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?

  • $\begingroup$ 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 '13 at 19:55

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.

  • 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 '13 at 11:09
  • $\begingroup$ @MattL.: Good point! $\endgroup$ – Peter K. Jun 30 '13 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 '13 at 10:36
  • $\begingroup$ Hmmm. Toughy. Yes, but... Let me update. $\endgroup$ – Peter K. Jul 3 '13 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 '13 at 14:53

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