There's a mistake in one of my calculations, and I'm just wondering if I've messed this up.Am I correct in saying that f(t) is a deterministic function, thenandare independent? (where W is a brownian motion, obviously)Specifically, my function is f(t) = e^{kt}; I'm messing around with the vasicek model.Cheers.

i was thinking of a couples way you could prove this. one might be to use the monotone class theorem, since its clearly true when f(s) is constant. second might be to try and prove it directly using an approximating sequence, and then have independence being preserved in an L^2 limit.this is the best way i came up: since f(s) is deterministic, both integrals are normal random variables, and in fact they're jointly normal (compute the characteristic function). So they are independent if and only if theyre uncorrelated. in the first integral, replace it with , where . do a similar thing with the second integral. compute their covariance and you end up doing a lebesgue integral of 0, which is zero.

Cheers, I'll think about that. The second method requires the joint distribution to be gaussian as well, I guess you'd have to show that too.

haha yeah i just realized that. but you can compute the joint characteristic function with an exponential martingale using novikov's condition.

I'm 99% sure it's true, the proof should follow along the lines of the proof of the Ito isometry. It's just that niggling 1%...

(deleted: was about to say the same thing... should really read other replies before replying!)

ehremo remark is correct and simple but is not the simplest. If f is deterministic then 2 integrals are not actually Ito integrals. The if you recall that stochastic integral is limit of a stepwise sequence that converge to the f then you fist check the statement for stepwise functions. Then you need to make limit transition for an arbitrary f from specified class of functions, say for f integrable in the second power on [ 0 , T ]

regarding what RDK says about the independence being preserved, all that's really needed is that the L^2([0,T]xOmega) limit preserves measurability with respect to a certain sub-sigma-algebra. Because for deterministic approximating processes, the second integral is F-measurable, where F=sigma(W over (t,T]) whereas its a standard property that the first integral is G-measurable, where G=sigma(W over [0,t]). Then by the properties of brownian motion, these two sigma algebras should be indepedent.All of which needs to be proven more carefully, of course.

As f ( t ) is detyerministic you do not need ito construction as well as you could ommit any sigma flows.