1},\ldots,X_{t_n})$ and $Y:=(Y_{t_1},\ldots,Y_{t_n})$ are independent for any $t_1<\ldots<t_n$. By assumption, $(X,Y)$ is a Gaussian random vector and $X$ and $Y$ are Gaussian as well. Recall that Gaussian random vectors are independent if and only if their components are uncorrelated. Thus,

$$\text{cov}(X_s,Y_t) = \mathbb{E}((X_s-\mathbb{E}X_s) \cdot (Y_t-\mathbb{E}Y_t))=0$$

implies the independence of the processes. If $\mathbb{E}X_s = \mathbb{E}Y_t=0$, the latter condition is equivalent to

$$\mathbb{E}(X_s \cdot Y_t) = 0$$

$\endgroup$
2
  • 3
    $\begingroup$ "Recall that Gaussian random vectors are independent if and only if their components are uncorrelated"... Not in the way you are using it afterwards. The condition would be that the whole vector (X,Y) of size 2n is gaussian, not that both vectors X and Y are. $\endgroup$
    – Did
    Dec 9 '13 at 15:32
  • $\begingroup$ @Did You are right. $\endgroup$
    – saz
    Dec 9 '13 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.