Definition of orthogonality and independence for a stochastic processes - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2019-08-22T22:22:04Z https://quant.stackexchange.com/feeds/question/10679 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://quant.stackexchange.com/q/10679 5 Definition of orthogonality and independence for a stochastic processes Probilitator https://quant.stackexchange.com/users/7279 2014-03-23T19:42:25Z 2014-03-23T22:45:41Z <p>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.</p> <p>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.</p> <p><strong>Thus the questions:</strong></p> <ul> <li>are independence and orthogonality equivalent for a stochastic process?</li> <li>what are the metric and the space used to define orthogonality of stochastic processes ?</li> </ul> https://quant.stackexchange.com/questions/10679/-/10680#10680 4 Answer by quasi for Definition of orthogonality and independence for a stochastic processes quasi https://quant.stackexchange.com/users/3415 2014-03-23T21:04:25Z 2014-03-23T22:45:41Z <p>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)</p> <p>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.</p> <p>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. </p> <p>A good reference for this topic is Protter's book. Can look up the exact section if you're interested.</p> <p><strong>edit</strong>, in response to Probilitator's questions:</p> <p>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 $Y_i$.</p> <p>You can also start from the point of view of a mapping into $\mathbb{R}^n$.</p> <p>As sigma algebras, $\mathcal{F} \perp \mathcal{G}$ if for any $A \in \mathcal{F}, B \in \mathcal{G}$, $\mu(A \cap B) = \mu(A) \cdot \mu(B)$.</p> <p>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.</p>