# independent vs. uncorrelated

If I have two stochastic processes $X_t$ and $Y_t$ that are dependent however $dX_tdY_t=0$ where $dX_t=a(z,x,t)dt$ and $dY_t=b(z,x,t)dt$ and $dZ_t=dW_t$(Brownian motion) then from Ito formula $du(Xt,Yt,Zt)=u_tdt+u_XdX+u_Ydt+u_ZdZ+0.5u_{ZZ}dZdZ$, which is an SDE and if I want to solve it for all terms of dt be zero: it implies I have a pde: $u_t+a*u_x+b*u_y+u_z=0$. Here is the question: this pde in three space variables which is to be solved assumes x,z and y are independent but it came from considering dependent variables. Where is the paradox? So when we write a pde which follows from Ito we care only about the fact that they are uncorrelated? Because I don't see any dependence embedded into the pde...Please clarify that for me.

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@ Medan : first observe that in your case $u_t=0$, as long as $u$ doesn't depend explicitely on time as a function. Second I'm not sure about what you mean when you say that this pde is to be solved assuming that $x,y$ and $z$ are independent, could you clarify this further ? Maybe you are mixing two distinct notions, first independence in variables in differential calculus context, second independence in the probabilistic context. Best regards. – TheBridge Apr 16 '12 at 20:09
I meant my function u(t,x,y,z), sorry about it. And yes, this is exactly my question: from the class in PDE I have learned that x,y,z have no randomness and they are independent. that is for each x,y,z fixed the function should satisfy a pde. But I am trying to get a connection with probabilistic context. and if they were independent in probability then everything is fine, but what if not? – Medan Apr 16 '12 at 20:19
@ Medan : Independent in the "pde context" means IMO that you can express $u$ as an explicit function of $x,y,z$ not that $X_t,Y_t,Z_t$ are independent in probability. Regarding the fact that your "not random" pde is originating from a probabilistic context, it only expresses the structure that you need for $u$ seen as a stochastic process to be a (local) martingale nothing else nothing more. Best regards – TheBridge Apr 17 '12 at 8:03
@TheBridge: I understand what you are saying but let me pose another question: say, my PDE comes from writing Ito formula and setting terms with dt to 0 where X_t and Y_t are dependent processes. So I get something like u_t+u_x+u_y=0. Now I treat x and y as independent in differential calculus context and solve numerically. On the other hand by Feymnann Kac I can compute an expectation which will take into account all probabilistic structure. Thus my question is, are they the same? – Medan Apr 17 '12 at 12:45
@ Medan : Your condition is here to guarantee that $u$ is a martingale, so that $u_0=u(0,x,y,z)=E[u(T,X_T,Y_T,Z_T)|t=0,X_0=x,Y_0=y,Z_0=z]$ so depending on your problem exact setting you might solve the "martingale condition" pde to get $u_0$ or use any other "lead" that can help you determine deterministacally $u_0$. If you are a probabilist used to Monte Carlo Simulations you might prefer simulating $X_T,Y_T,Z_T$ and use LLN or CLT to get a value (unless you can calculate analytically this integral). This is generally a matter of taste. Best regards. – TheBridge Apr 17 '12 at 15:43