Somehow I can't find the explicit definition of when two processes are supposed to be orthogonal or independent anywhere. I think orthogonality and independence should mean the same thing in this context.

Up to now I always assumed independence/orthogonality of two wiener processes meant $ dW_i(t) dW_j(t) = \delta_{ij} dt $ Or in a different notation $[W_i,W_j]_t=\delta_{ij}t$. Unfortunately this is just a consequence of independence not the actual definition.

Thus the questions:

  • are independence and orthogonality equivalent for a stochastic process?
  • what are the metric and the space used to define orthogonality of stochastic processes ?
  • 4
    $\begingroup$ As usual with Gaussianity; since you have in mind Gaussian processes, yes: correlations=0 is equivalent to independence. $\endgroup$
    – lehalle
    Mar 23, 2014 at 21:02

1 Answer 1


Orthogonality and independence are different concepts. The concepts are the same for Wiener processes because in the context of normal random variables, independence is equivalent to orthogonality (i.e. uncorrelatedness)

Independence is the standard definition for probability. Let $\mathcal{F}, \mathcal{G}$ be the sigma algebras generated by two processes, $X_\cdot, Y_\cdot$. Then $X$ and $Y$ are independent if $\mathcal{F}, \mathcal{G}$ are.

For orthogonality, the condition is actually only defined for things similar to continuous square-integrable martingales. By similar, I mean that things can be extended to processes which are locally square integrable, locally martingales, have discontinuities, semimartingales. But, for $X$ and $Y$ continuous square integrable martingales, then $X$ is orthogonal to $Y$ if $XY$ is a martingale.

A good reference for this topic is Protter's book. Can look up the exact section if you're interested.

edit, in response to Probilitator's questions:

Independence for random vectors: Let $\Omega, \mathcal{H}, \mu$ be a probability space, on which $(X_i)$ and $(Y_i)$ are all defined. $X_i$ is a measurable map from $\Omega$ to $\mathbb{R}$, so it induces a subsigma algebra $\mathcal{F}_i \subset \mathcal{H}$. You can then take $\mathcal{F}$ as the sigma algebra generated by the $\mathcal{F}_i$. You can similarly get $\mathcal{G}$ from the $Y_i$.

You can also start from the point of view of a mapping into $\mathbb{R}^n$.

As sigma algebras, $\mathcal{F} \perp \mathcal{G}$ if for any $A \in \mathcal{F}, B \in \mathcal{G}$, $\mu(A \cap B) = \mu(A) \cdot \mu(B)$.

Next, for the statement about normal random variables, if $X,Y$ are jointly normal and uncorrelated, you calculate the function $f(s,t) = E \left[ \exp(sX + tY) \right]$. You can evaluate this exactly using a bare hands Riemann integral of the normal density. Then you show that $f$ can be factored into functions of $s$ and $t$, which is the equivalent characterization of independence.

  • $\begingroup$ I also found the following defintion of independence math.stackexchange.com/questions/22360/… - do you know haw the independence of two random vectors is defined exactly ? - perhaps you can also edit it into you answer - to make it more comprehensive - cheers :) $\endgroup$ Mar 23, 2014 at 22:08
  • $\begingroup$ Which process-generated-sigma-Algebra do you mean exactly? Do you mean the terminal one $\mathcal{F}_{\infty}=\lim \mathcal{F}_i$ ? $\endgroup$ Mar 23, 2014 at 22:16
  • $\begingroup$ Yeah if you had the filtration defined already you would do that. $\endgroup$
    – quasi
    Mar 23, 2014 at 22:42
  • $\begingroup$ where did you find this definitions ? google was not really helpful and my favourite prob book wasn't much help $\endgroup$ Mar 23, 2014 at 22:56
  • $\begingroup$ I'm not sure where I learned them, that's just how I remember them. I think they're both pretty standard. $\endgroup$
    – quasi
    Mar 24, 2014 at 0:14

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