0
$\begingroup$

For a multivariate time series

$$\underline{y}(t)=[y_1(t) \dots y_n(t)]^T$$

What is the difference between the components $y_i(t)$ being spatially independent and instantaneously independent?

$\endgroup$
1
  • $\begingroup$ I believe I know the answer/have an idea, but I will leave it to see whether any answers agree with this first. $\endgroup$
    – rwolst
    Jun 29, 2013 at 19:55

1 Answer 1

2
$\begingroup$

The components are instantaneously uncorrelated if $$ E[y_i(t_1) y_j(t_2)] = 0 $$ for all $i \not= j$ and for all $t_1,t_2$ and when $i = j$ : $$ E[y_i(t_1) y_i(t_2)] = 0 $$ for all $t_1 \not= t_2$.

The components are spatially uncorrelated if $$ E[y_i(t) y_j(t)] = 0 $$ for all $i \not= j$ (assuming a change in index corresponds to a spatial change).

Note that correlation is not quite the same thing as statistical independence.

$\endgroup$
6
  • 1
    $\begingroup$ I agree, but I would like to point out that you implicitly assume that the components $y_i(t)$ have zero mean. If this were not the case, then the definitions you gave refer to orthogonality instead of uncorrelatedness, because $X$ and $Y$ are uncorrelated if $E[XY]=E[X]E[Y]$, and they're orthogonal if $E[XY]=0$. Obviously, if either $E[X]=0$ or $E[Y]=0$, orthogonality and uncorrelatedness are equivalent. $\endgroup$
    – Matt L.
    Jun 30, 2013 at 11:09
  • $\begingroup$ @MattL.: Good point! $\endgroup$
    – Peter K.
    Jun 30, 2013 at 14:52
  • $\begingroup$ Should it not say 'instantaneously uncorrelated ... for all $i \neq j$ and for all $t_1 , t_2$? $\endgroup$
    – rwolst
    Jul 3, 2013 at 10:36
  • $\begingroup$ Hmmm. Toughy. Yes, but... Let me update. $\endgroup$
    – Peter K.
    Jul 3, 2013 at 11:54
  • $\begingroup$ Can you explain the line $E[y_i(t_1),y_i(t_2))]=0$ for all $t_1 \neq t_2$. Why is this necessary? Would this be omitted if the components were instantaneously $mutually$ uncorrelated. $\endgroup$
    – rwolst
    Jul 3, 2013 at 14:53

Your Answer

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

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